Re: [PHP] Weird problem with is_file()
On 25 April 2010 22:14, Michelle Konzack linux4miche...@tamay-dogan.net wrote:
Hi,
I have a code sniplet which does not work and I do not know why:
8--
$isfile=shell_exec(ls /tmp/tdphp-vserver/SESSION_ .
$_SERVER['REMOTE_ADDR'] . _ . $_COOKIE['VSERVER_AUTHUSER'] . _* |head
-n1);
if (is_file($isfile)) {
snip
8--
nothing special, and the file is there, but the stuff with
is_file($isfile)
is not working... If I enter the file in place of $isfile, then it is
working. Quoting of $isfile does not work too.
What have a overseen?
var_dump($isfile);
Don't make assumptions of what the value is, just check it.
Regards
Peter
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Re: [PHP] Weird problem with is_file()
2010/4/25 Michelle Konzack linux4miche...@tamay-dogan.net: Hi, I have a code sniplet which does not work and I do not know why: 8-- $isfile=shell_exec(ls /tmp/tdphp-vserver/SESSION_ . $_SERVER['REMOTE_ADDR'] . _ . $_COOKIE['VSERVER_AUTHUSER'] . _* |head -n1); Hi Michelle, I would recommend not to let any user input to your shell. This piece of code is very insecure as any client may manipulate the shell command at will. You don't want people to take over your server that easily. See http://www.php.net/escapeshellcmd and alike. Regards -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Weird problem with is_file()
Hi Michelle,
Maybe try adding an error catch to see what you get.
if(is_file($isFile)) or die(mysql_error());
On Apr 25, 2010, at 3:14 PM, Michelle Konzack wrote:
if (is_file($isfile)) {
Karl DeSaulniers
Design Drumm
http://designdrumm.com
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Re: [PHP] Weird problem with is_file()
Oops, my apologies. That is a PHP MySQL solution. Wrong list.
Please disregard that solution.
Maybe find a way to get an error response from your session to see
what the problem is.
Karl
On Apr 25, 2010, at 3:24 PM, Karl DeSaulniers wrote:
Hi Michelle,
Maybe try adding an error catch to see what you get.
if(is_file($isFile)) or die(mysql_error());
On Apr 25, 2010, at 3:14 PM, Michelle Konzack wrote:
if (is_file($isfile)) {
Karl DeSaulniers
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Re: [PHP] Weird problem with T_gettext()
Michelle Konzack wrote: Now selecting de as prefered language is working, but if I select de_DE in Firefox, the second term About Us is something like Ãber Uns insteed Über Uns because the Website is UTF-8. Indeed. I'll venture a guess and say it's because your dictionary file is bad or not actually UTF8. This is realy weird, since the MO file is the same for both locale versions. Ah. Very weird, yes. Are you doing a setlocale() anywhere? Also it is weird, that php-gettext does not do a fallback, if I have only de and not de_DE. Fallback to what? I only have 'de' in my browser language preferences, and your site came up fine in German. /Per -- Per Jessen, Zürich (11.7°C) -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Weird problem with HTML form and $_POST
On Wed, Nov 5, 2008 at 8:47 AM, Oscar Gosdinski [EMAIL PROTECTED] wrote: Name the select tag as sub_projects[], then in PHP you can read the $sub_project variable as an array. Oops, my error... you have to use the $_POST['sub_projects'] variable as an array. -- Saludos Oscar -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Weird problem with HTML form and $_POST
On Sat, Nov 1, 2008 at 6:20 PM, Michelle Konzack [EMAIL PROTECTED] wrote: select class=setupProjectAddSelect name=sub_projects multiple=multiple size=10 option value=6 selectedATX 120/option option value=4ATX 60/option option value=5ATX 90/option option value=11Device 2/option option value=12Device 4/option option value=13Lowvoltage/option option value=7P4 Module/option option value=8SATA 2/option option value=9 selectedSATA 4/option option value=10 selectedSATA 8/option /select Hell, why is $_POST['sub_projects'] eating the other three items and take only the last one? Name the select tag as sub_projects[], then in PHP you can read the $sub_project variable as an array. -- Saludos Oscar -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] weird problem in php
On 12/06/06, weetat [EMAIL PROTECTED] wrote:
Hi all ,
I have using php 4.3.2 and mysql database.
