Re: [PHP-DB] mysql table join
You may try to create a VIEW. In this case the tables will be dynamically linked, that is when some values change in both tales the views automatically update itself each time you call it. CPT John W. Holmes [EMAIL PROTECTED] ha scritto nel messaggio news:[EMAIL PROTECTED] From: Roger Miranda (Sumac) [EMAIL PROTECTED] Is there a way to permanently join/link two mysql tables? Not without creating another table. CREATE TABLE MyTable SELECT ... FROM Table1 JOIN Table2 ON ... WHERE ... Although I have to wonder about your schema if you need to do this. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] LEFT joins
Jochem Maas ha scritto: Han wrote: Hmm, still no luck. Thanks for the help. I think I'll have to break the you mean that it still times out? crashes mysql? maybe the table (tc_countries) is corrupt? try doing a repair. select up into 2 selects and throw the results of the first into arrays. Didn't want to do it like that, but it's gonna be quicker now. if at first you don't succeed... hack ;-) snip try removing the ORDER BY Clause: SELECT b.fldName, b.fldEmail, b.fldCountryCode, d.fldCode as FCode, b.fldMobile, a.fldTime as Time, c.fldUsername FROM tblSubscribersChoices a LEFT JOIN tblUser c ON c.fldClientID = a.fldChoice LEFT JOIN tblSubscribers b ON b.fldID = a.fldClientID LEFT JOIN tc_countries d ON d.fldCode = b.fldCountryCode ORDER BY c.fldUsername ASC OR SELECT b.fldName, b.fldEmail, b.fldCountryCode, d.fldCode as FCode, b.fldMobile, a.fldTime as Time, c.fldUsername as Username FROM tblSubscribersChoices a LEFT JOIN tblUser c ON c.fldClientID = a.fldChoice LEFT JOIN tblSubscribers b ON b.fldID = a.fldClientID LEFT JOIN tc_countries d ON d.fldCode = b.fldCountryCode ORDER BY Username ASC and if that works but doesn't give you repeat records try adding the DISTINCT keyword after SELECT. snip Why are you so intrested in LEFT JOIN? I cn't help much since i don't know the schema of the db and i can't see clear the meaning of the tables in the query. Anyway try this: SQL SELECT b.fldName, b.fldEmail, b.fldCountryCode, d.fldCode as FCode, b.fldMobile, a.fldTime as Time, c.fldUsername as Username FROM tblSubscribersChoices a, tblUser c, tblSubscribers b, tc_countries d WHERE c.fldClientID = a.fldChoice AND b.fldID = a.fldClientID AND d.fldCode = b.fldCountryCode ORDER BY Username ASC /SQL Bye -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Cannot load MySQL extension (mysql.so issues)
Eve Atley ha scritto: Platform: Redhat Linux Enterprise WS 3 PHP installed: 4.3.2 MySQL installed: 4.0.21 Apache installed: 2.0.46 When setting up PhpMyAdmin today, I got the error: Cannot load mysql extension, Please check PHP configuration My phpinfo() shows: 'with-mysql=shared,/usr' (yes, the comma is not a mistake) - This looks like a glitch; should I fix, and if so where? Mysql.so was found in: /usr/lib/php4/ My php.ini has been editted to read: extension=mysql.so ...and I also tried: extension=/usr/lib/php4/mysql.so Any clues on remedying this problem, without upgrading MySQL? I have read the problem is solved by installing a PHP-MySQL package, but when I attempted to do so via up2date, I was notified that a dependency was required for a MySQL-client. But Mysql is already installed and running. Any clues out there? Thanks, Eve Wrong thread and group. Bye -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Returns Blank
Squeakypants ha scritto: From a forum that recommended this to me, I changed the query execution to this: $query = SELECT credits FROM krypto WHERE user=$user AND pass=$pass; echo query = . $query .\n; $result = mysql_query($query); echo result = . $result .\n; So now it outputs this: query = SELECT credits FROM krypto WHERE user=admin AND pass= result = Could not run query: Unknown column 'admin' in 'where clause' Your query is missing quotes near the values for the attributes user ad pass, so mysql try to compare the value user to some value inside an ipotetic admin field Try this: $query = SELECT credits FROM krypto WHERE user=\$user\ AND pass=\$pass\; Bye -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: MySQL: Random select with specific count of a column
