how to add $_POST['submit'] in ajax jquery or what is the best way to do this..?

Question

index.php

<form id="myForm" action="test2.php" method="POST">
    <input type="submit" value="Print 1" name="submit" id="submit">
    <input type="submit" value="Print 2" name="submit2" id="submit2">
</form> 
<div id="ack"></div>

here's my script

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
<script type="text/javascript" src="scripts/login_script.js"></script>

test2.php

<?php
if(isset($_POST['submit'])){ 
echo "1"; // i want to show this everytime i click submit (not working)
}          // or what is the proper way to do this..

if(isset($_POST['submit2'])){
echo "2"; // i want to show this everytime i click submit2 (not working)
}         // or what is the proper way to do this..
?>

login_script.js

$("#submit, #submit2").click( function() {
  $.post( $("#myForm").attr("action"),
         $("#myForm :input").serializeArray(),

         function(data) {
                $("#ack").empty();
                $("#ack").html(data);
    });


$("#myForm").submit( function() {
   return false;    
});
});

if i remove the if(isset($_POST['submit'])).... its working... so what is the proper way to do this..? sorry im noob in ajax and jquery

Make sure the `name=` of your second input is also `submit2` — RichardBernards, Dec 02 '14 at 09:42
still not working sir, i think the problem there is if(isset($_POST['submit2'])){...? do you have any idea? — Paulo B. Dela Cruz, Dec 02 '14 at 09:46
see this answer http://stackoverflow.com/questions/4007942/jquery-serializearray-doesnt-include-the-submit-button-that-was-clicked?answertab=votes#tab-top — Girish, Dec 02 '14 at 09:49

score 2 · Answer 1 · edited Dec 02 '14 at 10:05

2

No, you have to realize that <input type="submit" and <button> submit is not included in .serializeArray(), you'll have to just build the data object to be passed:

    $("#submit, #submit2").click( function() {
        var data = null;
        if($(this).attr('id') == 'submit') {
            values = {'submit': true};
        } else {
            values = {'submit2': true};
        }

            $.post( $("#myForm").attr("action"), values,

         function(data) {
                $("#ack").empty();
                $("#ack").html(data);
    });
});

Sample Demo

edited Dec 02 '14 at 10:05

Suchit kumar

11,809
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answered Dec 02 '14 at 09:52

Kevin

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thank you sir for this `if($_SERVER['REQUEST_METHOD'] == 'POST') {` but i also want to pass data like this `mysql_real_escape_string( $_POST["name"] ); and i found the answer above.. thank you for your time.. :) – Paulo B. Dela Cruz Dec 02 '14 at 17:16
can i ask one more question? @Ghost? because i have 3 days before i can ask question again. – Paulo B. Dela Cruz Dec 03 '14 at 14:17
@PauloB.DelaCruz yes and what is it? – Kevin Dec 03 '14 at 14:25
its not related,sometimes my external css is not reading my new codes,sometimes working ,sometimes not? and i found that when i press f12 and look at my css codes , still the same,the old code is still there, even i change it and save. – Paulo B. Dela Cruz Dec 04 '14 at 02:19
@PauloB.DelaCruz try to clear cookies, cache, etc. of the browser, and use a hard refresh on the browser – Kevin Dec 04 '14 at 02:20
ah hard refresh, my chrome has no hard refresh.. hehe thank you for your time sir.. – Paulo B. Dela Cruz Dec 04 '14 at 03:04
@PauloB.DelaCruz i think its ctrl + F5 or ctrl + shift + F5 not sure though – Kevin Dec 04 '14 at 03:13
@PauloB.DelaCruz sure tol no prob – Kevin Dec 04 '14 at 03:31
1

pinoy karin pala whahha – Paulo B. Dela Cruz Dec 04 '14 at 06:09

Suchit kumar · Accepted Answer · 2014-12-02T10:13:04.387

0

try with this:

$("#submit, #submit2").click( function() {
        var formData = $(this).closest('form').serializeArray();
          formData.push({ name: this.name, value: this.value });
      $.post( $("#myForm").attr("action"),formData,


             function(data) {
                    $("#ack").empty();
                    $("#ack").html(data);
        });


    $("#myForm").submit( function() {
       return false;    
    });
    });

edited Dec 02 '14 at 10:13

answered Dec 02 '14 at 09:56

Suchit kumar

11,809
3
22
44

thank you sir :) because i also passing a data like this $username = mysql_real_escape_string( $_POST["name"] ); i used your codes... – Paulo B. Dela Cruz Dec 02 '14 at 17:06

score 0 · Answer 3 · answered Dec 02 '14 at 09:58

Yes I just try it and below code is working for me please check.

index.php

<form id="myForm" action="test2.php" method="POST">
<input type="submit" value="Print 1" name="submit" id="submit">
<input type="submit" value="Print 2" name="submit2" id="submit2">
</form> 
<div id="ack"></div>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
<script type="text/javascript">
$("#submit, #submit2").click( function() {
   $.post( $("#myForm").attr("action"),
      $("#myForm :input").serializeArray(),
       function(data) {
            $("#ack").empty();
            $("#ack").html(data);
});
$("#myForm").submit( function() {
return false;    
});
});
</script>

test.php

<?php
if(isset($_POST['submit'])){  
echo "1"; // i want to show this everytime i click submit (not working)
}          // or what is the proper way to do this..
if(isset($_POST['submit2'])){
echo "2"; // i want to show this everytime i click submit2 (not working)
}         // or what is the proper way to do this..
?>

score -1 · Answer 4 · answered Dec 02 '14 at 09:45

In the post it user name instead of the ID. Please try code below hope it will help you.

<form id="myForm" action="test2.php" method="POST">
<input type="submit" value="Print 1" name="submit" id="submit">
<input type="submit" value="Print 2" name="submit2" id="submit2">
</form>

how to add $_POST['submit'] in ajax jquery or what is the best way to do this..?

4 Answers4