Use jquery variable in php

Problem in passing Jquery variable to php .

When i pass variable id to B.php i am getting this error

"Notice: Undefined index: id1 in C:\xampp  \htdocs\PhpProject1\OtherUsableItems\B.php on line 2 

How to solve this problem ???? This A.php

This B.php

I want id variable to pass on to B.php

Mr. Alien

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asked Apr 5, 2013 at 11:39

3

Try this one:

Then B.php

answered Apr 5, 2013 at 11:48

Try changing the input type to button instead of submit

answered Apr 5, 2013 at 11:44

The problem is that your form is submited ,$.post is for ajax sumit. For that you need to prevent form submit.

Add event.preventDefault(); in your button click function

Lodder

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answered Apr 5, 2013 at 11:48

Shijin TRShijin TR

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2

Use a hidden field to set the value you want to send across. Since you are using submit, your entire form is submitted. The hidden field is also sent and this can be accessed directly using the name of the hidden field.

answered Apr 5, 2013 at 11:49

AashrayAashray

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2

$(document).ready(function(){
  var id = 23; 
  $("#submit").click(function(e){
     $.post("B.php", {id1: id} );     
     e.preventDefault();
   });
 });

Try code above. First: JS does not allow to do something like this: {id1: id,} (see comma at the end). That is a feature of PHP. Actually, your JS code should not be executed at all because that is a syntax error (suppose you will find it in console of your browser)

Another problem: you must prevent default submit of a form. e.preventDefault() is needed for that. Without it, ajax post will be started and than usual form submit will happen. And your form has no element with name id1, suppose that second request produce that error message.

EDIT: This is to post a value with regular submit, so browser is redirected to B.php

$(document).ready(function(){
      var id = 23; 
      $("#submit").click(function(){
              $(this).closest("form")
                     .append($("", {"type":"hidden", "name":"id1", "value":id}));
       });
     });

answered Apr 5, 2013 at 11:44

Viktor S.Viktor S.

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2

Try this code,

Here you can pass value of 'id' to B.php,And will be access as $_POST['id1']

answered Apr 5, 2013 at 12:08

Shijin TRShijin TR

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Try this:

Since you are using ajax feature so you have to stop the form submit feature. You have a trailing comma here {id1 : id,} this is an issue in IE browsers but latest other browsers will just work fine.

Note:

You posted this error:

"Notice: Undefined index: id1 in C:\xampp \htdocs......

Don't run the page in filesystem $.post() won't work this way.

Instead try using http://localhost/yourapplication/page.php

answered Apr 5, 2013 at 11:44

JaiJai

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4

Or you could use submit() this way

$(function(){
    var id = 23;
    $("#myform").submit(function() {
        $.post("B.php", { id1: id }).done(function(data) {
            alert( data );
        });
        return false; //true(post form data)
    });
});

And form

EDIT: Oh and yeah for B.php isset check or error_reporting(0);

if(isset($_POST['id1'])){
    print_r( $_POST );
}

answered Apr 5, 2013 at 12:28

ZhamaZhama

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