I suggest to first analize the problem and then go through the query, php coding ...etc etc In the table u have only one superkey which is also a key and it is made of the three fields (attributes) 'category', 'language' and ' name'. You should consider to look for a prymary key randomly, toghether with some restriction in the search. Something like select the pryimary key from the table where 'category' equals a number(1, 2,3,n) and 'name' equals some random criterium generated number. A random criteria could be generating a random variables between the ascii code representing the alphabet letters (from 97 to 122 to cover a,b,c,...z). You need to do some work before u can get an executable code here. Hope can open a road Bye Torsten Roehr [EMAIL PROTECTED] ha scritto nel messaggio news:[EMAIL PROTECTED] Hi, I've got the following table: categorylanguagename 1 de a 1 de b 1 de c 2 de a 2 de b 2 de c 3 de a 3 de b 3 de c ... 1 en a 1 en b 1 en c I would like to select 6 random rows where the language is 'de' AND make sure that I will always have 2 rows of EACH category in my result set: 1 de b 1 de c 2 de a 2 de c 3 de a 3 de b Any help greatly appreciated. Thanks and best regards, Torsten Roehr -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Erroneous date and time
Have you tried with a insted of A for am/pm indication? Hope can help Bye Franciccio Philip Thompson [EMAIL PROTECTED] ha scritto nel messaggio news:[EMAIL PROTECTED] Hi all. Maybe there's something I'm doing incorrectly, but I cannot get the date() function to return the appropriate time. It's always 12 hours off. I've tried using 24-hour time and 12-hour time (with AM/PM), but they both give the wrong time. I even emailed the server admin to make sure that the server clock was set correctly, and he said that it was. I don't know... Here's my PHP: $problemDateTime = date(Y-m-d g:i:s A); which returns: 2004-06-13 11:24:21 PM To the untrained eye ;), that might appear correct. But that was actually submitted at 11:24 AM, not PM. So, that's my code. Any suggestions?? Thanks a bunch, ~Philip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: [PHP] update count
I don't think you can have a relation (table) without a unique key. Either a single field or a unique relation between 2 or more fields is necessary for the table. Franciccio Bob Lockie [EMAIL PROTECTED] ha scritto nel messaggio news:[EMAIL PROTECTED] On 06/16/04 09:53 John Nichel spoke: Bob Lockie wrote: What is the best way to only do an update if it going to update only one row? I want to protect my code so that it won't accidentally update more than one row. I can do a select first but there must be an easier way. :-) UPDATE thisDB.thisTable SET thisTable.thisColumn = 'thisValue' WHERE thisTable.uniqueColumn = 'someUniqueValue' Ok, a primary key is unique, right? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: anonymous select error
Here is the lean, corrected code:
?php
$link = mysql_connect( $site, $id, $pass ) or die (error connecting);
echo 'connection okbr';
mysql_select_db( $dbname, $link) or die (eror selecting db);
echo 'selected db ok'br;
$result = mysql_query( 'Select * From newsletter_subscribers',$link) or die
(error query);
// some here code to disply result
?
It is easier to evaluate each time the return of the functions using the
expression or die
Also some sintax error:
resource mysql_query(string query [, connection id], int result mode])
you used the mysql_select_db() return instead of mysql_connect() return
value (altough it is optional), also it is not necessary to write ; at the
end of the query string inside the function.
