mysql_fetch_array() error in php

Question

This is my code:

<?php
include('admin/connect.php');
$applink = $_GET['app'];
$fetchapps = mysql_fetch_array('SELECT * FROM apps WHERE link = "$applink"');
$title = $fetchapps['title'];
?>

And i receive this error: Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in /home/trupu8o/public_html/apps.php on line 4

How i can solve this problem ?

Please send me the code that work , because i don't know how to do — user3570695, Apr 18 '15 at 14:42
Variables dont process in single quotes as well. use double quotes for query and single for actual query. — chris85, Apr 18 '15 at 14:42
See this doc, http://php.net/manual/en/function.mysql-query.php has examples for how to query. — chris85, Apr 18 '15 at 14:44
@user3570695 i hope you have a username and a password? or are you just using 'username' as a username and 'password' as a password? — Parvesh, Apr 18 '15 at 14:44

Roland Seuhs · Answer 1 · 2015-04-18T14:51:14.820

0

You need to first create a statement, then fetch from that statement:

<?php
include('admin/connect.php');
$applink = $_GET['app'];
$statement = mysql_db_query($database, 'SELECT * FROM apps WHERE link = "'.mysql_real_escape_string($applink).'"');
$fetchapps = mysql_fetch_array($statement);
$title = $fetchapps['title'];
?>

It is assumed that "admin/connect.php" provides the database handle in $database.

Usually you would use that in a loop:

<?php
include('admin/connect.php');
$applink = $_GET['app'];
$statement = mysql_db_query($database, 'SELECT * FROM apps WHERE link = "'.mysql_real_escape_string($applink).'"');
while ($fetchapps = mysql_fetch_array($statement))
{
     $title = $fetchapps['title'];
     // Do something with $title
}
?>

edited Apr 18 '15 at 14:51

answered Apr 18 '15 at 14:47

Roland Seuhs

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This wont work. Still trying to select for literal `$applink`. Use double quotes. Also missing closing semi-colon on query execution line. – chris85 Apr 18 '15 at 14:49
If i use this code i receive this error : Parse error: syntax error, unexpected '$fetchapps' (T_VARIABLE) in /home/s/public_html/apps.php on line 5 – user3570695 Apr 18 '15 at 14:50
Sorry, made a mistake - it should work now – Roland Seuhs Apr 18 '15 at 14:53
I also added mysql_real_escape_string() to sanitize data input – Roland Seuhs Apr 18 '15 at 14:55
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/crazyapps/public_html/apps.php on line 5
i receive this error – user3570695 Apr 18 '15 at 14:59
Is $database an open database connection? – Roland Seuhs Apr 18 '15 at 15:05

mysql_fetch_array() error in php

1 Answers1