Post method not working in PHP

Question

I've tried some other posts, but never made it.

My php form inside html file is :

<form action="test.php" method="POST">
VAR:<input type="text" name="var"><br>
<input type="submit">
</form>

I'm receiving the variable in test.php as <br>

$_POST['var'] ---- > doesn't work! <br>

Tried, $_REQUEST['var'] ----> doesn't work!

I tried the form using get request, it works like a charm on the other end at test.php by using both $_GET['var'] and $_REQUEST['var'] Any way I can parse the request in the form of post request from that form to test.php ..

Appreciate the help! Thanks.

The tag is closed. Can you be more specific? Like line number etc.. — Ninja Boy, Dec 25 '15 at 07:47
No, it is not. Input tags must be closed by `/>`. Note the slash. — marekful, Dec 25 '15 at 07:49
Also, in the provided code your input has no value. Is that intentional? — marekful, Dec 25 '15 at 07:52
It is working... You are saying that it works like the charm? What doesn't work actually? — Rehmat, Dec 25 '15 at 08:22
When I use method POST and try to have the variable using $_POST, I can't have the variable on test.php — Ninja Boy, Dec 25 '15 at 08:24
Have a look in the network tab of your browser (F12) and see if you actually made a post request with the specified data. — flec, Dec 25 '15 at 09:14

score 0 · Answer 1 · answered Dec 25 '15 at 07:51

0

Did you check the value of 'submit' whether it is set or not like

if(isset($_POST['submit'])){
$_POST['var']
}

answered Dec 25 '15 at 07:51

should I add in test.php – Ninja Boy Dec 25 '15 at 08:08
it says nothing, even replacing submit with var in line 1 to see if the var is being parsed or not, it also says nothing. I replaced second line with echo "Working"; to check – Ninja Boy Dec 25 '15 at 08:28

score 0 · Accepted Answer · answered Dec 26 '15 at 04:49

0

Updated to PHP 7.0 and it's working now!

Thanks everyone for their help!

answered Dec 26 '15 at 04:49

Ninja Boy

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fugitive · Answer 3 · 2015-12-25T08:06:49.933

-1

Your html tag is wrong. You need to set name to it like this

<input type= "submit" name= "submit">

Then when you submit your form like this:

if(isset($_POST['submit'])) {
// you will set your post var, try to print it with
print_r($_POST);

}

And you don't need to add "/" to the input, it will work without it.

And sorry, ive made syntax mistake, forgot to close if statement. Now is correct.

edited Dec 25 '15 at 08:06

answered Dec 25 '15 at 07:52

fugitive

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Sorry for being so lame, but should i add thatif(isset_ .. in test.php? If yeah, where? Sorry again, I'm very new to php.. – Ninja Boy Dec 25 '15 at 08:01
Well.. yes.. that way you basicly check is submit button pressed.. And that way you submit form, and process it to "test.php" – fugitive Dec 25 '15 at 08:03
Also @pbssubhash you access every form value in php with name atribute in html. On the side.. Who the hell gave me -1 for answer, because it is correct?? – fugitive Dec 25 '15 at 08:06
Oh, I didn't mark -1 , BTW, The post works like charm with GET method. Does that change anything? – Ninja Boy Dec 25 '15 at 08:07
@pbssubhash well if you wanna use get, you need to change html, so method="GET" Difference between POST and GET, is that, get vals will be in url, while post wont. Sorry Im little in hurry so this is rough explanation. Cheers. – fugitive Dec 25 '15 at 08:11
Yeah I know, When I'm using method as get , I was able to parse the parameters correctly. But when I try with POSt changing the method to post, it doesn't work. In a secure point of view,I don't want sensitive info in the url as params – Ninja Boy Dec 25 '15 at 08:12

Alfiza malek · Answer 4 · 2015-12-25T09:11:50.103

-1

Try this what is post on same page

<?php
if(isset($_POST['var_name'])) {
 print_r($_POST);
}
exit();
?>

<form action="" method="POST">
<input type="text" name="var_name"><br>
<input type="submit" name="Save">
</form>

edited Dec 25 '15 at 09:11

answered Dec 25 '15 at 09:06

Alfiza malek

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Post method not working in PHP

4 Answers4