Bye
Joshuah Goldstein [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
To: [EMAIL PROTECTED]
MIME-Version: 1.0
Content-Type: text/plain; charset=us-ascii
I'm trying this query:
$link = mysql_connect( $site, $id, $pass );
if (!$link) {
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfullybr';
$db = mysql_select_db( $dbname, $link);
if( !$db ) {
echo DB falsebr;
exit;
}
$result = mysql_query( 'Select * From
newsletter_subscribers;', $db );
if (!$result) {
echo DB Error, could not list tablesbr;
echo 'MySQL Error: ' . mysql_error();
exit;
}
with this output:
Connected successfully
DB Error, could not list tables
MySQL Error:
I dont understand why there is an error, but it prints
nothing for the error. Any ideas? Thanks, Josh
__
Do you Yahoo!?
Friends. Fun. Try the all-new Yahoo! Messenger.
http://messenger.yahoo.com/
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[PHP-DB] Re: Pass into the title meta-tag a variable from an URL Directory Manager
Wathever you are ging to do with meta tag they still are html tag, so as
long as you embend the tag in php code it shuold just work fine.
?php
echo any_html_tag attribute01=\$myattribute01\
attribute02=\$myzttribute02\
/any_html_tag;
?
Any text editor should help you modifying the code by find replace
tools.
Hope can help
ps. FORZA ITALIA AGLI EUROPEI
Franciccio
Alessandro Folghera [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
I hope that phpers will be not angry if I'm posting that maybe stupid
question.
I have the following script, an URL Directory Manager (PHP4 from Wrox
Press).
I'd like to pass the variable category into the TITLE META TAG from
php_category when I select the corrispondent category_id into the URL
Directory
Manager, i.e. to display the right category everytime I call it from the
script. As I'm an absolute beginner, I'd like someone could suggestme the
most
rapid way to do it whitout rewriting all the code. Thanks for all,
Alessandro
Tables of MySQL sample_db #
Table
structure for table 'php_category' CREATE TABLE php_category ( category
varchar(30) NOT NULL, category_id varchar(15) NOT NULL, num_item int(5)
NOT
NULL, PRIMARY KEY (category_id) ); # Table structure for table
'php_directory'
CREATE TABLE php_directory ( url_id int(10) NOT NULL auto_increment,
category_id varchar(15) NOT NULL, title varchar(150) NOT NULL, url
varchar(150)
NOT NULL, description text NOT NULL, registerdate date, hit int(5) NOT
NULL,
lastaccesstime timestamp(14), password varchar(20) binary, email
varchar(100)
NOT NULL, approved char(1) NOT NULL, PRIMARY KEY (url_id), UNIQUE url
(url,
category_id) ); xxx php_directory.inc xx
xx php_directory.php x ; $MYSQL_ERRNO =
'';
$MYSQL_ERROR = ''; function directory_header() { global $new_window_width,
$new_window_height; ? alert(\Error: $msg\);history.go(-1);
directory_footer(); exit; } function user_message($msg, $url='') {
directory_header(); if(empty($url)) echo alert(\$msg\);history.go(-1);
else
echo alert(\$msg\);self.location.href='$url'; directory_footer();
exit; }
function get_category_info($category_id) { global $default_dbname,
$category_tablename, $root_category_id, $category_id_length,
$welcome_message,
$PHP_SELF; global $link_id; if(!$link_id) $link_id =
db_connect($default_dbname); if($category_id == $root_category_id) {
$category_info_array[category] = Top; $category_info_array[num_item]
= 0;
$query = SELECT max(category_id), count(*) FROM $category_tablename WHERE
length(category_id) = $category_id_length; $result = mysql_query($query);
if(!$result) echo sql_error(); $query_data = mysql_fetch_row($result);
$sibling_id = $query_data[0]; $num_child = $query_data[1];
$category_info_array[num_child] = $num_child; if(!$sibling_id)
$next_category_id = '001'; else { $sibling_length = strlen($sibling_id);
$next_category_id = $sibling_id + 1; for($i = strlen($next_category_id);
$i
.
