PHP and JQuery ajax url

Question

I am using a ready booking form source code and I would like to make some changes according to our needs.

I quote below the source code that is important to be seen and not the whole source code of the file. I would like to execute, as soon as a button is clicked, a mysqli_query to update variables on the database. So, I am trying to use Jquery and AJAX to make this happen.

The code below shows a button Check Availability already defined which executes Javascript code and I added also my button "Book Now" and I would like to run also my code. See the code below:

<form id="bsSearchForm_<?php echo $_GET['index'];?>" action="<?php echo PJ_INSTALL_URL; ?>index.php?controller=pjFront&amp;action=pjActionCheck" method="post" >
<div class="row ">
      <div class="col-lg-6 col-md-6 col-sm-6 col-xs-6 col-xss-12">
           <div class="form-group">
                <label for="address">Pickup Address</label>
                <input type="text" class="form-control" name="pickup" placeholder="Enter a location">
           </div>
      </div>

      <div class="form-group pjBsFormActions">
           <button type="submit" class="btn btn-primary"><?php echo("Check Availability"); ?></button>
           <button type="submit" class="btn btn-primary" onclick="checkClicked()"><?php echo("Book Now"); ?></button>
      </div><!-- /.form-group pjBsFormActions -->
</div>
</form>

Now at the same php file at the beginning I have defined this source code:

<script>
function checkClicked() {
    $.ajax({
        url: "test.php",
        type: 'POST',
        dataType: 'html'
    });
}
</script>

So, I would like to run my own PHP source code at an external php file like test.php and get the input field data from "pickup" and perform a mysqli_query on database.

However the code at test.php file is not executed at all. I think that the problem is the form action parameter <?php echo PJ_INSTALL_URL; ?>index.php?controller=pjFront&action=pjActionCheck. How could I find this file(a lot of source files) and maybe place the source code there?

Or should I have to define the new button differently and so I would be able to call my own PHP file at any directory?

The AJAX URL parameter should be a relative or absolute path to the test.php file? Where should I create the test.php file at my directories?

Please help me find a quick solution to my issue.

Vignesh · Accepted Answer · 2018-05-01 12:23:45Z

1

This very easy and simple

1. Create an html page with the form like.

    <form id="sample">
//some thing here....
                <input type="submit" class="btn btn-success" name="submit" value="Create" id="sample">
    </form>

2. Create an Js page like this.

//click function
          $("#sample").click(function(event) {
              sample();
          });

//ajax function here
          function sample(){
            $.ajax({
                url: '/path/to/file',
                type: 'GET/POST',
                dataType: 'default: Intelligent Guess (Other values: xml, json, script, or html)',
                data: {
                    param1: 'value1'
                },
                success:function(result){
                    alert(result);

                }
            })
            .done(function() {
                console.log("success");
            })
            .fail(function() {
                console.log("error");
            })
            .always(function() {
                console.log("complete");
            });

          }

3. Create an php script file.

add the actual path in directory like if you are using the localhost example: public/script/test.php then the ajax path is ../script/test.php

4. the javascript link to the the html page like

example:

<script type="text/javascript" src="sample.js"></script>

answered May 1, 2018 at 12:23

Vignesh

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3 Comments

pc-pacman Over a year ago

Although it seems very logical your solution.....it's not working until now. The code for the click event seems that either os not being called at all or the button click event does not match with the Javascript code. Are you sure that the declaration for the button should be something like that <input type="submit" class="btn btn-success" name="submit" value="Create" id="sample"> ??

Vignesh Over a year ago

Yes this is exactly what i use in my application

Vignesh Over a year ago

Post your code here or specific part so that it will be easy for us to give solution

Pawan Developers · Accepted Answer · 2018-05-01 12:26:35Z

0

you need to create test.php on same directory where your form's file is created.

answered May 1, 2018 at 12:26

Pawan Developers

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Comments

Alexander Weixelbaumer · Accepted Answer · 2018-05-01 12:29:01Z

0

If you have a submit-Button the form will be submitted with the action you defined and not with the onclick-event. For that to work you have to simply replace it with <button type="button"

answered May 1, 2018 at 12:29

Alexander Weixelbaumer

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