PHP URL parameter isn't passed using form action with GET method

Question

I am facing the problem in WordPress. I passed the value from 1st page to 2nd page (here app-2) which shows the result (by using inspector) but when the form is submitted URL shows only

"app-2/?Screens=30"

But

"app-2/?type=app&Screens=30"

is meant to show. Why does this happen? Is it something to do with 'GET' method?

N.B: $_GET['type'] works perfectly.

<form action="app-2/?
type=<?php $type=$_GET['type']; echo $type;?>
&Screens=<?php echo $_POST['Screens'];?>" 
method="GET">

  <input type="radio" name="Screens" value="3 - 5" required/>
  <input type="radio" name="Screens" value="16 - 30" />

 <input type="submit" value="submit">
</form>

Is it a typo with form action: $_POST['Screens'] and you mean $_GET['Screens'] instead? — BenRoob
– BenRoob, Commented Jun 19, 2017 at 9:54
@BenRoob No. He is using $_POST['Screens'] and appends it to the URL, so afterwards he can access it like $_GET['...']. It doesn't make too much sense though because he has a field with the name 'Screens' aswell that is required. — xate
– xate, Commented Jun 19, 2017 at 9:55
actually, I used $_POST['Screens'] to get the form radio button value and used $_GET['type'] to get the previous URL parameter 'type' — Ashiqur Rahman
– Ashiqur Rahman, Commented Jun 19, 2017 at 10:07

Nirav Joshi · Accepted Answer · 2017-06-19 10:03:52Z

1

$(".submit-form").submit(function(e){
   e.preventDefault;
   window.location.href = 'app-2/'+$(".submit-form").serialize();
   return false;
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form action="" method="GET" class='submit-form'>
   
  <input type="hidden" name="type" value=<?php echo $_GET['type']; ?> 
  <input type="radio" name="Screens" value="3 - 5" required/>
  <input type="radio" name="Screens" value="16 - 30" />

 <input type="submit" value="submit">
</form>

You can try with javascript Please check answer it will helps you :)

answered Jun 19, 2017 at 10:03

Nirav Joshi

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1 Comment

Nirav Joshi Over a year ago

i just tried this example and it is working good.. url will be like xxxxxxxxxx/app-2/type=test&Screens=16+-+30

lalithkumar · Accepted Answer · 2017-06-20 07:41:38Z

1

Try this one:

<?php
$parameters = "";
$parameters .= "?type=".(isset($_GET["type"])?$_GET["type"]:'')."&Screens=".isset($_GET["Screens"])?$_GET["Screens"]:'';

?>
<form action="app-2/<?=parameters?>" 
method="GET">

edited Jun 20, 2017 at 7:41

answered Jun 19, 2017 at 9:59

lalithkumar

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2 Comments

lalithkumar Over a year ago

this is result for the above script?i'm pretty sure you are showing me some other result with your code.try the above one and show me the that url.

Ashiqur Rahman Over a year ago

1st of all your code has syntax error then I try to resolve that and get that output. Here's the error of your code

Error: There is 1 more closing curly braces '}' found. This count is unaware if curly braces are inside of a string. PHP Syntax Check: Parse error: syntax error, unexpected '<' in your code on line 3 $parameters .= "?type=".(isset($_GET["type"])?$_GET["type"]:'')."&Screens=".<?=isset($_GET["Screens"])?$_GET["Screens"]:''?>;} PHP Syntax Check: Errors parsing your code

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