$query_data[0] .; $parent_array[$j] = $query_data[0]; $j++; }
$category_info_array[href_fullname] = implode(-, $parent_href_array);
$category_info_array[fullname] = implode(-, $parent_array); } return
$category_info_array; } function get_url_info($url_id) { global $link_id,
$default_dbname, $directory_tablename; if(!$link_id) $link_id =
db_connect($default_dbname); $query = SELECT url_id, category_id, title,
url,
description,registerdate, date_format(registerdate, '%M, %e, %Y') as
formatted_registerdate, hit, lastaccesstime, date_format(lastaccesstime,
'%M,
%e, %Y %r') as formatted_lastaccesstime, password, email, approved FROM
$directory_tablename WHERE url_id = '$url_id'; $result =
mysql_query($query);
if(!$result) echo sql_error(); return mysql_fetch_array($result); }
function
search_form($category_id) { global $root_category_id, $PHP_SELF;
if(!isset($category_id)) $category_id = $root_category_id; ?
Search: Current Category All Categories OR AND
$category_href_fullname ($num_child_str, $num_item_str)
\n; else echo
$category_href_fullname ($num_child_str)
\n; search_form($category_id); $query = SELECT category, category_id
FROM
$category_tablename WHERE (length(category_id) = $category_depth *
$category_id_length) AND category_id LIKE '$category_id%' ORDER BY
category;
$result = mysql_query($query); if(!$result) echo sql_error();
if(mysql_num_rows($result) 0) { ?
\n; echo \n; echo \$PHP_SELF?action=show_list
category_id=$my_category_id\ $my_category ($num_child_str,
$num_item_str)\n;
echo \n; if(!($i % 2)) echo \n; $i++; } if(!($i % 2)) { echo \n ;
echo
; } ?
$found_str - Keyword(s): $keywords
\n; if($total_num 0) echo
Displaying page $page_num out of $last_page_num
\n; $query = SELECT url_id, category_id, title, url, description,
date_format(registerdate, '%M, %e, %Y') as formatted_registerdate, hit,
date_format(lastaccesstime, '%M, %e
[PHP-DB] Re: mysql results, arrays, and for loops
Hi, here is my solution (one of the possible) it is tested so it should work
fine:
?php
$recordset=mysql_query($query);
$num_righe=mysql_num_rows($recordset);
// here starts the dinamyc table
echo TABLE align=\left\ border=\1\ cellspacing=\2\ cellpadding=\2\
width=\80%\\ntr;
$record= mysql_fetch_array($recordset, MYSQL_ASSOC);
foreach ($record as $k =$val)
{
echo tdb$k/b/td\n; // print the field name in the first row of
the table
for ($i=0;$i$num_righe;$i++)
{
echo /tr;
mysql_data_seek($recordset,$i);
$record = mysql_fetch_array($recordset, MYSQL_ASSOC);
foreach ($record as $val){
echo td$val/td\n; // print the result of query, starting under
1st row
}
}
mysql_free_result($recordset);
mysql_close();
echo /tr/tablebr;
?
Let me know if it fits your problem.
Best regards
Francesco Basile
Philip Thompson [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
Hi all!
I am using a select statement to obtain all the dates whenever someone
submitted a problem in a database. Well, I want to get the result
(which could be multiple dates) and then print that in a table format
with some other information on a webpage. So I want to use a FOR loops
and go through each of the dates and dynamically create a table.
My question is: how do I store the results of the select query? Would I
want to store them in an array, and then just parse through each
element of the array, and what is the syntax for that? Or is there a
better way?
Thanks,
~Philip
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[PHP-DB] Re: how to reuse DB results
Sorry for mystakes, (shame on me, i said i was new) using Object array,
here is a working version (tested) with use of array.
