RE: [PHP] How to output a NULL field?
Paul,
This all started because when I try this:
?php echo $rs-Fields(22);?
It work fine, as long as there is a non-null value there, otherwise it
produces an error.
Also, I'm working with a Microsoft SQL 2000 database, not MySQLnot
sure if that matters
But echo $rs-Fields(22) works perfectly for dumping values out of my
$rs recordset...that is, unless the value is NULL is the database - then
I get:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176
-Original Message-
From: Paul M Foster [mailto:pa...@quillandmouse.com]
Sent: Tuesday, August 25, 2009 4:39 PM
To: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted
to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
Just a thought... do you really mean $rs-Fields(22) or do you mean
$rs-Fields[22]? The former is a function call and the latter is an
array variable.
Paul
--
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Re: [PHP] How to output a NULL field?
use is_null() check it
2009/8/26 David Stoltz dsto...@shh.org:
Paul,
This all started because when I try this:
?php echo $rs-Fields(22);?
It work fine, as long as there is a non-null value there, otherwise it
produces an error.
Also, I'm working with a Microsoft SQL 2000 database, not MySQLnot
sure if that matters
But echo $rs-Fields(22) works perfectly for dumping values out of my
$rs recordset...that is, unless the value is NULL is the database - then
I get:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176
-Original Message-
From: Paul M Foster [mailto:pa...@quillandmouse.com]
Sent: Tuesday, August 25, 2009 4:39 PM
To: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted
to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
Just a thought... do you really mean $rs-Fields(22) or do you mean
$rs-Fields[22]? The former is a function call and the latter is an
array variable.
Paul
--
Paul M. Foster
--
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RE: [PHP] How to output a NULL field?
I tried that -it's in the first part of my message
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 7:39 AM
To: David Stoltz
Cc: Paul M Foster; php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
use is_null() check it
2009/8/26 David Stoltz dsto...@shh.org:
Paul,
This all started because when I try this:
?php echo $rs-Fields(22);?
It work fine, as long as there is a non-null value there, otherwise it
produces an error.
Also, I'm working with a Microsoft SQL 2000 database, not MySQLnot
sure if that matters
But echo $rs-Fields(22) works perfectly for dumping values out of my
$rs recordset...that is, unless the value is NULL is the database - then
I get:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176
-Original Message-
From: Paul M Foster [mailto:pa...@quillandmouse.com]
Sent: Tuesday, August 25, 2009 4:39 PM
To: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted
to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
Just a thought... do you really mean $rs-Fields(22) or do you mean
$rs-Fields[22]? The former is a function call and the latter is an
array variable.
Paul
--
Paul M. Foster
--
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Re: [PHP] How to output a NULL field?
Could you post your database's class to here?
I'm use mssql with php for several years and read NULL Fields is never
appear your case.
2009/8/26 David Stoltz dsto...@shh.org:
I tried that -it's in the first part of my message
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 7:39 AM
To: David Stoltz
Cc: Paul M Foster; php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
use is_null() check it
2009/8/26 David Stoltz dsto...@shh.org:
Paul,
This all started because when I try this:
?php echo $rs-Fields(22);?
It work fine, as long as there is a non-null value there, otherwise it
produces an error.
Also, I'm working with a Microsoft SQL 2000 database, not MySQLnot
sure if that matters
But echo $rs-Fields(22) works perfectly for dumping values out of my
$rs recordset...that is, unless the value is NULL is the database - then
I get:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176
-Original Message-
From: Paul M Foster [mailto:pa...@quillandmouse.com]
Sent: Tuesday, August 25, 2009 4:39 PM
To: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted
to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
Just a thought... do you really mean $rs-Fields(22) or do you mean
$rs-Fields[22]? The former is a function call and the latter is an
array variable.
Paul
--
Paul M. Foster
--
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RE: [PHP] How to output a NULL field?
Sorry, didn't follow this whole thread but have you tried print_r() or
var_dump() to see what it looks like when it's type variant? I'd be
curious to see what you'd find in it.
Also, I forget about SQL Server, but MySQL has a function ifnull() which
can make sure you never return a null value. This is also handy for
checking for empty fields where you don't know if it's going to be an
empty string or a null:
SELECT * FROM table WHERE ifnull(somefield, '') = ''
If you know which columns are nullable, try seeing if there's an ifnull()
function in SQL Server to adjust the data before it even gets to your DB
class.
-TG
- Original Message -
From: David Stoltz dsto...@shh.org
To: Paul M Foster pa...@quillandmouse.com, php-general@lists.php.net
Date: Wed, 26 Aug 2009 07:29:53 -0400
Subject: RE: [PHP] How to output a NULL field?
Paul,
This all started because when I try this:
?php echo $rs-Fields(22);?
It work fine, as long as there is a non-null value there, otherwise it
produces an error.
Also, I'm working with a Microsoft SQL 2000 database, not MySQLnot
sure if that matters
But echo $rs-Fields(22) works perfectly for dumping values out of my
$rs recordset...that is, unless the value is NULL is the database - then
I get:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176
-Original Message-
From: Paul M Foster [mailto:pa...@quillandmouse.com]
Sent: Tuesday, August 25, 2009 4:39 PM
To: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted
to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
Just a thought... do you really mean $rs-Fields(22) or do you mean
$rs-Fields[22]? The former is a function call and the latter is an
array variable.