I have a form which have select tag which have the value for example
-New York.
When use submit the form , i need to find the first occurence of - ,
i use strpos function as shown below :
$country = $_POST['country'];
$findme = '-';
$pos = strpos($country, $findme);
if ($pos === false) {
$_logger-logdebug(starting searchChassisTblDB() not found);
} else {
$_logger-logdebug(starting searchChassisTblDB() found);
}
however it always give false . I did not know why.
Then i did a testing , added below code in the php file without form
submission, it give true value which what i wanted. Anybody have ideas
or suggestion what happening ? Any difference when value submitted from
form ? Thanks
$mystring = '-New York';
$findme = '-';
$pos = strpos($mystring, $findme);
if ($pos === false) {
echo The string '$findme' was not found in the string '$mystring';
} else {
echo The string '$findme' was found in the string '$mystring';
echo and exists at position $pos;
}
why don't you echo $_POST['country'] to see what you get?
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Re: [PHP] weird problem in php
On 6/12/06, weetat [EMAIL PROTECTED] wrote:
I have a form which have select tag which have the value for example
-New York.
When use submit the form , i need to find the first occurence of - ,
i use strpos function as shown below :
$country = $_POST['country'];
$findme = '-';
$pos = strpos($country, $findme);
if ($pos === false) {
$_logger-logdebug(starting searchChassisTblDB() not found);
} else {
$_logger-logdebug(starting searchChassisTblDB() found);
}
however it always give false . I did not know why.
Then i did a testing , added below code in the php file without form
submission, it give true value which what i wanted. Anybody have ideas
or suggestion what happening ?
Perhaps there is a problem with the HTML for your select box.
var_dump($_POST) in your PHP code to check if the value is
being sent correctly.
Rabin
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Re: [PHP] weird problem with index page
loll wrote: Hi, I am not sure if this is a PHP or Apache problem, but I am hoping someone can tellme what is wrong. When I go to www.domain.com on my server it shows the page as text instead of parsing the php code. If I go to www.domain.com/index.php (same page) it runs as it should. if I go to www.domain.com/subdir it sya s page not found even though there is a index.php file in that directory. I am at a loss as to what is wrong. index.php is listed inthe directoryindex for apache so I dont understand why it is doing all this. Interesting. Do you use virtual hosts? How many DirectoryIndex directives in your configuration file? What are their contexts? Got any error in your log file? Best Bao If anyone can help me it would be appreciated. Thanks Loll -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] weird problem with index page
Well, I managed to part fix the problem, now I have a problem where there is a page that all it does is redirect to another page, it wont redirect unless I hit the refresh button. When I go to index.php it redirects to login.php, this is an old page that I just added a header(Location: blah); to so it would go to the new page. now it seems to jsut show a blank page instead of doing the second redirect, if I hit refresh on the blank page it will redirect correctly. I wondered if there was some restriction somewhere on only allowing 1 redirect where this page needs to be able to do more than one? is htis something that would be in the httpd.conf or php.ini file? Thanks! Loll -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Weird Problem with INPUT tag
