PHP: Make an Array value into a variable

Question

I´m making a form, and I want to change the value of the POST into a variable. But I'm doing it wrong somehow.

Check the sample:

$_POST['name'] = $name;
$_POST['age'] = $age;
$_POST['country'] = $country;

This error pops: Parse error: syntax error, unexpected T_VARIABLE on the first $_POST

You want to store the posted values in separate variabels? You have to write it in the oposite direction: $name = $_POST['name']; — strauberry
– strauberry, Commented Apr 13, 2011 at 19:46
The error is on the previous line from the first $_POST. Update with that code please. — Khez
– Khez, Commented Apr 13, 2011 at 19:47
Why would be assigning variables to the post array? Not sure what you are trying to do here. — Mike D
– Mike D, Commented Apr 13, 2011 at 19:47
@all Has anyone actually read the error? The compiler is telling him he put a variable right smack in the middle of something else. That variable is the first $_POST. Doesn't matter if he does a=b or b=a. The error is unrelated to your solutions. — Khez
– Khez, Commented Apr 13, 2011 at 19:53
@Khez: It's probably also useful to do this for unit testing. — anon
– anon, Commented Apr 13, 2011 at 20:15

anon · Accepted Answer · 2011-04-13 19:58:37Z

3

While everyone else is entirely correct to point that you shouldn't be assigning values to the $_POST superglobal, it is possible for you to make such an assignment. The $_POST superglobal is just an array, after all, and so it acts like one.

The error you're seeing is because PHP is recognizing $_POST['name'] as being part of the previous statement. Check to make sure that you have properly ended the previous statement (i.e. the line before $_POST['name'] = $name ends with a ;).

You probably do want to be assigning $_POST['name'] to a variable, rather than the other way around as you have it now, but that's not what's causing the error.

answered Apr 13, 2011 at 19:58

anon

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2 Comments

anon Over a year ago

@Khez: Oddly, I was responding to your post-level comment while you were responding to my answer.

Gabriel Meono Over a year ago

Thanks so much, that was it a simple ; missing in the previous line. I also inverted to the A = B order that everybody stated.

Mike Lewis · Accepted Answer · 2011-04-13 19:46:37Z

1

You don't programmatically set $_POST variables. These are set by the server based on what was POST'ed to that page(via forms or otherwise).

So I'm fairly sure you want:

$name = $_POST['name'];
$age = $_POST['age'];
$country = $_POST['country'];

This is because the assignment operator works as such:

a = b

Set a to the value of b.

answered Apr 13, 2011 at 19:46

Mike Lewis

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Comments

Naftali · Accepted Answer · 2011-04-13 19:46:38Z

1

go the other way:

$name = $_POST['name'];
$age = $_POST['age'];
$country = $_POST['country'];

answered Apr 13, 2011 at 19:46

Naftali

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Comments

halfdan · Accepted Answer · 2011-04-13 19:46:49Z

1

Assignment works right to left, so to get the values from the into a variable you'd have to do:

$name = $_POST['name'];
...

Your code above does not contain any syntax error, it must be from somewhere else.

answered Apr 13, 2011 at 19:46

halfdan

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DaOgre · Accepted Answer · 2011-04-13 19:48:14Z

0

You have this backwards, it should be:

$name = $_POST["name"];  
$age= $_POST["age"];  
$country= $_POST["country"];

The value to the right of the = gets assigned to the left. As is you are trying to replace the post variable with an unassigned variable.

answered Apr 13, 2011 at 19:48

DaOgre

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2 Comments

Gabriel Meono Over a year ago

Thanks but I'm still getting the same error at: $name = $_POST['name']; Parse error: syntax error, unexpected T_VARIABLE

Khez Over a year ago

@Gabriel check @AgentConundrum's response or my comment in your OP.

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