Paul
--
Paul M. Foster
--
PHP General Mailing List (http://www.php.net/)
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Re: [PHP] How to output a NULL field?
2009/8/26 David Stoltz dsto...@shh.org:
Paul,
This all started because when I try this:
?php echo $rs-Fields(22);?
It work fine, as long as there is a non-null value there, otherwise it
produces an error.
Also, I'm working with a Microsoft SQL 2000 database, not MySQLnot
sure if that matters
But echo $rs-Fields(22) works perfectly for dumping values out of my
$rs recordset...that is, unless the value is NULL is the database - then
I get:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176
That error indicates that you're not getting NULL since NULL is easily
converted to a string.
Try this...
$var = $rs-Fields(22);
var_dump($var);
...and tell us what you get.
-Stuart
--
http://stut.net/
-Original Message-
From: Paul M Foster [mailto:pa...@quillandmouse.com]
Sent: Tuesday, August 25, 2009 4:39 PM
To: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted
to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
Just a thought... do you really mean $rs-Fields(22) or do you mean
$rs-Fields[22]? The former is a function call and the latter is an
array variable.
Paul
--
Paul M. Foster
--
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RE: [PHP] How to output a NULL field?
This:
$var = $rs-Fields(22);
var_dump($var);
Produces this:
object(variant)#3 (0) { }
-Original Message-
From: Stuart [mailto:stut...@gmail.com]
Sent: Wednesday, August 26, 2009 8:47 AM
To: David Stoltz
Cc: Paul M Foster; php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
2009/8/26 David Stoltz dsto...@shh.org:
Paul,
This all started because when I try this:
?php echo $rs-Fields(22);?
It work fine, as long as there is a non-null value there, otherwise it
produces an error.
Also, I'm working with a Microsoft SQL 2000 database, not MySQLnot
sure if that matters
But echo $rs-Fields(22) works perfectly for dumping values out of my
$rs recordset...that is, unless the value is NULL is the database - then
I get:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176
That error indicates that you're not getting NULL since NULL is easily
converted to a string.
Try this...
$var = $rs-Fields(22);
var_dump($var);
...and tell us what you get.
-Stuart
--
http://stut.net/
-Original Message-
From: Paul M Foster [mailto:pa...@quillandmouse.com]
Sent: Tuesday, August 25, 2009 4:39 PM
To: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted
to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
Just a thought... do you really mean $rs-Fields(22) or do you mean
$rs-Fields[22]? The former is a function call and the latter is an
array variable.
Paul
--
Paul M. Foster
--
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Re: [PHP] How to output a NULL field?
My code for mssql
please enable the php's mssql extentions.
it used like so many mysql class that you can find by google
--
?php
if(!defined('IN_WEB')) {
exit('Access Denied');
}
ini_set('mssql.datetimeconvert',0);//php4.2.0 disable php's automatic
datetime convert
Class DB {
var $querynum=0;
var $mssql_link;
var $conn_link;
var $sp_link;
var $sp_name='';
var $error_stop=0;
var $show_error=0;
var $dbhost;
var $dbuser;
var $dbpw;
var $dbname;
var $pconnect;
var $var_type=array();
var $fields_name=array();
var $last_error_msg='';
var $phprunversion='';
function DB() {
//define type for sp
$this-var_type['sp_bit']=SQLBIT;
$this-var_type['sp_tinyint']=SQLINT1;
$this-var_type['sp_smallint']=SQLINT2;
$this-var_type['sp_int']=SQLINT4;
$this-var_type['sp_bigint']=SQLVARCHAR;
$this-var_type['sp_real']=SQLFLT4;
$this-var_type['sp_float']=SQLFLT8;
$this-var_type['sp_float-null']=SQLFLTN;
$this-var_type['sp_smallmoney']=SQLFLT8;
$this-var_type['sp_money']=SQLFLT8;
$this-var_type['sp_money-null']=SQLFLT8;
$this-var_type['sp_char']=SQLCHAR;
$this-var_type['sp_varchar']=SQLVARCHAR;
$this-var_type['sp_text']=SQLTEXT;
$this-var_type['sp_datetime']=SQLINT4;
$this-phprunversion=phpversion();
//end
}
/*=php4.4.1,=php5.1.1
a new paramate for if use newlink for connect,pconnect
*/
function rconnect($newlink=false){//2007.03.01 by hack988 fix
phpversion check
if($this-phprunversion = '4.4.1' $this-phprunversion
'5.0.0'
|| $this-phprunversion = '5.1.1'){
return $this-rconnect4p($newlink);
}else{
return $this-rconnect3p();
}
}
function rconnect3p(){
$this-mssql_link = $this-pconnect==0 ?
mssql_connect($this-dbhost, $this-dbuser, $this-dbpw) :
mssql_pconnect($this-dbhost, $this-dbuser, $this-dbpw);
if(!$this-mssql_link){
$this-halt(connect
($this-pconnect)MSSQL($this-dbhost,$this-dbuser)failed!);
return false;
}else{
$this-conn_link=$this-mssql_link;
if($this-dbname) {
if
(!...@$this-select_db($this-dbname,$this-conn_link)){
$this-halt('can not use database
'.$this-dbname);
return false;
}else{
return true;
}
}else{
return true;
}
}
}
function rconnect4p($newlink=false){
$this-mssql_link = $this-pconnect==0 ?