Bye
$recordset=mysql_query('SELECT * FROM my_tabella'); // or whatever query is
/* define an array with fields to be shown on top of page. Attention
...values name must be the same of table field names you want to show on top
of the page*/
$my_key = array('key01','key02','key03'); /* u can use $_REQUEST variables
instead of
fixed values*/
$i=0;
while($record=mysql_fetch_array($recordset, MYSQL_ASSOC))
{
foreach ($record as $k=$val)
if ($k==$my_key[$i])
{
// echo the top of the page
echo p$k :$val/p;
/* if u need a top_array to store result shown on top of the
page, comment this line only
$top_array[$i]=$val; // loose the matching key-field!! */
$i++;
}
}
//reset the resource
mysql_data_seek($recordset,0);
// some middle-page code here
echo 'pthis is the bottom page and displays the whole result/p';
while ($record=mysql_fetch_array($recordset, MYSQL_ASSOC))
foreach ($record as $k=$val)
echo p$k : $val/p; // make your choice for html code
mysql_free_result($recordset);
// Bye
Aaron Wolski [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
Hi All,
Got this logic problem I don't know how to solve.
Is there any way I can make a call to the DB for some records.
Display some info from a column or two say at the top of the page and
then display the full result set in a while() loop?
Right now, I am making two calls to the Db to get the data I need to
display at the top of the page and then a second query to retrieve the
full result set.
I'm confused!
Thanks for any help!
Aaron
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[PHP-DB] Re: Retrieve data from a table, edit/add it and enter it in a new table
I see one bug in your code , that is you never rewind the pointer of
mysql_fetch_array($result), so at the end of the first cycle ...while
($r=mysq...)... the pointer is at the end of the query resource. You should
use mysql_data_seek($result,0) to rewind before doing another while cycle.
Hope it can help
Bye
Francesco Basile (PHP_newbee)
Justin [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
Hi,
I am trying to do the following:
Retrieve some information from a table, edit it by appending some further
information to it (a few more fields) and then enter the new data record
into a new table, and delete the old data in the original table. Sounds
confusing I know.
The code is below (I apologise for its poor style etc as I am very new to
php), but when I click on the 'Enter Information' button, nothing happens.
The existing data is retrieved without any problems and I can select it
using the radio button. But when I try and add data to the new fields
('option_close_price' and 'notes'), nothing happens.
Any help appreciated.
form action=? echo $PHP_SELF ? method=post
?
mysql_pconnect(localhost,root,password);
mysql_select_db(options);
if(!$cmd)
{
$result = mysql_query(select * from open_trades);
while($r=mysql_fetch_array($result))
{
$open_date=$r[open_date];
$share=$r[share];
$code=$r[code];
$short_long_trade=$r[short_long_trade];
$id=$r[id];
$expiry=$r[expiry];
$exercise=$r[excercise];
$option_price=$r[option_price];
$no_purchased=$r[no_purchased];
$no_sold=$r[no_sold];
$income_in=$r[income_in];
$income_out=$r[income_out];
print table border=\1\ cellpadding=\3\ cellspacing=\0\\n;
print td/tdtdOpen Date/tdtdShare/tdtdCode/tdtdShort
orbr Long Trade/tdtdExpiry/tdtdExcercise/tdtdOption
Price/tdtdNumberbr Purchased/tdtdNumber Sold/tdtdIncome
In/tdtdIncome Out/tdtd
/tr;
while ($row = mysql_fetch_array($result))
{
print trtd;
print INPUT TYPE='RADIO' NAME='id' VALUE='echo $id';
print /tdtd;
print $row[open_date];
print /tdtd;
print $row[share];
print /tdtd;
print $row[code];
print /tdtd;
print $row[short_long_trade];
print /tdtd;
print $row[expiry];
print /tdtd;
print $row[excercise];
print /tdtd;
print $row[option_price];
print /tdtd;
print $row[no_purchased];
print /tdtd;
print $row[no_sold];
print /tdtd;
print $row[income_in];
print /tdtd;
print $row[income_out];
print /td/tr\n;
}
print /table\n;
?
