simple PHP AJAX : How to call a php script

Question

I'm trying to get my javascript function to use ajax to call a php script. The php script will update a MYSQL table but I've tested the PHP script and it works fine.

This is my function:

    function rate()
{

 $.ajax({
    data: '' ,       
    url: 'update.php',
    method: 'GET', 
    success: function(msg) {
        alert(msg);
    }
});

}

and the function is called later with:

rate();

The php script doesn't need any information given to it, it just needs to be called, can anyone point out where I'm going wrong.

Learning to use the Network tab in modern browsers is going to aid you significantly with AJAX. — Marty
– Marty, Commented Apr 24, 2014 at 11:33
firefox, install firebug module and check in the console what's wrong — maxime1992
– maxime1992, Commented Apr 24, 2014 at 11:34
Does the url exist? If not, the 'success' function will never be called. I also recommend to install firebug (Maxime). Your code should work. — Kai
– Kai, Commented Apr 24, 2014 at 11:43
What is the error in console.? which script is not running the ajax script or the PHP script..? I am not clear here. — shashank
– shashank, Commented Apr 24, 2014 at 12:24

szpal · Accepted Answer · 2014-04-24 12:01:34Z

1

Just use like in this example:

<script>
  function rate() {
    $.get("update.php");
  }
</script>

edited Apr 24, 2014 at 12:01

answered Apr 24, 2014 at 11:41

szpal

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1 Comment

user3568576 Over a year ago

I don't understand sorry, this would go instead of the rate function?

Mad Angle · Accepted Answer · 2014-04-24 11:41:05Z

0

If you dont need to pass the data to the PHP page, you can omit 'Data' from the ajax parameters.

<script>
    function rate(){
        $.ajax({      
            url: 'update.php',
            method: 'POST', 
            success: function(msg) {
               alert(msg);
            }
        });
    }

    rate();
</script>

answered Apr 24, 2014 at 11:41

Mad Angle

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1 Comment

Marty Over a year ago

That's true, but isn't relevant to the issue.

Chetan Gawai · Accepted Answer · 2014-04-24 11:41:27Z

0

Have you included jquery library Try this

<script src="http://code.jquery.com/jquery-latest.min.js"
        type="text/javascript"></script>
<script>
      function rate()
{

 $.ajax({
    data: '' ,       
    url: 'update.php',
    method: 'GET', 
    success: function(msg) {
        alert(msg);
    }
});

}


    rate();


</script>

answered Apr 24, 2014 at 11:41

Chetan Gawai

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1 Comment

Marty Over a year ago

How does waiting for the DOM to load affect this?

Borza Adrian · Accepted Answer · 2014-04-24 11:51:45Z

Here is an example of how I use the ajax call:

$.ajax({
                               type: "POST",
                               url: "page_url",
                               dataType: 'json',
                               data: {
                                   'date1' : date1,
                                   'call': 'function_name'
                               },
                               beforeSend: function(){
                                   $("#loading").show();
                               },
                               success : function(response){    
                                  },
                                complete: function(){ 
                                   $("#loading").hide();
                               }
                      })

and on php part I add:

function function_name($request){
  your code here
}

if (!empty($_POST['call'])) {
    die($_POST['call']($_POST));
}

szpal · Accepted Answer · 2014-04-24 12:33:51Z

0

i show my all working code, try it:

<html>
  <head>
    <script type="text/javascript" src="/js/jquery-1.8.2.js"></script>
    <script>
      function rate() {
        $.get("update.php");
      }
    </script>
  </head>

  <body>
    <button onclick="rate()">Click me</button>
  </body>
</html>

answered Apr 24, 2014 at 12:33

szpal

6479 silver badges27 bronze badges

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