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Bug #24879 Some data lost in returned array by mysql_fetch_array
Submitted: 2003-07-31 05:26 UTC Modified: 2003-08-04 10:37 UTC
From: sm at grand-prix dot ru Assigned:
Status: Not a bug Package: MySQL related
PHP Version: 4.3.2 OS: Linux RedHat 9
Private report: No CVE-ID: None
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From: sm at grand-prix dot ru
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 [2003-07-31 05:26 UTC] sm at grand-prix dot ru 
Description:
------------
In some cases mysql_fetch_array returns numbered array element, but does'nt returns it's named copy.

This script was tested as a part of vbulletin's code.

Bug: in the first case $user[userid] is empty, but $user[0] contains correct data. For other user - 2'nd case - all is OK. Behavior depends from data, returned by mysql.


Reproduce code:
---------------
$query_string = "SELECT *,FROM_UNIXTIME(joindate) AS joindate,FROM_UNIXTIME(lastvisit) AS lastvisit,FROM_UNIXTIME(lastactivity) AS lastactivity,FROM_UNIXTIME(lastpost) AS lastpost,
                               avatar.avatarpath,NOT ISNULL(customavatar.avatardata) AS hascustomavatar
                               FROM user
                               LEFT JOIN avatar ON avatar.avatarid=user.avatarid
                               LEFT JOIN customavatar ON customavatar.userid=user.userid
                               WHERE user.userid=412";

$res = mysql_query($query_string,$link_id);
$user = mysql_fetch_array($res);
print_r($user);


$query_string = "SELECT *,FROM_UNIXTIME(joindate) AS joindate,FROM_UNIXTIME(lastvisit) AS lastvisit,FROM_UNIXTIME(lastactivity) AS lastactivity,FROM_UNIXTIME(lastpost) AS lastpost,
                               avatar.avatarpath,NOT ISNULL(customavatar.avatardata) AS hascustomavatar
                               FROM user
                               LEFT JOIN avatar ON avatar.avatarid=user.avatarid
                               LEFT JOIN customavatar ON customavatar.userid=user.userid
                               WHERE user.userid=287";

$res = mysql_query($query_string,$link_id);
$user = mysql_fetch_array($res);
print_r($user);


Expected result:
----------------
 Array (
 [0] => 412
 [userid] => 412
 [1] => 6
 [usergroupid] => 6
 [2] => SM_1
 [username] => SM_1
 .......
 
 Array (
 [0] => 287
 [userid] => 287
 [1] => 6
 [usergroupid] => 6
 [2] => SM
 [username] => SM
 .......

Actual result:
--------------
 Array (
 [0] => 412
 [userid] =>
 [1] => 6
 [usergroupid] => 6
 [2] => SM_1
 [username] => SM_1
 .......
 
 Array (
 [0] => 287
 [userid] => 287
 [1] => 6
 [usergroupid] => 6
 [2] => SM
 [username] => SM
 .......

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 [2003-07-31 13:33 UTC] sniper@php.net 
Please try using this CVS snapshot:

  http://snaps.php.net/php4-STABLE-latest.tar.gz
 
For Windows:
 
  http://snaps.php.net/win32/php4-win32-STABLE-latest.zip

Also let us know what mysql version you're using,
and what configure line was used to configure PHP.
 [2003-08-01 03:59 UTC] sm at grand-prix dot ru 
I have installed recommended version (last stable, 4.3.2). Problem persist.

Additional info: 
 
MySQL v3.23.56
Linux kernel v2.4.20 SMP
 [2003-08-01 04:02 UTC] sm at grand-prix dot ru 
./configure :
 --with-mysql=/home/mysql --with-apache=../apache_1.3.27rusPL30.17_match --enable-track-vars
 [2003-08-01 08:45 UTC] sniper@php.net 
Please provide a complete but short example script and mysql db schema with which we can reproduce this ourselves.
(we're not going to install vbulletin just to test this)
 [2003-08-01 08:45 UTC] sniper@php.net 
And I asked you to try the SNAPSHOT, NOT the latest stable release..
 [2003-08-01 12:08 UTC] sm at grand-prix dot ru 
I can provide more accurate test script and tables only on monday... And I can't install RC version, I'm not a hoster...

