$variable = header location

Question

I can find hundreds of examples of "insert into header location", but I have not found any exaples where the $variable = header location.

Is this possible:

$error=header('Location: http://www.example.com/error.php');

Basicly I had 3 conditions if row not found in mysql.

So I set up the 3 conditions:

if condition1 $error="a statement"

if condition2 $error="SELECT data FROM table WHERE data= 'default'";

if condition3 $error=header('Location: example.com/error.php');

and then if($result->num_rows =0) $error

In my case $error could be 3 different things, one of them being "redirect to the error page" but $error=header('Location:blah'); causes 500 server error.

I altered my code so that I could use header location in the conventional manner, but would still like to know if $var = header location is possible.

The original code that prompted the question has been overwritten, I no longer have it to post.

Thanks Tom

What would you expect that to do? And yes, it's possible, it just may not do whatever you think this should do. — deceze
– deceze ♦, Commented Jun 2, 2013 at 20:52
Please post more code form this file for us to ID what the problem might be. — Dave Chen
– Dave Chen, Commented Jun 2, 2013 at 20:59
I have completely rewritten the code, so I no longer have it to post. I could reconstruct it but since I write code at a snails pace, you would have to wait a couple days. My php knowledge is very limited. — Tom
– Tom, Commented Jun 2, 2013 at 21:20

Dave Chen · Accepted Answer · 2013-06-02 20:53:46Z

2

Header returns type of void. You can not expect $error to be anything besides null.

If you are getting a 500 error, it is caused by something else (syntax error?).

answered Jun 2, 2013 at 20:53

Dave Chen

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Tommy Naidich · Accepted Answer · 2013-06-02 21:14:43Z

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You can't assign the 'header' function to anything, as it would always return void. A bit useless.

answered Jun 2, 2013 at 21:14

Tommy Naidich

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Tom Over a year ago

Must be why I couldn't find any examples of it. In my searching I also couldn't find any post like this saying it can't be done. Thank you to all who replied.

kelunik · Accepted Answer · 2013-06-03 04:43:16Z

0

http://php.net/manual/de/function.header.php

header() returns void, so, you can write $error=header('Location: http://www.example.com/error.php');, but it wont do much.

After header('Location: ...'); you should call exit;.

edited Jun 3, 2013 at 4:43

answered Jun 2, 2013 at 20:55

kelunik

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6 Comments

Dave Chen Over a year ago

exit; (it's a keyword, so you don't need the ())

kelunik Over a year ago

I know, but I usually add ()

deceze Over a year ago

You can write $error = header(...), it just won't do much.

Dave Chen Over a year ago

It's possible that the OP rewrote the header function in which it does return something.

deceze Over a year ago

@Dave Now that's pretty darn hypothetical. :P

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