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Making AJAX call to PHP program: How to get input parameters??

Posted on 2013-06-12
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Last Modified: 2013-06-12
Hey all.

I've tried using Google but I can't get this to drop.

I am using the following call to a PHP program. All I want to do is simply extract the two fields in order to use them inside the PHP program.

var postData = {
                "user": "test",
                "passwd": "9999"
                };

      php_url="/phptarget.php";
   
    reqGroups = $.ajax({url: php_url, type: "POST", data: postData, error: err, timeout: 50000, dataType: 'json'});

The PHP program is a new program so any example of getting the parms would be helpful. Here is a starting point.

<?php

ini_set('error_reporting', E_ALL);
ini_set('display_errors', 1);

$w_user = $_POST['w_user'];
$w_passwd = $_POST['w_passwd'];


......

Another question is whether or not this is needed: header('Content-type: application/json');

If so, do I just put it at the top of the program?
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Question by:Webspeeder
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11 Comments
 
LVL 58

Accepted Solution

by:
Gary earned 500 total points
ID: 39241155
Why are you passing json data to a php page? Just use a normal querystring, using Json you then have to parse it out again in the php page.

$.ajax({
    type: 'POST',
    url: 'http://phptarget.php/',
    data: { 
         "user": "test",
          "passwd": "9999"
    },
    success: function(msg){
        // Add actions to perform if the post was successful
    }
});

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Then in your php page you can just use
$user = $_POST['user']
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Expert Comment

by:Jagadishwor Dulal
ID: 39241156
0
 
LVL 82

Expert Comment

by:leakim971
ID: 39241162
$w_user = $_POST['user'];
$w_passwd = $_POST['passwd'];

Open in new window


If so, do I just put it at the top of the program?
Put it before any output (echo, print, die "reason",...)
and be sure your JSON object is valid (double quotes, not simple  quote for example) :
http://jsonlint.com/
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LVL 58

Expert Comment

by:Gary
ID: 39241166
url: 'http://phptarget.php/',

Should be
url: '/phptarget.php',
0
 

Author Comment

by:Webspeeder
ID: 39241254
Gary123, you're right, there is no need to be using JSON. But when I change it over, I'm still not able to see the two parameters in the PHP file.
0
 

Author Comment

by:Webspeeder
ID: 39241266
This is what I'm echoing out from PHP.

$userid = $_POST['user'];
$pass = $_POST['passwd'];

echo '<br>'.'ID: '.$userid.'<br>'.'Pass: '.$pass.'<br>';

I get blank for both.
0
 

Author Comment

by:Webspeeder
ID: 39241277
I am getting an error, but is is "undefined index" which can be resolved with the ISSET() but that would only allow a bypass to the error, doesn't resolve the variables being blank.
0
 

Author Comment

by:Webspeeder
ID: 39241280
I removed the w_ from the names.
0
 
LVL 58

Expert Comment

by:Gary
ID: 39241293
Paste your ajax code
0
 
LVL 82

Expert Comment

by:leakim971
ID: 39241301
Try this :
var postData = { "user": "test", "passwd": "9999" };
php_url = "/phptarget.php";
$.post(php_url, postData, function(){ alert("ok"); }, "json");

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0
 

Author Comment

by:Webspeeder
ID: 39241332
I started completely over. I don't know where the code was jacked up, but it works now. Thanks.
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