Hello Ahbaid, Sunday, February 29, 2004, 6:33:45 AM, you wrote: AG this uses a FORM with INPUT tags, what is weird is that under IE6 some AG of the INPUT fields are showing up with yellow AG backgrounds Sounds to me like you have the Google toolbar installed with the AutoFill option enabled. See the preferences to turn the yellow background off. -- Best regards, Richard Davey http://www.phpcommunity.org/wiki/296.html -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] (WEIRD problem) Not following directories!!! (0t)
Hey Andrew, First, note that that is an apache problem, not PHP. Yep, thats why I put the Ot in the subject line but since you cant have PHP without a webserver and apache is the most popular on the list or the net...I was hopeing someone else had this problem and could guide me In httpd.conf add Option Indexes to the folder (root for whole site). That will produce a file list of the directory. Thanks, will try that right now. Or, put in a 'default' document (ie: index.html) that will produce the output when people browse to the directory. Will try that too. Directory browsing is normaly disabled in web servers (from experience, apache and IIS). See? didnt know that. Learned something new. HTH Sure did. Thanks. Cheers, -Ryan Andrew Hi, Heres a totally weird problem, I created some folders on our server (which we are accessing only via ip as the domain has not been resolved) but I get an error when i try to access the folder(s)! eg: (I created this folder) ryan and i try to access it like this: http://208.234.29.220/ryan/ and it gives me this error: You don't have permission to access /ryan/ on this server. Which is totally weird coz I have NOTHING in that folder right now... not even a .htaccess file What could be wrong? I have looked in my httpd.conf file and on google but cant seem to -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Weird problem with objects
Mark Cubitt wrote: $this-$database = new database(); $this-$accessory = new accessory(); $this-$order = new order(); These should be... $this-database = new database(); $this-accessory = new accessory(); $this-order = new order(); You don't want the $ unless $database, $accessory and $order are strings containing the class attribute names. Additionally, you might want to assign the reference of the new objects otherwise PHP will create a new object and then make a copy of it - not very efficient. Like so... $this-database = new database(); $this-accessory = new accessory(); $this-order = new order(); -- Stuart -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: RE : [PHP] Weird problem with objects
Vincent DUPONT wrote: what do you mean by this :Additionally, you might want to assign the reference of the new objects otherwise PHP will create a new object and then make a copy of it - not very efficient. Like so... $this-database = new database(); $this-accessory = new accessory(); $this-order = new order(); The = operator makes a copy of the right side and puts it into the left side. So the new database() bit will create a new object of type database. The = operator will then make a copy of that object and put it into $this-database. So it's actually creating 2 database objects one of which will get deleted at the end of the function. By assigning the reference (that's what the gets) only one object is created. Does that make sense now? -- Stuart -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Weird problem when creating a db connection and trying to reference it in a function
Matt Babineau wrote:
I found another strange problem. I am creating a mysql database
connection like this:
@ $db = mysql_pconnect(host, user, pass);
mysql_select_db(dbname);
you forgot the $ ... should be mysql_select_db($dbname);
If you had turned up your php error reporting level
(error_reporting(E_ALL); for example) it whould have given you a notice
about undeclared constant for that line.
A php-aware syntax highlighting editor would have also given you a
visual cue about the possible error on that mysql_select_db() line.
