php.net |  support |  documentation |  report a bug |  advanced search |  search howto |  statistics |  random bug |  login
Doc Bug #41111 parse error in example code (translation: ru)
Submitted: 2007-04-17 03:40 UTC Modified: 2007-04-17 05:17 UTC
From: theshut at gmail dot com Assigned:
Status: Closed Package: Documentation problem
PHP Version: Irrelevant OS: Linux Gentoo
Private report: No CVE-ID: None
Welcome back! If you're the original bug submitter, here's where you can edit the bug or add additional notes.
If you forgot your password, you can retrieve your password here.
Password:
Status:
Package:
Bug Type:
Summary:
From: theshut at gmail dot com
New email:
PHP Version: OS:

 

 [2007-04-17 03:40 UTC] theshut at gmail dot com
Description:
------------
http://www.php.net/manual/ru/function.mysql-result.php
Parse error in example.
Must be a:
<?php
    $link = mysql_connect("localhost", "mysql_user", "mysql_password")
            or die("Could not connect: " . mysql_error());
 
    $result = mysql_query("SELECT name FROM work.employee")
            or die("Could not query: " . mysql_error());
           
    echo mysql_result($result,2); // outputs third employee's name
   
    mysql_close($link);
?> 


Patches

Pull Requests

History

AllCommentsChangesGit/SVN commitsRelated reports
 [2007-04-17 05:17 UTC] philip@php.net
This has been fixed in CVS and will show up after the next RU build, thank you for helping make the PHP Manual better :)
 
PHP Copyright © 2001-2026 The PHP Group
All rights reserved.
Last updated: Sat Feb 07 04:00:01 2026 UTC