CODEWITHSUNDEEP
phpajaxjquery
zach
zach

Reputation: 87

Using Ajax to generate an alert box

I want to pop up an alert box after checking whether some data is stored in the database. If stored, it will alert saved, else not saved.

This is my ajax function:

      AjaxRequest.POST(
              {
                'url':'GroupsHandler.php'
                ,'onSuccess':function(creategroupajax){ alert('Saved!'); }
                ,'onError':function(creategroupajax){ alert('not saved');}
              }
            );

but now it show AjaxRequest is undefined.

How can I fix this?

Upvotes: 1

Views: 9775

Answers (7)

Alfred
Alfred

Reputation: 21396

First, decide whether to use POST or GET (I recommend POST) to pass AJAX data. Make a php file (ajax.php) such that it echos true or false after checking whether some data is stored in the database. You may test with a variable $your_variable = "some_data_to_check"; having a data inside and once you are finished, you may replace it with $your_variable = $_POST["ajaxdata"];.

Then in your page, set up AJAX using jQuery plugin like:

var your_data_variable = "data_to_send";
$.ajax({
  type: "POST",
  url: "ajax.php",
  data: 'ajaxdata=' + your_data_variable,
  success: function(result){
    if(result == "true"){
      alert("saved");
    }else{
      alert("not saved");
    }
}

You may have a look at jQuery AJAX Tutorial, Example: Simplify Ajax development with jQuery.

Upvotes: 0

ind
ind

Reputation: 43

function Ajax(data_location){
    var xml;
    try { 
        xml = new XMLHttpRequest();
    } catch (err){
        try { 
            xml = new ActiveXObject("Msxml2.XMLHTTP");
        } catch (error){
            try {
                xml = new ActiveXObject("Microsoft.XMLHTTP");   
            } catch (error1){
                //
            }

        }
    }
    xml.onreadystatechange = function(){
        if(xml.readyState == 4 && xml.status == 200){
            alert("data available");
        }
    }
    xml.open("GET", data_location, true);
    xml.send(null);

}


window.onload = function(){
    Ajax("data_file_location");
}

Upvotes: 1

Mj1992
Mj1992

Reputation: 3504

use the ajax to call the script and check values in the database through the script. If data present echo success else not.lets look an example of it.

Assuming databasename = db

Assuming tablename = tb

Assuming tableColumn = data

Assuming server = localhost

Ajax:

$.ajax({
  url: 'GroupsHandler.php',
  success:function(data){
     if(data=="saved")
     {
          alert("success");
     }
  }
});

Now in the myphpscript.php :

<?php


    $Query = "select data from table";

    $con = mysql_connect("localhost","user","pwd");     //connect to server
    mysql_select_db("db", $con);                        //select the appropriate database
    $data=mysql_query($Query);                         //process query and retrieve data
    mysql_close($con);                                //close connection


   if(!$empty(mysql_fetch_array($data))
   {
      echo "saved";  
   }
   else
   {
      echo " not saved ";
   }

?>

EDIT: You must also include jquery file to make this type of ajax request.Include this at the top of your ajax call page.

<script src='ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js' type='text/javascript'></script>

Upvotes: 0

Abid Hussain
Abid Hussain

Reputation: 7762

Below code based on jQuery.

Try it

    $.ajax({
      type: 'POST',
      url: 'http://kyleschaeffer.com/feed/',
data: { postVar1: 'theValue1', postVar2: 'theValue2' },
      beforeSend:function(){
        // this is where we append a loading image
        $('#ajax-panel').html('<div class="loading"><img src="/images/loading.gif" alt="Loading..." /></div>');
      },
      success:function(data){
        // successful request; do something with the data
        $('#ajax-panel').empty();
        $(data).find('item').each(function(i){
          $('#ajax-panel').append('<h4>' + $(this).find('title').text() + '</h4><p>' + $(this).find('link').text() + '</p>');
        });
      },
      error:function(){
        // failed request; give feedback to user
        $('#ajax-panel').html('<p class="error"><strong>Oops!</strong> Try that again in a few moments.</p>');
      }
    });

Upvotes: 0

verisimilitude
verisimilitude

Reputation: 5108

This of course is possible using Ajax.

Consider the below sample code for the same.

Ajax call : 

$.ajax({
  url: 'ajax/example.php',
  success: function(data) {
    if(data == "success")
      alert('Data saved.');
  }
});

example.php's code 

<?php 
           $bool_is_data_saved = false;
          #Database processing logic here i.e
          #$bool_is_data_saved is set here in the database processing logic
          if($bool_is_data_saved) {
             echo "success";
          }
     exit;  
?>

Upvotes: 1

Asciiom
Asciiom

Reputation: 9975

This is possible using ajax. Use jQuery.ajax/pos/get to call the php script that saves the data or just checks if the data was saved previously (depends on how you need it exactly) and then use the succes/failure callbacks to handle its response and display an alert if you get the correct response.

Upvotes: 0

kelsar
kelsar

Reputation: 108

You can create an addtitional table with date(time) of last update database and check if this date is later. You can use standard setInterval function for it.

Upvotes: 0

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