mssql_connect($this-dbhost, $this-dbuser, $this-dbpw , $newlink) :
mssql_pconnect($this-dbhost, $this-dbuser, $this-dbpw, $newlink);
if(!$this-mssql_link){
$this-halt(reconect($this-pconnect)MSSQL($this-dbhost,$this-dbuser)failed);
return false;
}else{
$this-conn_link=$this-mssql_link;
if($this-dbname) {
if
(!...@$this-select_db($this-dbname,$this-conn_link)){
$this-halt('can not use database
'.$this-dbname);
return false;
}else{
return true;
}
}else{
return true;
}
}
}
function connect($dbhost, $dbuser, $dbpw, $dbname, $pconnect =
0,$auto_conn=0 ,$newlink=false) {
$this-dbhost=$dbhost;
$this-dbuser=$dbuser;
$this-dbpw=$dbpw;
$this-dbname=$dbname;
$this-pconnect=$pconnect;
if($auto_conn){
return $this-rconnect($newlink);
}else{
return true;
}
}
function close() {
if($this-conn_link){
$result=mssql_close($this-conn_link);
}else{
RE: [PHP] How to output a NULL field?
Sorry - I don't know what you mean by DB class?
I'm using Microsoft SQL 2000with this code:
?php
//create an instance of the ADO connection object
$conn = new COM (ADODB.Connection)
or die(Cannot start ADO);
//define connection string, specify database driver
$connStr = PROVIDER=SQLOLEDB;SERVER=;UID=xxx;PWD=;DATABASE=;
$conn-open($connStr); //Open the connection to the database
$query = SELECT * FROM eval_evaluations WHERE id = .$_POST[eval];
$rs = $conn-execute($query);
echo $rs-Fields(22); //this is where that particular field is NULL, and
produces the error
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 8:08 AM
To: David Stoltz
Cc: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
Could you post your database's class to here?
I'm use mssql with php for several years and read NULL Fields is never
appear your case.
2009/8/26 David Stoltz dsto...@shh.org:
I tried that -it's in the first part of my message
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 7:39 AM
To: David Stoltz
Cc: Paul M Foster; php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
use is_null() check it
2009/8/26 David Stoltz dsto...@shh.org:
Paul,
This all started because when I try this:
?php echo $rs-Fields(22);?
It work fine, as long as there is a non-null value there, otherwise it
produces an error.
Also, I'm working with a Microsoft SQL 2000 database, not MySQLnot
sure if that matters
But echo $rs-Fields(22) works perfectly for dumping values out of my
$rs recordset...that is, unless the value is NULL is the database - then
I get:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176
-Original Message-
From: Paul M Foster [mailto:pa...@quillandmouse.com]
Sent: Tuesday, August 25, 2009 4:39 PM
To: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted
to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
Just a thought... do you really mean $rs-Fields(22) or do you mean
$rs-Fields[22]? The former is a function call and the latter is an
array variable.
Paul
--
Paul M. Foster
--
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Re: [PHP] How to output a NULL field?
my god you use ado connect sorry I'm use php_mysql extentions for all
mssql function.
I'm never use ado connect before.
2009/8/26 David Stoltz dsto...@shh.org:
Sorry - I don't know what you mean by DB class?
I'm using Microsoft SQL 2000with this code:
?php
//create an instance of the ADO connection object
$conn = new COM (ADODB.Connection)
or die(Cannot start ADO);
//define connection string, specify database driver
$connStr = PROVIDER=SQLOLEDB;SERVER=;UID=xxx;PWD=;DATABASE=;
$conn-open($connStr); //Open the connection to the database
$query = SELECT * FROM eval_evaluations WHERE id = .$_POST[eval];
$rs = $conn-execute($query);
echo $rs-Fields(22); //this is where that particular field is NULL, and
produces the error
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 8:08 AM
To: David Stoltz
Cc: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
Could you post your database's class to here?
I'm use mssql with php for several years and read NULL Fields is never
appear your case.
2009/8/26 David Stoltz dsto...@shh.org:
I tried that -it's in the first part of my message
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 7:39 AM
To: David Stoltz
Cc: Paul M Foster; php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
use is_null() check it
2009/8/26 David Stoltz dsto...@shh.org:
Paul,
This all started because when I try this:
?php echo $rs-Fields(22);?
It work fine, as long as there is a non-null value there, otherwise it
produces an error.
Also, I'm working with a Microsoft SQL 2000 database, not MySQLnot
sure if that matters
But echo $rs-Fields(22) works perfectly for dumping values out of my
$rs recordset...that is, unless the value is NULL is the database - then
I get:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176
-Original Message-
From: Paul M Foster [mailto:pa...@quillandmouse.com]
Sent: Tuesday, August 25, 2009 4:39 PM
To: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted
to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
Just a thought... do you really mean $rs-Fields(22) or do you mean
$rs-Fields[22]? The former is a function call and the latter is an
array variable.