Also, I don't see where you are setting $dbname to something. You might
have just not included it in your excerpt, but it may also be missing
from your script :)
Now, I have a function that inside uses mysql_list_fields()
function getFields($table, $dbname) {
$fields = mysql_list_fields($dbname, $table);
return $fields;
}
you can just do this return mysql_list_fields($dbname, $table);
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Re: [PHP] Weird Problem
Hello, This is a reply to an e-mail that you wrote on Tue, 29 Jul 2003 at 20:40, lines prefixed by '' were originally written by you. I have the following chunk of code: $sql = SELECT setting from settings where name='display_rows'; include(connect.inc.php); print $sql; $row = mysql_fetch_row($result); $path = $row[0]; print $path; It always prints out 1 But if I run the code at the sql command prompt, it prints out 25, which is the correct value. Does anyone have any ideas why im having this problem?? - --- Unless it is done within connect.inc.php you have not executed your query. Maybe the last query you executed has a row containing the value 1 ready to be fetched? Try adding $queryid = mysql_query($sql); before $row = mysql_fetch_row($result); David. -- phpmachine :: The quick and easy to use service providing you with professionally developed PHP scripts :: http://www.phpmachine.com/ Professional Web Development by David Nicholson http://www.djnicholson.com/ QuizSender.com - How well do your friends actually know you? http://www.quizsender.com/ (developed entirely in PHP) -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Weird Problem
I have the following chunk of code: $sql = SELECT setting from settings where name='display_rows'; include(connect.inc.php); print $sql; $row = mysql_fetch_row($result); $path = $row[0]; print $path; It always prints out 1 But if I run the code at the sql command prompt, it prints out 25, which is the correct value. Does anyone have any ideas why im having this problem?? Are you actually executing the query? I don't see a mysql_query() anywhere and it would be anti-intuituve to have that in your connection include. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Weird Problem
[snip] $sql = SELECT setting from settings where name='display_rows'; include(connect.inc.php); print $sql; $row = mysql_fetch_row($result); $path = $row[0]; print $path; [/snip] Because $result is not being gotten properly... $sql = SELECT setting from settings where name='display_rows'; include(connect.inc.php); print $sql; $result = mysql_query($sql); $row = mysql_fetch_row($result); $path = $row[0]; print $path; HTH! -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Weird Problem
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Yes, connect.inc.php does the following:
@$connect=mysql_connect(localhost,$dbuser,$dbpass);
if (!$connect) {
echo Could not connect to mysql!;
exit();
}
@$db=mysql_select_db($dbname);
if (!$db){
echo Could not connect to database;
exit ();
}
@$result=mysql_query($sql);
if (!$result){
echo Invalid Sql Command;
exit ();
}
- ---
Aaron Axelsen
AIM: AAAK2
Email: [EMAIL PROTECTED]
Want reliable web hosting at affordable prices?
www.modevia.com
Web Dev/Design Community/Zine
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- -Original Message-
From: Jennifer Goodie [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 29, 2003 2:46 PM
To: Aaron Axelsen; [EMAIL PROTECTED]
Subject: RE: [PHP] Weird Problem
I have the following chunk of code:
$sql = SELECT setting from settings where name='display_rows';
include(connect.inc.php);
print $sql;
$row = mysql_fetch_row($result);
$path = $row[0];
print $path;
It always prints out 1
But if I run the code at the sql command prompt, it prints out
25, which is the correct value.
Does anyone have any ideas why im having this problem??
Are you actually executing the query? I don't see a mysql_query()
anywhere and it would be anti-intuituve to have that in your
connection include.
-BEGIN PGP SIGNATURE-
Version: PGPfreeware 7.0.3 for non-commercial use http://www.pgp.com
iQA/AwUBPybRBrrnDjSLw9ADEQJrmACg70sEqN0efwGDg4o958tPx9s6K+0AoJKK
3WZTtCIVXsszKlGA/jOzrmMR
=LwnA
-END PGP SIGNATURE-
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RE: [PHP] Weird Problem