? }?
input type=submit name=cmd value=Close/form
? }
?
?
if($cmd==Close)
{
if (!$submit)
{
$result = mysql_query(select * from open_trades);
while($myrow=mysql_fetch_array($result))
?
input type=hidden name=id value=?php echo $myrow[id] ?
Option Close PriceINPUT TYPE=TEXT NAME=option_close_price VALUE=?PHP
echo $myrow[option_close_price]? SIZE=7 br
Notes:INPUT TYPE=TEXT NAME=notes VALUE=?php echo $myrow[notes] ?
SIZE=60br
input type=hidden name=cmd value=edit
input type=Submit name=submit value=Enter information
/form
? } ?
?
if($submit)
{
$query = INSERT INTO closed_trades SET code='$code',
option_price='$option_price', option_close_price='$option_close_price',
no_sold='$no_sold', open_date='$open_date',
short_long_trade='$short_long_trade, share='$share', expiry='$expiry',
excercise='$excercise', no_purchased='$no_purchased',
income_in='$income_in', income_out='$income_out', notes='$notes', id=$id';
$result = mysql_query($ql);
$query = DELETE FROM open_trades WHERE id=$id;
$result = mysql_query($sql);
echo Thank you! Information updated.;
}
}
?
/td
/table
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Re: [PHP-DB] Re: HTTP header information
You may try with this:
?php
ob_start();/* buffer the output so no header is sent until the script is
complete*/
session_start();
if (!session_is_registered('_isLoggedIn')) {
header (Location:viewer.php?type=login);
} else {
// do the stuff to submit a problem if they have
// logged in and have pressed the submit button
// in the html
}
?
// then here is all the html for that page
?php
ob_end_flush();// send the output now
?
Bye
Basile Francesco
Philip Thompson [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
On Jun 10, 2004, at 9:44 AM, Torsten Roehr wrote:
Philip Thompson [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Hi all.
I am running a website to where a user needs to login to authenticate
themselves to perform certain tasks. So a user logs in, and I start a
session (in PHP, of course). Well, the catch is, I am doing this all
from one page, 'viewer.php', and I just tack on the specific view/page
that I want them to see, depending on the link selected on that page.
Meaning, they click on the 'submit problem' link and it goes to
'viewer.php?type=submitproblem'.
The problem comes whenever I load the view 'submitproblem' and I start
a session with session_start(), which carries over the variable to
tell
whether or not the user is logged in. If they have not logged in
whenever they click on 'submitproblem' then it will redirect them to
'viewer.php?type=login'. So I log in, and then go to 'submitproblem'.
This is where I get the error: Warning: session_start(): Cannot send
session cookie - headers already sent. Essentially, I understand why
this is occurring, but is there an easy way to get around it without
creating a new page, such as 'submitproblem.php' instead of
'viewer.php?type=submitproblem'???
Make sure that NO output is done before session_start() is called. Can
you
post some of your code?
Regards,
Torsten Roehr
See, that's the case. Because I'm essentially just changing the content
within the page, it never leaves the page 'viewer.php' - it just
changes the content by tacking on '?type=login, submitproblem, etc'.
But my code in the beginning of the file 'submitproblem.view' is:
?php
session_start();
if (!session_is_registered('_isLoggedIn')) {
header (Location:viewer.php?type=login);
} else {
// do the stuff to submit a problem if they have
// logged in and have pressed the submit button
// in the html
}
?
// then here is all the html for that page
So, does that help any?
~Philip
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[PHP-DB] Re: how to reuse DB results
Hi here is my logic, i haven't tested but it should work (i hope)!!!
i'm not sure if it works only on php5...!?! Let me know if it is a complete
crap i've just started learning php.
