ajax using php and mysql

Question

<html>
<head>
<script>
$('patientlist').click(function showpatient()
{ 
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
 {// code for IE6, IE5
 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
 }
   xmlhttp.open("POST","ajaxlistpatient.php",true);
  xmlhttp.send();         
 })
  </script>
  </head>
   <body>
  <form>
  <input type="button" id="patientlist" name="patientlist" value="List Patient"/>
  </form>
 </body>
 </html>

please help, i want to list my patientlist using a button in the same page without reloading my mainpage.

ajaxlistpatient.php contains my sqlquery ..

Are you using jQuery here? It looks like you are, but somehow you've sort of re-implemented the ajax feature. The documentation for that library is quite good, so it's worth searching through it before creating your own solutions. — tadman
– tadman, Commented Sep 5, 2013 at 14:55

rAjA · Accepted Answer · 2013-09-05 15:20:07Z

1

You can't access the DOM like this $('patientlist'). It has to be either $('.patientlist') or $('#patientlist')

Assuming 'patientlist' is a class,

$('.patientlist').click(function (){ 

   console.log("Clicked");  //Check you got here without any problem

   $.ajax({
            type: "POST",
            url: "ajaxlistpatient.php",
            dataType: 'json',
            success: function (data) 
            {
                console.dir(data); 
            },
            error: function (jqXHR,textStatus,errorThrown)
            {
                //Check for any error here
            }
        });

});

edited Sep 5, 2013 at 15:20

answered Sep 5, 2013 at 14:52

rAjA

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6 Comments

John Christian De Chavez Over a year ago

still i cant see my result =(

rAjA Over a year ago

@JohnChristianDeChavez no worries. You just need spend little more time on this. Can you show where is ajaxlistpatient.php is located? Is it placed at the root directory of your project?

John Christian De Chavez Over a year ago

yes sir.. i saved your code as use.js in the root so my root folder contains this use.js ajaxlistpatient.php index.php index.php now my code is <script type="text/javascript" src="use.js"></script><html> <head> </head> <body> <form> <input type="button" id="patientlist" name="patientlist" value="List Patient"/> </form> </body> </html>

rAjA Over a year ago

Cool. so clearly you are missing the Jquery library in your html. Take a look at this learn.jquery.com/about-jquery/how-jquery-works and also go through every single page of w3schools.com/jquery.. Happy Coding!!

John Christian De Chavez Over a year ago

oops i forgot to add the library this now my new code for index.php <script type="text/javascript" src="js/bootstrap.js"></script> <script type="text/javascript" src="js/bootstrap-tooltip.js"></script> <script type="text/javascript" src="js/bootstrap-popover.js"></script> <script type="text/javascript" src="js/jquery-1.7.2.min.js"></script> <script type="text/javascript" src="use.js"></script><html> <head> </head> <body> <form> <input type="button" id="patientlist" name="patientlist" value="List Patient"/> </form> </body> </html>

|

Joon · Accepted Answer · 2013-09-05 14:48:29Z

1

try using jQuery library for this, because ajax operation is quirky and complicated.

Take a look at this document:

http://api.jquery.com/jQuery.ajax/

answered Sep 5, 2013 at 14:48

Joon

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1 Comment

tadman Over a year ago

Absolutely. Writing AJAX code by hand is a super bad idea. Use a library or you will get it wrong.

azzy81 · Accepted Answer · 2013-09-05 14:53:22Z

Even though you havn't tagged Jquery in your question it looks like Jquery is being used for the click event anyway. So here you go:

$('#patientlist').on('click', function(e){
      e.preventDefault();
      e.stopPropagation();
      $.ajax({
           url : "ajaxlistpatient.php",
           type: "post",
           complete: function(d){
              //do something with the return data'd'
           }
      });
});

You will need to decide what you want to do with the response. Make sure you have access to something like firebug in firefox which will help you out massivly so you can see the ajax call being sent out and see the response via the firebug console. Chrome also has a console and debug tools but I prefer firefox/firebug.

Busychild · Accepted Answer · 2013-09-05 14:54:11Z

0

You could juse use the jQuery ajax method:

$('#patientlist').click(function showpatient() {
    $.ajax({
        type: 'POST',
        dataType: "xml",
        url: 'ajaxlistpatient.php',
        success: function (data) {
            $('#myContentContainer').html(data);
        }
    });
});

answered Sep 5, 2013 at 14:54

Busychild

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1 Comment

John Christian De Chavez Over a year ago

what do you mean by the container? sorry if beginner

Hassan Sardar · Accepted Answer · 2013-09-05 14:57:07Z

0

Here is the general and simplest syntax of using ajax:

$('patientlist').click(function ()
{

        $.post('ajax_files/ajaxlistpatient.php', {var_id:anyData},
        function(data) 
        {
            $('#yourDataHolder').html(data); 
        });
});

answered Sep 5, 2013 at 14:57

Hassan Sardar

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2 Comments

John Christian De Chavez Over a year ago

what should i do the dataholder? sorry beginner here

Hassan Sardar Over a year ago

This is the div or span etc where you want to list your data coming from ajax file . e.g, <div id="yourDataHolder"></div>

Community · Accepted Answer · 2017-05-23 12:12:52Z

0

All responses are in the right direction, Jquery+php is your best bet, and to further extend those responses, here are some previous similar posts that might help you to have a reference:

AJAX to call PHP file which removes a row from database?

AND

How to pass jQuery variables to PHP variable?

are good sample resources,

hope this helps.

edited May 23, 2017 at 12:12

CommunityBot

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answered Sep 5, 2013 at 15:03

maurix suarez

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