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 The only setting contained in that table is the one im calling for, and its value is 25. - --- Aaron Axelsen AIM: AAAK2 Email: [EMAIL PROTECTED] Want reliable web hosting at affordable prices? www.modevia.com Web Dev/Design Community/Zine www.developercube.com - -Original Message- From: Aaron Axelsen [mailto:[EMAIL PROTECTED] Sent: Tuesday, July 29, 2003 2:40 PM To: [EMAIL PROTECTED] Subject: [PHP] Weird Problem - -BEGIN PGP SIGNED MESSAGE- Hash: SHA1 I have the following chunk of code: $sql = SELECT setting from settings where name='display_rows'; include(connect.inc.php); print $sql; $row = mysql_fetch_row($result); $path = $row[0]; print $path; It always prints out 1 But if I run the code at the sql command prompt, it prints out 25, which is the correct value. Does anyone have any ideas why im having this problem?? - - --- Aaron Axelsen AIM: AAAK2 Email: [EMAIL PROTECTED] Want reliable web hosting at affordable prices? www.modevia.com Web Dev/Design Community/Zine www.developercube.com - -BEGIN PGP SIGNATURE- Version: PGPfreeware 7.0.3 for non-commercial use http://www.pgp.com iQA/AwUBPybNTLrnDjSLw9ADEQI4LQCeP+vaCk+UtDJSeDU9H5yTt+r12JEAniaI LVwLVog89WOCsXxzIHbGsIEb =g4P4 - -END PGP SIGNATURE- - -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -BEGIN PGP SIGNATURE- Version: PGPfreeware 7.0.3 for non-commercial use http://www.pgp.com iQA/AwUBPybRWLrnDjSLw9ADEQJMrQCeIFkRzjnN2zoOhtS+JEO3tcVxiJ0AnR2v JHRpp/8B2Gb581ntxzS+5OGL =85R7 -END PGP SIGNATURE- -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Weird Problem
The only setting contained in that table is the one im calling for, and its value is 25. see what print_r($row) outputs, then you'll at least know what is in the array. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Weird Problem
Isn't there a known problem with variables and included files? Like you have
to create $result in the first file, or something... Just try to have it all
in one file... Also, change mysql_fetch_row($result); to
mysql_fetch_row($result) or print(mysql_error());... Something also tells
me you are using mysql_fetch_row... Why not array? You should try to just
switch between them and try once... I didn't notice it at first...
mysql_fetch_array($result) is what I would use... Will probably return same
thing, but who knows?
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Mail: [EMAIL PROTECTED]
Aaron Axelsen [EMAIL PROTECTED] skrev i meddelandet
news:[EMAIL PROTECTED]
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Yes, connect.inc.php does the following:
@$connect=mysql_connect(localhost,$dbuser,$dbpass);
if (!$connect) {
echo Could not connect to mysql!;
exit();
}
@$db=mysql_select_db($dbname);
if (!$db){
echo Could not connect to database;
exit ();
}
@$result=mysql_query($sql);
if (!$result){
echo Invalid Sql Command;
exit ();
}
- ---
Aaron Axelsen
AIM: AAAK2
Email: [EMAIL PROTECTED]
Want reliable web hosting at affordable prices?
www.modevia.com
Web Dev/Design Community/Zine
www.developercube.com
- -Original Message-
From: Jennifer Goodie [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 29, 2003 2:46 PM
To: Aaron Axelsen; [EMAIL PROTECTED]
Subject: RE: [PHP] Weird Problem
I have the following chunk of code:
$sql = SELECT setting from settings where name='display_rows';
include(connect.inc.php);
print $sql;
$row = mysql_fetch_row($result);
$path = $row[0];
print $path;
It always prints out 1
But if I run the code at the sql command prompt, it prints out
25, which is the correct value.
Does anyone have any ideas why im having this problem??
Are you actually executing the query? I don't see a mysql_query()
anywhere and it would be anti-intuituve to have that in your
connection include.
-BEGIN PGP SIGNATURE-
Version: PGPfreeware 7.0.3 for non-commercial use http://www.pgp.com
iQA/AwUBPybRBrrnDjSLw9ADEQJrmACg70sEqN0efwGDg4o958tPx9s6K+0AoJKK
3WZTtCIVXsszKlGA/jOzrmMR
=LwnA
-END PGP SIGNATURE-
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Re: [PHP] Weird Problem
* Thus wrote DvDmanDT ([EMAIL PROTECTED]): Isn't there a known problem with variables and included files? Like you have to create $result in the first file, or something... Just try to have it all in one file... Also, change mysql_fetch_row($result); to mysql_fetch_row($result) or print(mysql_error());... Something also tells me you are using mysql_fetch_row... Why not array? You should try to just switch between them and try once... I didn't notice it at first... mysql_fetch_array($result) is what I would use... Will probably return same thing, but who knows? For the record, there are no problems with variables and include files. Include files are expanded into the parent file as if they file is in the file that included it. Curt -- I used to think I was indecisive, but now I'm not so sure. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Weird Problem
Aaron Axelsen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Yes, connect.inc.php does the following:
@$connect=mysql_connect(localhost,$dbuser,$dbpass);
if (!$connect) {
echo Could not connect to mysql!;
exit();
}
@$db=mysql_select_db($dbname);
if (!$db){
echo Could not connect to database;
exit ();
}
@$result=mysql_query($sql);
if (!$result){
echo Invalid Sql Command;
exit ();
}
- ---
Aaron Axelsen
AIM: AAAK2
Email: [EMAIL PROTECTED]
Want reliable web hosting at affordable prices?