But some important information - this problem persist only for such database records, for which left joined records does'nt exist.
 [2003-08-04 03:13 UTC] sm at grand-prix dot ru 
As i mean, bug occurs when both tables (main and left joined) contains columns with equal names and left-joined table does'nt contain needed row. In the such case named array elements with names, equal to such column names are cleared.

Here is test script :
------------------------------------------------------------

<?php
  error_reporting(7);

  $srv = "mysql.server.ru";
  $usr = "username";
  $pwd = "password";
  $db = "database";

  $link_id=mysql_connect($srv,$usr,$pwd);  

  if (!$link_id) {
    echo "Can't connect to mysql server<br>";
    exit;
  }

  if (!mysql_select_db($db, $link_id)) {
    echo "Can't connect to mysql server<br>";
    exit;
  }

  if ($action == "create_bases") {

    $query = "CREATE TABLE test1table (
              userid int(10) unsigned DEFAULT '0' NOT NULL auto_increment,
              usergroupid smallint(5) unsigned DEFAULT '0' NOT NULL,
              username varchar(50) NOT NULL,
              password varchar(50) NOT NULL,
              PRIMARY KEY(userid) )";

    $query_id = mysql_query($query,$link_id);
    if (!$query_id) {
      echo "invalid SQL:<br>$query<br>";
      exit;
    }


    $query = "CREATE TABLE test2table (
              userid int(10) unsigned DEFAULT '0' NOT NULL,
              field1 char(250) NOT NULL,
              field2 char(250) NOT NULL,
              field3 char(250) NOT NULL,
              field4 char(250) NOT NULL,
              PRIMARY KEY (userid) )";

    $query_id = mysql_query($query,$link_id);
    if (!$query_id) {
      echo "invalid SQL:<br>$query<br>";
      exit;
    }

    $query = "INSERT INTO test1table VALUES (1,6,'user1','password1')";
    $query_id = mysql_query($query,$link_id);
    if (!$query_id) {
      echo "invalid SQL:<br>$query<br>";
      exit;
    }
    $query = "INSERT INTO test1table VALUES (2,6,'user2','password2')";
    $query_id = mysql_query($query,$link_id);
    if (!$query_id) {
      echo "invalid SQL:<br>$query<br>";
      exit;
    }

    $query = "INSERT INTO test2table VALUES (1,'dummy1','dummy2','dummy3','dummy4')";
    $query_id = mysql_query($query,$link_id);
    if (!$query_id) {
      echo "invalid SQL:<br>$query<br>";
      exit;
    }

    echo "<br>tables created and initialized<br>";
    exit;

  }

  $query = "SELECT * FROM test1table
            LEFT JOIN test2table ON test2table.userid=test1table.userid
            WHERE test1table.userid=1";
  $query_id = mysql_query($query,$link_id);
  if (!$query_id) {
     echo "invalid SQL:<br>$query<br>";
     exit;
  }
  $array = mysql_fetch_array($query_id);
  print_r($array); 
  echo "<br><br>";
  mysql_free_result($query_id); 

  $query = "SELECT * FROM test1table
            LEFT JOIN test2table ON test2table.userid=test1table.userid
            WHERE test1table.userid=2";
  $query_id = mysql_query($query,$link_id);
  if (!$query_id) {
     echo "invalid SQL:<br>$query<br>";
     exit;
  }
  $array = mysql_fetch_array($query_id);
  print_r($array); 
  echo "<br><br>";
  mysql_free_result($query_id); 


?>
 [2003-08-04 08:22 UTC] georg@php.net 
see #21589
 [2003-08-04 08:33 UTC] sniper@php.net 
Bug #21589 is bogus -> bogus.
 [2003-08-04 10:37 UTC] sm at grand-prix dot ru 
Hmmm... But why numbered element is not equal to named?
 
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