Paul
--
Paul M. Foster
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Re: [PHP] How to output a NULL field?
On Wed, 2009-08-26 at 22:00 +0800, hack988 hack988 wrote:
my god you use ado connect sorry I'm use php_mysql extentions for all
mssql function.
I'm never use ado connect before.
2009/8/26 David Stoltz dsto...@shh.org:
Sorry - I don't know what you mean by DB class?
I'm using Microsoft SQL 2000with this code:
?php
//create an instance of the ADO connection object
$conn = new COM (ADODB.Connection)
or die(Cannot start ADO);
//define connection string, specify database driver
$connStr = PROVIDER=SQLOLEDB;SERVER=;UID=xxx;PWD=;DATABASE=;
$conn-open($connStr); //Open the connection to the database
$query = SELECT * FROM eval_evaluations WHERE id = .$_POST[eval];
$rs = $conn-execute($query);
echo $rs-Fields(22); //this is where that particular field is NULL, and
produces the error
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 8:08 AM
To: David Stoltz
Cc: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
Could you post your database's class to here?
I'm use mssql with php for several years and read NULL Fields is never
appear your case.
2009/8/26 David Stoltz dsto...@shh.org:
I tried that -it's in the first part of my message
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 7:39 AM
To: David Stoltz
Cc: Paul M Foster; php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
use is_null() check it
2009/8/26 David Stoltz dsto...@shh.org:
Paul,
This all started because when I try this:
?php echo $rs-Fields(22);?
It work fine, as long as there is a non-null value there, otherwise it
produces an error.
Also, I'm working with a Microsoft SQL 2000 database, not MySQLnot
sure if that matters
But echo $rs-Fields(22) works perfectly for dumping values out of my
$rs recordset...that is, unless the value is NULL is the database - then
I get:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176
-Original Message-
From: Paul M Foster [mailto:pa...@quillandmouse.com]
Sent: Tuesday, August 25, 2009 4:39 PM
To: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted
to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
Just a thought... do you really mean $rs-Fields(22) or do you mean
$rs-Fields[22]? The former is a function call and the latter is an
array variable.
Paul
--
Paul M. Foster
--
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That's impressive, but are you sure you don't use php_mssql extensions
instead? :p
Thanks,
Ash
http://www.ashleysheridan.co.uk
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RE: [PHP] How to output a NULL field?
Wow - thanks for the code, but it's over my head at this point.
I'm a PHP newbieI typically use ASP Classic, but I realize I need to learn
PHP for ongoing development. Problem is, we don't have MySQL here, so I have to
fumble my way through with MS SQL.
Thanks!
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 10:13 AM
To: a...@ashleysheridan.co.uk
Cc: David Stoltz; php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
My code for mssql
please enable the php's mssql extentions.
it used like so many mysql class that you can find by google
--
?php
if(!defined('IN_WEB')) {
exit('Access Denied');
}
ini_set('mssql.datetimeconvert',0);//php4.2.0 disable php's automatic
datetime convert
Class DB {
var $querynum=0;
var $mssql_link;
var $conn_link;
var $sp_link;
var $sp_name='';
var $error_stop=0;
var $show_error=0;
var $dbhost;
var $dbuser;
var $dbpw;
var $dbname;
var $pconnect;
var $var_type=array();
var $fields_name=array();
var $last_error_msg='';
var $phprunversion='';
function DB() {
//define type for sp
$this-var_type['sp_bit']=SQLBIT;
$this-var_type['sp_tinyint']=SQLINT1;
$this-var_type['sp_smallint']=SQLINT2;
$this-var_type['sp_int']=SQLINT4;
$this-var_type['sp_bigint']=SQLVARCHAR;
$this-var_type['sp_real']=SQLFLT4;
$this-var_type['sp_float']=SQLFLT8;
$this-var_type['sp_float-null']=SQLFLTN;
$this-var_type['sp_smallmoney']=SQLFLT8;
$this-var_type['sp_money']=SQLFLT8;
$this-var_type['sp_money-null']=SQLFLT8;
$this-var_type['sp_char']=SQLCHAR;
$this-var_type['sp_varchar']=SQLVARCHAR;
$this-var_type['sp_text']=SQLTEXT;
$this-var_type['sp_datetime']=SQLINT4;
$this-phprunversion=phpversion();
//end
}
/*=php4.4.1,=php5.1.1
a new paramate for if use newlink for connect,pconnect
*/
function rconnect($newlink=false){//2007.03.01 by hack988 fix
phpversion check
if($this-phprunversion = '4.4.1' $this-phprunversion
'5.0.0'
|| $this-phprunversion = '5.1.1'){
return $this-rconnect4p($newlink);
}else{
return $this-rconnect3p();
}
}
function rconnect3p(){
$this-mssql_link = $this-pconnect==0 ?
mssql_connect($this-dbhost, $this-dbuser, $this-dbpw) :
mssql_pconnect($this-dbhost, $this-dbuser, $this-dbpw);
if(!$this-mssql_link){
$this-halt(connect
($this-pconnect)MSSQL($this-dbhost,$this-dbuser)failed!);
return false;
}else{
$this-conn_link=$this-mssql_link;
if($this-dbname) {
if
(!...@$this-select_db($this-dbname,$this-conn_link)){
$this-halt('can not use database
'.$this-dbname);
return false;
}else{
return true;
}
}else{
return true;
}
}
}
function rconnect4p($newlink=false){
$this-mssql_link = $this-pconnect==0 ?
