gettingerror as Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource

Question

<?php
{
session_start();
include "dbconnect.php";
$target1=$_SESSION['target1'];
echo $target1;
$query = "SELECT * FROM userpictures where pictures = $target1";

$result = mysql_query($query);

var_dump($result); 
    while($row=mysql_fetch_object($result)) 
echo $row;  
    {
    $_SESSION['picid1']=$row['picid'];
    //$_SESSION['picid1']=$row->picid;

    echo $_SESSION['picid1'];

    }
}
?>

it is returning me the output as images/2101.jpgbool(false) Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource

it is giving the target1 but not the picid please help.

You need to check the right-hand bar, quote your variable or - even better - switch to PDO or mysqli and prepared statements. — jeroen, Jan 19 '14 at 17:13

Krish R · Answer 1 · 2014-01-19T18:07:52.783

0

Try this, Added ' - pictures = '$target1'

$query = "SELECT * FROM userpictures where pictures = '$target1' ";
$result = mysql_query($query) or die(mysql_error());
while($row=mysql_fetch_object($result)) 
{
   $_SESSION['picid1']=$row->picid;       
   echo $_SESSION['picid1'];

}

Use,

$row->picid;

instead of

$row['picid']

Note: Use mysqli_* fucntions or PDO instead of mysql_* functions(deprecated)

edited Jan 19 '14 at 18:07

answered Jan 19 '14 at 17:13

Krish R

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but i dont need the whole array, i just need current picid single picid, thats why i am using where clause mr.dragon – Yogendra Karekar Jan 19 '14 at 17:53
In you have only one row which is associated with `pictures = '$target1'` then you get only one row! – Krish R Jan 19 '14 at 17:59
`echo $query;` after query statment and copy and paste into phpmyadmin or command line in mysql and check what is the results you have ? – Krish R Jan 19 '14 at 18:02
Update my answer, please check – Krish R Jan 19 '14 at 18:08

gettingerror as Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource

1 Answers1