www.modevia.com
Web Dev/Design Community/Zine
www.developercube.com
The resource identifier returned by mysql_connect() is an optional input to
the mysql_query() function. You should test it out to see if your system
isn't passing the value automatically.
$result = mysql_query($sql, $connect); //-- like this
The reason you may be seeing 1 instead of the value you expected is that the
mysql_query() function can return True even if it doesn't succesfully
execute the query. In that case you may see the output 1 becuase the
boolean True evaluates to 1 in the parser.
--
Kevin
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Re: [PHP] Weird Problem
Curt Zirzow wrote: Include files are expanded into the parent file as if they file is in the file that included it. Ah... clear as mud. heh... ;) -- ---John Holmes... Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/ PHP|Architect: A magazine for PHP Professionals www.phparch.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Weird Problem
On Tue, 27 Feb 2001 05:53, Steve Segarra wrote:
Hi everyone,
First, let me explain I am a very knowledgable php developer. I'm not
someone trying to write their first script. I have been working on a
very large and complex problem for some time now when it suddenly
stopped working. I did not change anything in any configurations or
edit any code that would have affected how my database functionality
class would function. But, all of my connections started to bomb out.
Follows are the entire contents of a file simp.php (minus the line
numbers):
1 ?
2
3 $db = mysql_connect("localhost", "web", "pw") || print "error
connecting ".mysql_error()." ".mysql_errno();
4 mysql_select_db("surveys");
5
6 $result = mysql_query("select * from attract") || print "error
querying ".mysql_error()." ".mysql_errno();
7
8 $i = 0;
9 while ($row = mysql_fetch_array(mysql_query("select * from
attract"))){ // ($i++ 5)){
10 echo $row[0]." ".$row[1]."br";
11 }
12 ?
Please note that the user web@localhost identified by pw has
permissions to select from the table attract.
Here is what I have identified so far:
A - If I use $result in mysql_fetch_array() line 9, I get the error:
Warning: Supplied argument is not a valid MySQL result resource
in simp.php3 on line 9
But if I use the statement mysql_query("select * from attract")
in its
place, the loop (although infinite) will produce expected results, ie
print out the first and second columns of the first row of the table
infinitely. So basically, I would conclude the life of the variable
result is terminiating before I can use it. Why, I don't have a clue.
B - Being tired of an infinite loop in my investigation, I added line 8
and I placed the statement ($i++ 5) inside the while loop
condition. Now, only the br and spaces from line 10 will print, not
the actual values of the $row array. The output is br br br br
br, with no values from the table attract.
This is where I stand now, completely flabberghasted and nearly insane.
Can anyone shed any light on this before I have to commit to bringing
down the webserver while I reinstall php4. (oh, btw I am running
Apache/1.3.14 (Unix) PHP/4.0.4pl1 on Red Hat 6.2 (I hate redhat, but
the bossman says its better supported than slakware. bah.))
Thank you all for your help.
steve
Is it possible that the || construct doesn't work in the way you are
using it and thus you aren't seeing mysql_error? The most common examples
I have seen use the actual string 'or'. Try with a query that you _know_
will trip an error.
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RESEARCH CENTRE FOR INJURY STUDIES | http://www.nisu.flinders.edu.au/
AusEinet| http://auseinet.flinders.edu.au/
Flinders University, ADELAIDE, SOUTH AUSTRALIA
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