mssql_connect($this-dbhost, $this-dbuser, $this-dbpw , $newlink) :
mssql_pconnect($this-dbhost, $this-dbuser, $this-dbpw, $newlink);
if(!$this-mssql_link){
$this-halt(reconect($this-pconnect)MSSQL($this-dbhost,$this-dbuser)failed);
return false;
}else{
$this-conn_link=$this-mssql_link;
if($this-dbname) {
if
(!...@$this-select_db($this-dbname,$this-conn_link)){
$this-halt('can not use database
'.$this-dbname);
return false;
}else{
return true;
}
}else{
return true;
}
}
}
function connect($dbhost, $dbuser, $dbpw, $dbname, $pconnect =
0,$auto_conn=0 ,$newlink=false
RE: [PHP] How to output a NULL field?
On Wed, 2009-08-26 at 10:50 -0400, David Stoltz wrote:
Wow - thanks for the code, but it's over my head at this point.
I'm a PHP newbieI typically use ASP Classic, but I realize I need to
learn PHP for ongoing development. Problem is, we don't have MySQL here, so I
have to fumble my way through with MS SQL.
Thanks!
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 10:13 AM
To: a...@ashleysheridan.co.uk
Cc: David Stoltz; php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
My code for mssql
please enable the php's mssql extentions.
it used like so many mysql class that you can find by google
--
?php
if(!defined('IN_WEB')) {
exit('Access Denied');
}
ini_set('mssql.datetimeconvert',0);//php4.2.0 disable php's automatic
datetime convert
Class DB {
var $querynum=0;
var $mssql_link;
var $conn_link;
var $sp_link;
var $sp_name='';
var $error_stop=0;
var $show_error=0;
var $dbhost;
var $dbuser;
var $dbpw;
var $dbname;
var $pconnect;
var $var_type=array();
var $fields_name=array();
var $last_error_msg='';
var $phprunversion='';
function DB() {
//define type for sp
$this-var_type['sp_bit']=SQLBIT;
$this-var_type['sp_tinyint']=SQLINT1;
$this-var_type['sp_smallint']=SQLINT2;
$this-var_type['sp_int']=SQLINT4;
$this-var_type['sp_bigint']=SQLVARCHAR;
$this-var_type['sp_real']=SQLFLT4;
$this-var_type['sp_float']=SQLFLT8;
$this-var_type['sp_float-null']=SQLFLTN;
$this-var_type['sp_smallmoney']=SQLFLT8;
$this-var_type['sp_money']=SQLFLT8;
$this-var_type['sp_money-null']=SQLFLT8;
$this-var_type['sp_char']=SQLCHAR;
$this-var_type['sp_varchar']=SQLVARCHAR;
$this-var_type['sp_text']=SQLTEXT;
$this-var_type['sp_datetime']=SQLINT4;
$this-phprunversion=phpversion();
//end
}
/*=php4.4.1,=php5.1.1
a new paramate for if use newlink for connect,pconnect
*/
function rconnect($newlink=false){//2007.03.01 by hack988 fix
phpversion check
if($this-phprunversion = '4.4.1' $this-phprunversion
'5.0.0'
|| $this-phprunversion = '5.1.1'){
return $this-rconnect4p($newlink);
}else{
return $this-rconnect3p();
}
}
function rconnect3p(){
$this-mssql_link = $this-pconnect==0 ?
mssql_connect($this-dbhost, $this-dbuser, $this-dbpw) :
mssql_pconnect($this-dbhost, $this-dbuser, $this-dbpw);
if(!$this-mssql_link){
$this-halt(connect
($this-pconnect)MSSQL($this-dbhost,$this-dbuser)failed!);
return false;
}else{
$this-conn_link=$this-mssql_link;
if($this-dbname) {
if
(!...@$this-select_db($this-dbname,$this-conn_link)){
$this-halt('can not use database
'.$this-dbname);
return false;
}else{
return true;
}
}else{
return true;
}
}
}
function rconnect4p($newlink=false){
$this-mssql_link = $this-pconnect==0 ?
mssql_connect($this-dbhost, $this-dbuser, $this-dbpw , $newlink) :
mssql_pconnect($this-dbhost, $this-dbuser, $this-dbpw, $newlink);
if(!$this-mssql_link){
$this-halt(reconect($this-pconnect)MSSQL($this-dbhost,$this-dbuser)failed);
return false;
}else{
$this-conn_link=$this-mssql_link;
if($this-dbname) {
if
(!...@$this-select_db($this-dbname,$this-conn_link)){
$this-halt('can not use database
'.$this-dbname);
return false;
}else{
return true;
}
}else{
return true;
}
}
}
function connect($dbhost, $dbuser, $dbpw, $dbname, $pconnect =
0,$auto_conn=0 ,$newlink=false
Re: [PHP] How to output a NULL field?
