I am trying to show some text depending upon what a status
is set to in a db table.
See my code below:
$result=mysql_query("SELECT * FROM hr_recruitment_stages where vacancy_ref='$vacancyref' order by added_on DESC limit 0,1")or die('ERROR 315' );
$row = mysql_fetch_array($result);
$stage_name = $row ['stage_name'];
if($stage_name['stage_name'] == 'Shortlisting') { echo"Shortlisting"; } else { echo"Not Shortlisting"; } ?>
However this doesnt seem to be working properly as it is showing as Not Shortlisting
even when stage_name
equals Shortlisting.
Any ideas why?
Its variable type mistake. Check your assigned variable, you assigned the Array Element not the entire array. so try like below.
<?php
$result = mysql_query("SELECT * FROM hr_recruitment_stages where vacancy_ref='$vacancyref' order by added_on DESC limit 0,1") or die('ERROR 315' );
$row = mysql_fetch_array($result);
$stage_name = $row['stage_name'];
if($stage_name == 'Shortlisting') {
echo"Shortlisting";
} else {
echo"Not Shortlisting";
}
?>
Refer this Article for PHP Array understanding.
http://php.net/manual/en/language.types.array.php