If statement variable returning empty

Question

this is a test code i wrote to get a value set for $rectype to insert into a table. problem I'm having is $rectype is returning 0 or empty. the files do exist since I perform a similar is_file check to start the process that leads up to this section of the coding

$dirchk1 = "/temp/files/" . $data[0] . ".doc"; // exist
$dirchk2 = "/temp/files/" . $data[1] . ".doc"; // exist

$file_1 = (is_file($dirchk1));
$file_2 = (is_file($dirchk2));

if ($file_1) {
$rectype == ($file_2 ? '3' : '1');
}

echo $rectype . "\n";

you are comparing by '==' while you want to assign by '='

Javed Akram
– Javed Akram

2010-12-14 19:20:57 +00:00
Commented Dec 14, 2010 at 19:20 — Javed Akram
– Javed Akram, Commented Dec 14, 2010 at 19:20

Maerlyn · Accepted Answer · 2010-12-14 19:17:15Z

3

You are using == instead of =.

answered Dec 14, 2010 at 19:17

Maerlyn

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Comments

Alin P. · Accepted Answer · 2010-12-14 19:16:53Z

2

You have a typo:

$rectype = ($file_2 ? '3' : '1');

Also read something about operator precedence to avoid using so many useless parentheses.

answered Dec 14, 2010 at 19:16

Alin P.

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Comments

Saul · Accepted Answer · 2010-12-14 19:17:04Z

2

You need to do an assignment

$rectype = ($file_2 ? '3' : '1');

instead of a comparison

$rectype == ($file_2 ? '3' : '1');

answered Dec 14, 2010 at 19:17

Saul

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Comments

ssice · Accepted Answer · 2010-12-14 19:21:50Z

2

As typed everywhere, you are using a comparison $expression == $expression instead of an assignment $variable = $expression;.

They didn't tell you that your code, as exposed would drop a E_NOTICE saying that $rectype is uninitialized.

It is good practice to run your code with error_reporting(E_ALL) while debugging, and that will tell you these errors and the line where they are.

answered Dec 14, 2010 at 19:21

ssice

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