On Wed, Aug 26, 2009 at 10:52 AM, Ashley Sheridana...@ashleysheridan.co.uk wrote: You should try and see if you can get it installed there, as it will work on Windows servers. I've found it generally to be faster than MS SQL, and the choice of different database engines for each table gives you a LOT of flexibility for the future too. Also, MySQL offers a bit more functionality I've found. Thanks, Ash http://www.ashleysheridan.co.uk To be fair, I've not had found any performance problems when using SQL Server. And, while there is one main feature MySQL has that I really miss in SQL Server (the LIMIT clause), I miss enforcement of CHECK constraints in MySQL. (I suppose I could implement them via triggers, but that just seems messy to me.) I'm all for flexibility, though. Andrew -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] How to output a NULL field?
On Wed, Aug 26, 2009 at 9:51 AM, David Stoltzdsto...@shh.org wrote:
Sorry - I don't know what you mean by DB class?
I'm using Microsoft SQL 2000with this code:
?php
//create an instance of the ADO connection object
$conn = new COM (ADODB.Connection)
or die(Cannot start ADO);
//define connection string, specify database driver
$connStr = PROVIDER=SQLOLEDB;SERVER=;UID=xxx;PWD=;DATABASE=;
$conn-open($connStr); //Open the connection to the database
$query = SELECT * FROM eval_evaluations WHERE id = .$_POST[eval];
$rs = $conn-execute($query);
echo $rs-Fields(22); //this is where that particular field is NULL, and
produces the error
Because you are using COM, you can't use PHP's empty(), isset(), or
is_null() in your if(...) statement. I've not used COM for much in
PHP, but I think you'll have to do something like this:
switch (variant_get_type($rs-Fields(22)) {
case VT_EMPTY:
case VT_NULL:
$q4 = '';
break;
case VT_UI1:
--
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To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] How to output a NULL field?
Com function is just for Windows,I don't kown why some body like use it.:(
2009/8/27 Andrew Ballard aball...@gmail.com:
On Wed, Aug 26, 2009 at 9:51 AM, David Stoltzdsto...@shh.org wrote:
Sorry - I don't know what you mean by DB class?
I'm using Microsoft SQL 2000with this code:
?php
//create an instance of the ADO connection object
$conn = new COM (ADODB.Connection)
or die(Cannot start ADO);
//define connection string, specify database driver
$connStr = PROVIDER=SQLOLEDB;SERVER=;UID=xxx;PWD=;DATABASE=;
$conn-open($connStr); //Open the connection to the database
$query = SELECT * FROM eval_evaluations WHERE id = .$_POST[eval];
$rs = $conn-execute($query);
echo $rs-Fields(22); //this is where that particular field is NULL, and
produces the error
Because you are using COM, you can't use PHP's empty(), isset(), or
is_null() in your if(...) statement. I've not used COM for much in
PHP, but I think you'll have to do something like this:
switch (variant_get_type($rs-Fields(22)) {
case VT_EMPTY:
case VT_NULL:
$q4 = '';
break;
case VT_UI1:
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] How to output a NULL field?
On Wed, Aug 26, 2009 at 12:13 PM, Andrew Ballardaball...@gmail.com wrote:
On Wed, Aug 26, 2009 at 9:51 AM, David Stoltzdsto...@shh.org wrote:
Sorry - I don't know what you mean by DB class?
I'm using Microsoft SQL 2000with this code:
?php
//create an instance of the ADO connection object
$conn = new COM (ADODB.Connection)
or die(Cannot start ADO);
//define connection string, specify database driver
$connStr = PROVIDER=SQLOLEDB;SERVER=;UID=xxx;PWD=;DATABASE=;
$conn-open($connStr); //Open the connection to the database
$query = SELECT * FROM eval_evaluations WHERE id = .$_POST[eval];
$rs = $conn-execute($query);
echo $rs-Fields(22); //this is where that particular field is NULL, and
produces the error
Because you are using COM, you can't use PHP's empty(), isset(), or
is_null() in your if(...) statement. I've not used COM for much in
PHP, but I think you'll have to do something like this:
switch (variant_get_type($rs-Fields(22)) {
case VT_EMPTY:
case VT_NULL:
$q4 = '';
break;
case VT_UI1:
blast ... hit some key and Gmail just sent what I had typed so far.
At any rate, hopefully you can get the idea from that last post. Look
at this reference for handling different types returned from COM:
http://www.php.net/manual/en/com.constants.php
I would also suggest looking into a PHP library for querying SQL
Server rather than relying on COM. There are several, some work better
than others. I've found that the SQL Server Driver for PHP works best
for what I use. It is documented as being for 2005 and newer, but I
have been able to use it just fine with 2000 as well.
Andrew
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Re: [PHP] How to output a NULL field?
On Wed, Aug 26, 2009 at 12:06 PM, hack988 hack988hack...@dev.htwap.com wrote: Mysql,mssql has its own feature,you can't say Mysql is better than Mssql or Mssql it better than Mysql,Is'nt is? My the Way ,Mssql support Top n,m form mssql 2005 :) Perhaps, but the OP said he's using SQL Server 2000. Andrew -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] How to output a NULL field?
I'm sorry for my poor English,what is OP? I don't kown what is OP mean. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] How to output a NULL field?