Bye
$recordset=mysql_query($query); // whatever query is
$obj_array=new ArrayObject($recordset);
// get the iterator now
$interator=$obj_array-getIterator();
// you can now easly navigate in the recordset by the iterator
/* define an array with n fields to be shown on top of page. Attention
...values name must be the same of table field names you want to show on top
of the page*/
$my_key = array(0='key1',1= 'key2',,n-1='keyn');
// echo the top of the page
$i=0;
while($iterator-valid())
{
if ($iterator-key()==$my_key[$i] )
{
echo p.$iterator-key(). : .$iterator-current()./p;
/* if u need a top_array to store result shown on top of the page,
uncomment this line only
$top_array[$i]=$iterator-current(); // loose the matching
key-field!! */
$i++;
$iterator-next();
}
else
$iterator-next();
}
/* to display the whole recordset resulting from the query don't forget
you still have
$recordset free to use with mysql_fetch_array() or other way to display */
echo 'pthis is the bottom page and displays the whole result/p';
while ($res=mysql_fetch_array($recordset))
for each ($res as $k=$val)
echo p$k : $val/p; // make your choice for html code
Aaron Wolski [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
Hi All,
Got this logic problem I don't know how to solve.
Is there any way I can make a call to the DB for some records.
Display some info from a column or two say at the top of the page and
then display the full result set in a while() loop?
Right now, I am making two calls to the Db to get the data I need to
display at the top of the page and then a second query to retrieve the
full result set.
I'm confused!
Thanks for any help!
Aaron
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Re: [PHP-DB] Mysql not receiving the data
I agree, the slashes are killing the query. I would suggets doing this:
$add = INSERT INTO movies SET
movie_name=\$movie_name\,
genre=\$genre\,
director=\$director\,
star1=\$star1\,
star2=\$star2\,
star3=\$star3\,
brief_synopsis=\$brief_synopsis\,
imdb_link=\$imdb_link\;
Rich Hutchins [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
The apostrophe (') in your data is, most likely, killing the SQL statement
when it is sent to the server. Use addslashes() around all of your form
data
to prevent this and also to help guard against SQL injection attacks.
Ex:
$add = INSERT INTO movies SET
movie_name='.addslashes($movie_name).',
genre='.addslashes($genre).',
director='.addslashes($director).',
star1='.addslashes($star1).',
star2='.addslashes($star2).',
star3='.addslashes($star3).',
brief_synopsis='.addslashes($brief_synopsis).',
imdb_link='$imdb_link';
Hope this helped.
Rich
-Original Message-
From: Andrew Rothwell [mailto:[EMAIL PROTECTED]
Sent: Sunday, June 13, 2004 1:48 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Mysql not receiving the data
Hi Larry, Thank you very much for the very quick response, I set my
php.ini
file (located /etc/php.ini ) for the register_globals = On (it was off by
default)
Now however I get an error
Error adding entry: You have an error in your SQL syntax near 's spanish
driver is found shot dead, Inspector Jacques Clouseau is the first off' at
line 8
My Database is a movie database of my dvd's that I own (for insurance
reasons)
My addmovie.php is this
?
mysql_connect(localhost,username,password);
mysql_select_db(movies);
$add = INSERT INTO movies SET
movie_name='$movie_name',
genre='$genre',
director='$director',
star1='$star1',
star2='$star2',
star3='$star3',
brief_synopsis='$brief_synopsis',
imdb_link='$imdb_link';
if (@mysql_query($add))
{
echo(pYour entry has been added. br
$movie_name/p);
}
else
{
echo(pError adding entry: .
mysql_error() . /p);
}
?
