Using jquery $.ajax to PHP post GET values

Question

I have a PHP script which i want to pass a value in. So example.com/test.php?command=apple. If that's entered then do its code. That all works grand in the server side but how do i accomplish this using an ajax call?

My code:

PHP:

<?php
if ($_GET['command']=='apples')
{
    //do code
    echo "hello";
}       
?>

Jquery:

$.ajax({
    url : '../command/test.php?command=apples'
}).done(function(data) {
    console.log("result: " + data);
});

Poiz · Answer 1 · 2016-12-06T13:23:11.367

The Snippet below is Pretty straight-forward and self-explanatory... however should you have any questions, don't hesitate to use the comments functionality below this Post.... ;-)

JQUERY: AJAX....

        var request = $.ajax({
            url         : "../command/test.php",
            data        : {"command": "apples"},
            dataType    : "json",
            type        : "POST"
        });

        request.done(function(data, textStatus, jqXHR){
            if(data){
                console.log(data);
            }
        });

        request.fail(function(jqXHR, textStatus) {
            alert( "Request failed: " + textStatus );
        });

PHP....

<?php
    $response   = array();
    $command    = isset($_POST['command']) ? 
                  htmlspecialchars(trim($_POST['command'])) :
                  null;
    if($command){
        // BUILD UP YOUR RESPONSE HERE
        $response['command']  = $command;
        // ADD SOME ARBITRARY DATA - JUST FOR FUN
        $response['name']     = "Albert";
        $response['lastName'] = "Einstein";
     }

     die( json_encode($response) );

Michael Yousrie · Answer 2 · 2016-12-06T00:22:59.410

You can send the value in a POST request to the php page and get the returned value from there all using $.ajax

for example we will calculate 2 numbers and return the sum via ajax

Example Code

$.ajax({
   url: "/path/to/page.php",
   type: "POST",
   data: {number_1: 10, number_2: 5},
   success: function(response){
      // alert $result from the php page which is called response here.
      alert(response);
   }
});

Php page

// checking if data exists
if ($_POST['variable_1'] != null && $_POST['variable_2'] != null ){
    $variable_1 = $_POST['variable_1'];
    $variable_2 = $_POST['variable_2'];
    $result = $variable_1 + $variable_2;
}else{
    // either of the values doesn't exist
    $result = "No Data Was Sent !";
}
// returning $result which is either 15 or " no data was sent "
echo $result;

This will return 15. Hope this makes sense now.

This makes some sense to me but cant seem to get it going. Could you put an if statement into the php page in your answer? Or explain more? Thanks! — William, Dec 05 '16 at 16:55
You can do whatever you want in the php. I will edit my answer to add a real life situation. — Michael Yousrie, Dec 06 '16 at 00:17

Nytrix · Accepted Answer · 2016-12-07T17:51:24.623

2

You can put the data into a data section, I think that will solve your problem.

$.ajax({
    url : '../command/test.php',
    data: {command: "apples",login: "succes"}
}).done(function(data) {
    console.log("result: " + data);
});

Also, you can see what you get as a response, open developer tools and you can see under network your request appearing when made, click on that and you can see the response.

edited Dec 07 '16 at 17:51

answered Dec 05 '16 at 16:46

Nytrix

1,139
11
23

How do i pass multiple commands in? So: `$_GET['command']=='apples'` and `$_GET['login']=='success'`? – William Dec 07 '16 at 17:47
1

Like any `array`, just a `,` and then your next value. I've updated the answer. – Nytrix Dec 07 '16 at 17:51
1

I was doing exactly what you did but no luck. Copied yours and worked ._. Must of been asleep last night trying haha. Thanks a mill!! – William Dec 08 '16 at 10:46

score 1 · Answer 4 · edited Dec 05 '16 at 17:30

1

Can you try this

<script type="text/javascript">

        $.ajax({
            type: 'get',
            url: "../command/test.php",
            dataType: 'json',
            data: ({command: 'apples'}),
            success: function(data)
            {
                console.log("result: " + data);                           
            }
        });

</script>

edited Dec 05 '16 at 17:30

Nytrix

1,139
11
23

answered Dec 05 '16 at 16:46

Muya

142
1
15

score 1 · Answer 5 · answered Dec 05 '16 at 16:46

1

You need to post. You can use the url args or pass a data object.

$.ajax({
    method: "POST",
    url: "../command/test.php",
    data: {
        command: "apples"
    }
})

answered Dec 05 '16 at 16:46

Jeff

607
1
7
17

Using jquery $.ajax to PHP post GET values

5 Answers5

Example Code

Php page