PHP If variable equals

Question

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?

What have you done to debug this? What is the value of `$result`? What is the value of `$row`? What is the value of `$stage_name`? What is the value of `$stage_name['stage_name']`? `print` and `print_r` are your friends. — Quentin, Jun 30 '16 at 17:59
**Warning**: You are using [an **obsolete** database API](http://stackoverflow.com/q/12859942/19068) which has been **removed** entirely from the latest version of PHP. You should use a [modern replacement](http://php.net/manual/en/mysqlinfo.api.choosing.php). — Quentin, Jun 30 '16 at 18:00
How is this not working? Have you checked to see what's actually in `$row['stage_name']`? You might want to check `$stage_name == 'Shortlisting'`, too — andrewsi, Jun 30 '16 at 18:00
The value of stage_name is Shortlisting, so it should show as `Shortlisting` — Shane, Jun 30 '16 at 18:00
fetch_assoc() will return an associative array. Try that instead of mysql_fetch_array() — Radmation, Jun 30 '16 at 18:00
@user3092953 — No. Not what value *should* it be, what value is it actually? You need to inspect it (e.g. with `print_r`). — Quentin, Jun 30 '16 at 18:00
Are you _assuming_ it has that value, or have you _verified_ it? — CBroe, Jun 30 '16 at 18:00
Using `$stage_name['stage_name']` does not make sense. `$stage_name` should not be an array. — Don't Panic, Jun 30 '16 at 18:01
[**How do I get PHP Errors to display?**](http://stackoverflow.com/q/1053424/4577762) — FirstOne, Jun 30 '16 at 18:02
@Radmation, actually, [mysql_fetch_array](http://php.net/manual/en/function.mysql-fetch-array.php): _Fetch a result row as an associative array, a numeric array, or both_. _MYSQL_BOTH (default)_. It should work anyway... — FirstOne, Jun 30 '16 at 18:03

Venkat.R · Accepted Answer · 2016-06-30T18:03:12.603

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

PHP If variable equals

1 Answers1