On Thu, 2009-08-27 at 00:25 +0800, hack988 hack988 wrote: I'm sorry for my poor English,what is OP? I don't kown what is OP mean. It's basically the original person who asked the question. Thanks, Ash http://www.ashleysheridan.co.uk -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] How to output a NULL field?
I'm using COM because I don't know what else to use ;-)
Like I said, I'm new to PHP. Here is another way I communicate with the
database, let me know if this is a better way (it requires a stored procedure):
//Assign the server connection to a variable
$connect = mssql_connect(SERVER,1433', USER, 'PASSWORD');
//Select your database and reference the connection
mssql_select_db('DATABASE', $connect);
// Create a new stored prodecure
$stmt = mssql_init('StoredProcedureName');
// Bind the field names
mssql_bind($stmt, '@category',$cat,SQLINT4,false,false,4);
mssql_bind($stmt, '@userid',$userid,SQLINT4,false,false,4);
mssql_bind($stmt, '@passwordtitle',$pname,SQLVARCHAR,false,false,100);
mssql_bind($stmt, '@password',$encrypted_data,SQLVARCHAR,false,false,1000);
mssql_bind($stmt, '@alt',$alt,SQLVARCHAR,false,false,150);
mssql_bind($stmt, '@desc',$desc,SQLVARCHAR,false,false,100);
// Execute
mssql_execute($stmt);
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 12:18 PM
To: Andrew Ballard
Cc: David Stoltz; php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
Com function is just for Windows,I don't kown why some body like use it.:(
2009/8/27 Andrew Ballard aball...@gmail.com:
On Wed, Aug 26, 2009 at 9:51 AM, David Stoltzdsto...@shh.org wrote:
Sorry - I don't know what you mean by DB class?
I'm using Microsoft SQL 2000with this code:
?php
//create an instance of the ADO connection object
$conn = new COM (ADODB.Connection)
or die(Cannot start ADO);
//define connection string, specify database driver
$connStr = PROVIDER=SQLOLEDB;SERVER=;UID=xxx;PWD=;DATABASE=;
$conn-open($connStr); //Open the connection to the database
$query = SELECT * FROM eval_evaluations WHERE id = .$_POST[eval];
$rs = $conn-execute($query);
echo $rs-Fields(22); //this is where that particular field is NULL, and
produces the error
Because you are using COM, you can't use PHP's empty(), isset(), or
is_null() in your if(...) statement. I've not used COM for much in
PHP, but I think you'll have to do something like this:
switch (variant_get_type($rs-Fields(22)) {
case VT_EMPTY:
case VT_NULL:
$q4 = '';
break;
case VT_UI1:
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] How to output a NULL field?
On Thu, Aug 27, 2009 at 12:25:35AM +0800, hack988 hack988 wrote: I'm sorry for my poor English,what is OP? I don't kown what is OP mean. OP = Original Poster. That's usually the person who first started (posted) a thread on a list. Paul -- Paul M. Foster -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] How to output a NULL field?
Looking on Amazon and other book sites, I can't even find a book for PHP and MS SQL... It's all PHP and MySQL... Should I avoid developing PHP application with MS SQL databases? -Original Message- From: Paul M Foster [mailto:pa...@quillandmouse.com] Sent: Wednesday, August 26, 2009 12:38 PM To: php-general@lists.php.net Subject: Re: [PHP] How to output a NULL field? On Thu, Aug 27, 2009 at 12:25:35AM +0800, hack988 hack988 wrote: I'm sorry for my poor English,what is OP? I don't kown what is OP mean. OP = Original Poster. That's usually the person who first started (posted) a thread on a list. Paul -- Paul M. Foster -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] How to output a NULL field?
On Wed, 2009-08-26 at 12:48 -0400, David Stoltz wrote: Looking on Amazon and other book sites, I can't even find a book for PHP and MS SQL... It's all PHP and MySQL... Should I avoid developing PHP application with MS SQL databases? -Original Message- From: Paul M Foster [mailto:pa...@quillandmouse.com] Sent: Wednesday, August 26, 2009 12:38 PM To: php-general@lists.php.net Subject: Re: [PHP] How to output a NULL field? On Thu, Aug 27, 2009 at 12:25:35AM +0800, hack988 hack988 wrote: I'm sorry for my poor English,what is OP? I don't kown what is OP mean. OP = Original Poster. That's usually the person who first started (posted) a thread on a list. Paul -- Paul M. Foster -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php I'd definately recommend going down the MySQL route, but it may well depend on what sort of apps you are developing and where your target markets lie. Thanks, Ash http://www.ashleysheridan.co.uk -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] How to output a NULL field?
On Wed, Aug 26, 2009 at 12:48:12PM -0400, David Stoltz wrote: Looking on Amazon and other book sites, I can't even find a book for PHP and MS SQL... It's all PHP and MySQL... Should I avoid developing PHP application with MS SQL databases? My *opinion* is that you should avoid Microsoft products whenever possible. BTW, you can also find books on PHP and PostgreSQL, which, also in my *opinion* is a superior choice over MySQL. Paul -- Paul M. Foster -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] How to output a NULL field?