And the addmovie.htm page (atleast the form action is this)
body bgcolor=#FF
form method=post action=addmovie.php name=addmovies
table width=300 border=0 cellspacing=2 cellpadding=2
bordercolordark=#FF0033 bordercolorlight=#66
tr
td width=41% bgcolor=#99Movie Name /td
td width=59% bgcolor=#99FFCC
input type=text name=movie_name
/td
/tr
Andrew
-Original Message-
From: Larry E. Ullman [mailto:[EMAIL PROTECTED]
Sent: Sunday, June 13, 2004 11:22 AM
To: Andrew Rothwell
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Mysql not receiving the data
Online I could see everything, and the pages gave the appearance of
working, however when I went into the DB using PHPMYADMIN to check the
status of the new data entered, all I found was blank rows ( for the
new data since the rebuild, all the old data was there) There were the
correct number of new rows for the amount of records that I had
entered, which tells me (unless I am nistaken) that the PHP is talking
to the DB, and is atleast sending a insert command, but the rest of
the data is not getting in. -
Without seeing any code whatsoever and since this worked before but no
longer works on a new install, I can only assume that your code was
written
with the assumption that register_globals was turned on and it's not on in
your current configuration.
If that is the case, see the PHP manual or search the Web for the solution
($_POST, $_GET, etc.).
Larry
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[PHP-DB] Re: Problem in passing the necessary variable.
Ciao..hi, i would suggest to embed the html code in php like this:
?php
$_your_variable=$row[imgid];
echo
a href=\../images/image.php?img=$_your_variable\
target=\_blank\
onClick=\window.open(this.href, this.target,
\'width=200,height=250\');
return false;\img SRC=\../preview.jpg\ ALT=\Click to look at
the picture
you are charging\ border=\0\/a;
?
If it works let me know , bye
Francesco Basile
__
Alessandro Folghera [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
Hi,
I have developed a news system .. I can select to insert an image into the
articles.
The function to extract the image calling them by the imgid of images
table is the following:
function getImm() {
?
select name=image
?
connect();
$sql = select * from images;
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)) {
printf(option value=\%s\%s/option, $row[id_img], $row[imgid]);
}
?
/select
?
I'd like to be able to see the image before to publish the piece of news.
I think about a window pop up with java in order to create a preview of
the
image. The code follows:
a href=../images/image.php?img=XX target=_blank
onClick=window.open(this.href, this.target, 'width=200,height=250');
return false;img SRC=../preview.jpg ALT=Click to look at the picture
you are charging border=0/a
The problem is that I can't pass the imgid to the href inspite of the
May someone tell me how to pass the id number in order to show me the
image
or how may I solve the trouble?
I hope someone will help me ...
Sincerely,
Alexander
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Re: [PHP-DB] Mysql not receiving the data
Some of the reason to have register_global on is to easly use variables sent by post, get, cookie method of a form. Ex. form name=form1 method=post action=anypage.php?get_var=1 input type=text name=text_post value=Hello World input type=submit name=Submit value=Submit /form in your code at the page anypage.php you will have available the variables: $get_var==1 and $text_post==Hello World if register_global=TRUE in php.ini file - in this case u still have available $_POST;GET ecc... $_GET['get_var']==1 and $_POST['text_post']==Hello World if register_global=FALSE in php.ini file - in this case you don't have the $get_var e $text_post available, so code is safer I would suggest to leave register global=FALSE as to have safer code unless u have to rewrite the whole code. Think about having hacked variables value send by GET, COOKIE method. Bye Hans Lellelid [EMAIL PROTECTED] ha scritto nel messaggio news:[EMAIL PROTECTED] Hi Andrew, Andrew Rothwell wrote: Thank you everybody that responded so quickly - I used the suggestion of Franciccio - and the data is now gow into the db Thank you very much - I really appreciate the help. Another question - with this fix in place - do I still need the register_globals = On ? Or can I now turn it off? It seems like you should have kept your old php.ini file, as this other error you encountered was probably due to your old php.ini file having this setting: magic_quotes_gpc = 1 That INI var instructs PHP to automatically addslashes() to any GET/POST/COOKIE data. I would suggest turning this back on, unless you've thoroughly redesigned your code to not need it. This is unrelated to register_globals, which you will need to leave on unless you redesign your application. Hans -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
