Error in mysql_fetch_array

Question

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given on line 236

<?php
$select=mysql_query("select * from deptupload order by id desc");


while($row1=mysql_fetch_array($select) ){
    $name=$row1['name'];
    ?>
    <tr>
        <td width="300">
            <a href="download1.php?filename=<?php echo $name;?>"><?php echo $name ;?></a>
        </td>
    </tr> 
<?php }?>

i have one page having two divisions in that one is user file and the other is department file. the upload of user file should be displayed on other page and its displaying but department is not showing which is to be displayed on same page. and getting mysql_fetch_array.

Check your query result its not returning any data thats why you are facing such problem — Sunil Pachlangia, Jan 21 '15 at 12:01
`mysql_query("select * from deptupload order by id desc") or die(mysql_error())` Do this and you will see whats wrong with your query — Umair Ayub, Jan 21 '15 at 12:03

score 0 · Answer 1 · answered Jan 21 '15 at 12:02

Use die function for debuging:

$select=mysql_query("select * from deptupload order by id desc") or die(mysql_error());
                                                                 ^^

Note: mysql extension Officially deprecated (as of PHP 5.5. It's likely to be removed in the next major release.)

Error in mysql_fetch_array

1 Answers1