On Aug 26, 2009, at 12:31 PM, David Stoltz dsto...@shh.org wrote:
I'm using COM because I don't know what else to use ;-)
Like I said, I'm new to PHP. Here is another way I communicate with
the database, let me know if this is a better way (it requires a
stored procedure):
//Assign the server connection to a variable
$connect = mssql_connect(SERVER,1433', USER, 'PASSWORD');
//Select your database and reference the connection
mssql_select_db('DATABASE', $connect);
// Create a new stored prodecure
$stmt = mssql_init('StoredProcedureName');
// Bind the field names
mssql_bind($stmt, '@category',$cat,SQLINT4,false,false,4);
mssql_bind($stmt, '@userid',$userid,SQLINT4,false,false,4);
mssql_bind($stmt, '@passwordtitle',$pname,SQLVARCHAR,false,false,100);
mssql_bind($stmt, '@password',$encrypted_data,SQLVARCHAR,false,false,
1000);
mssql_bind($stmt, '@alt',$alt,SQLVARCHAR,false,false,150);
mssql_bind($stmt, '@desc',$desc,SQLVARCHAR,false,false,100);
// Execute
mssql_execute($stmt);
-Original Message-
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 12:18 PM
To: Andrew Ballard
Cc: David Stoltz; php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
Com function is just for Windows,I don't kown why some body like use
it.:(
2009/8/27 Andrew Ballard aball...@gmail.com:
On Wed, Aug 26, 2009 at 9:51 AM, David Stoltzdsto...@shh.org wrote:
Sorry - I don't know what you mean by DB class?
I'm using Microsoft SQL 2000with this code:
?php
//create an instance of the ADO connection object
$conn = new COM (ADODB.Connection)
or die(Cannot start ADO);
//define connection string, specify database driver
$connStr =
PROVIDER=SQLOLEDB;SERVER=;UID=xxx;PWD=;DATABASE=;
$conn-open($connStr); //Open the connection to the database
$query = SELECT * FROM eval_evaluations WHERE id = .$_POST
[eval];
$rs = $conn-execute($query);
echo $rs-Fields(22); //this is where that particular field is
NULL, and produces the error
Because you are using COM, you can't use PHP's empty(), isset(), or
is_null() in your if(...) statement. I've not used COM for much in
PHP, but I think you'll have to do something like this:
switch (variant_get_type($rs-Fields(22)) {
case VT_EMPTY:
case VT_NULL:
$q4 = '';
break;
case VT_UI1:
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MS has developed and released a new version of the mssql drivers for
php. It might be worth investigating that to see ifthat firs your needs.
Bastien
Sent from my iPod
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[PHP] How to output a NULL field?
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
--
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Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 2:00 PM, David Stoltzdsto...@shh.org wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
--
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To unsubscribe, visit: http://www.php.net/unsub.php
$q4 = '' . $rs-Fields(22);
Note that it's two single quotes
--
Bastien
Cat, the other other white meat
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RE: [PHP] How to output a NULL field?
if(empty($rs-Fields(22))){
$q4 = '';
}else{
$q4 = ''.$rs-Fields(22);
}
Still produces errors, whether using empty or is_nullwith single
quotes
???
-Original Message-
From: Bastien Koert [mailto:phps...@gmail.com]
Sent: Tuesday, August 25, 2009 2:17 PM
To: David Stoltz
Cc: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 2:00 PM, David Stoltzdsto...@shh.org wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
$q4 = '' . $rs-Fields(22);
Note that it's two single quotes
--
Bastien
Cat, the other other white meat
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Re: [PHP] How to output a NULL field?
2009/8/25 David Stoltz dsto...@shh.org:
if(empty($rs-Fields(22))){
Hi David,
You cannot call empty() on a function or class method like that. From
the manual:
Note: empty() only checks variables as anything else will result in
a parse error. In other words, the following will not work:
empty(trim($name)). - http://www.php.net/empty
You should assign the result of $rs-Fields(22) to a variable before
calling empty() on it, or use a class variable access to check it, or
check it first within the class and have Fields() return a suitable
value if needed.
Regards,
Torben
$q4 = '';
}else{
$q4 = ''.$rs-Fields(22);
}
Still produces errors, whether using empty or is_nullwith single
quotes
???
-Original Message-
From: Bastien Koert [mailto:phps...@gmail.com]
Sent: Tuesday, August 25, 2009 2:17 PM
To: David Stoltz
Cc: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 2:00 PM, David Stoltzdsto...@shh.org wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
$q4 = '' . $rs-Fields(22);
Note that it's two single quotes
--
Bastien
Cat, the other other white meat
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
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Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz wrote:
$rs-Fields(22) equals a NULL in the database
My Code:
if(empty($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
Produces this error:
Fatal error: Can't use method return value in write context in
D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
Line 32 is the if line...
If I switch the code to (using is_null):
if(is_null($rs-Fields(22))){
$q4 = ;
}else{
$q4 = $rs-Fields(22);
}
It produces this error:
Catchable fatal error: Object of class variant could not be converted to
string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
Line 196 is: ?php echo $q4;?
What am I doing wrong?
Thanks!
Just a thought... do you really mean $rs-Fields(22) or do you mean
$rs-Fields[22]? The former is a function call and the latter is an
array variable.
Paul
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