Pass data from ajax to php

Question

Ajax:

function check_user_country_prod(userId , countryCode  , testType )
{ //using alert to check that all data are correct
    $.ajax({
    type: "POST",

    url: "http://localhost/test/testForm.php",

    data: { userId: userId , 
            countryCode : countryCode  ,
             productCode:  testType
        },
    success:function(res) {
        if(res == "OK")
            return true;    
        else
            return false;
    }
});
}

PHP:

<?php
    require_once("Connections/cid.php");

    $userId= $_POST['userId'];
    $productCode= $_POST['productCode'];
    $countryCode= $_POST['countryCode'];

    $sql_check = "SELECT * FROM utc WHERE userId = '$userId' AND productCode = '$productCode' AND countryCode = '$countryCode'";

    $list = mysqli_query( $conn, $sql_check);
    $num  = mysqli_fetch_assoc($list);

    if($num >0)
        echo "OK";

    else
        echo "NOK";

?>

I am very sure that the data i had pass in to the php file are correct. However i cant seem to get my data to the php file and it keep return false value back to me. Anything i can do to make it works?

**Note: Somehow even if i change both result to return true in ajax, it will still return false. and i tried to change the link to another file with only echo "OK"; but it also doesn't work. So i think it is not the file problem. It just never run the ajax no matter what. Do i need to do any link for ajax to run? **

@NiranjanNRaju but it is under different folder. Does it matter? — yan, Sep 25 '15 at 09:13
and "http://localhost/test/testForm.php" is the full link of it. — yan, Sep 25 '15 at 09:27
my current folder is localhost/abc/bsd.php while the one that i am going is localhost/test/testForm.php — yan, Sep 25 '15 at 09:27
your `$num > 0`... I think the correct way is to check if it's `empty`.... or use `mysqli_num_rows`... or maybe the javascript in your browser is disabled. — Sam Teng Wong, Sep 28 '15 at 02:34
@ShiguriAnemone' i will try it later. Anyway how do i check if my javascript on my brower is disabled? — yan, Sep 28 '15 at 02:44
if you're using chrome... on the upper right corner.. see the 3 stack bar? click that go to `settings` and click the `show advance settings` and click the `content settings` button.. on `javascript` section you will see it there... — Sam Teng Wong, Sep 28 '15 at 02:48
@ShiguriAnemone' you are right for the mysqli_num_rows but my script can't even run the ajax. It just skip the ajax part completely — yan, Sep 28 '15 at 03:01
erase all your code in php file... and put a `echo test`. then go to your path `http://localhost/test/testForm.php` if test display... it means your path is okay... and one thing, are you sure about the path?. if not use `ip address`.. — Sam Teng Wong, Sep 28 '15 at 03:08
*"Somehow even if i change both result to return true in ajax, it will still return false."* Ajax is **asynchronous**. You cannot return from the callback. Read the duplicate question. — Felix Kling, Sep 28 '15 at 05:41

score 0 · Answer 1 · edited Sep 25 '15 at 09:52

0

function check_user_country_prod(userId , countryCode  , testType )
{ //using alert to check that all data are correct
    $.ajax({
        url: "test/testForm.php"  
        type: "POST",

        url: "http://localhost/test/testForm.php", // This needs to be "test/testForm.php"

        data: { userId: userId , 
            countryCode : countryCode  ,
             productCode:  testType
        },
        success:function(res) {
        if(res == "OK")
            return true;    
        else
            return false;
        }
    });
}

edited Sep 25 '15 at 09:52

Niranjan N Raju

12,047
4
22
41

answered Sep 25 '15 at 09:13

Puya Sarmidani

1,226
9
26

i dont think that http://localhost will work. if its in a different folder try ../ to go a folder up to see if that works – Puya Sarmidani Sep 25 '15 at 09:22
my current folder is localhost/abc/bsd.php while the one that i am going is localhost/test/testForm.php – yan Sep 25 '15 at 09:24
try ../test/testForm.php – Puya Sarmidani Sep 25 '15 at 09:24
Still cannot. Any other possible problem? – yan Sep 25 '15 at 09:25
change function(res) { if(res == "OK") return true; to function(data) { if(data == "OK") – Puya Sarmidani Sep 25 '15 at 09:31
It is still the same. – yan Sep 28 '15 at 00:48

score 0 · Answer 2 · answered Sep 25 '15 at 10:25

Adjust this code according your input type. I hope this will help you:

$(document).ready(function() {
$("#submit_btn").click(function() { 
    //get input field values
    var user_name       = $('input[name=name]').val(); 
    var user_email      = $('input[name=email]').val();
    var user_phone      = $('input[name=phone]').val();
    var user_message    = $('textarea[name=message]').val();

    //simple validation at client's end
    //we simply change border color to red if empty field using .css()
    var proceed = true;
    if(user_name==""){ 
        $('input[name=name]').css('border-color','red'); 
        proceed = false;
    }
    if(user_email==""){ 
        $('input[name=email]').css('border-color','red'); 
        proceed = false;
    }
    if(user_phone=="") {    
        $('input[name=phone]').css('border-color','red'); 
        proceed = false;
    }
    if(user_message=="") {  
        $('textarea[name=message]').css('border-color','red'); 
        proceed = false;
    }

    //everything looks good! proceed...
    if(proceed) 
    {
        //data to be sent to server
        post_data = {'userName':user_name, 'userEmail':user_email, 'userPhone':user_phone, 'userMessage':user_message};

        //Ajax post data to server
        $.post('contact_me.php', post_data, function(response){  

            //load json data from server and output message     
            if(response.type == 'error')
            {
                output = '<div class="error">'+response.text+'</div>';
            }else{
                output = '<div class="success">'+response.text+'</div>';

                //reset values in all input fields
                $('#contact_form input').val(''); 
                $('#contact_form textarea').val(''); 
            }

            $("#result").hide().html(output).slideDown();
        }, 'json');

    }
});

//reset previously set border colors and hide all message on .keyup()
$("#contact_form input, #contact_form textarea").keyup(function() { 
    $("#contact_form input, #contact_form textarea").css('border-color',''); 
    $("#result").slideUp();
}); });

Well.. I am using database to validate so i dont think it works. Thanks for trying to help — yan, Sep 28 '15 at 00:39
Ok.but you can remove this validation. use only that part which you need — Manisha Chaudhary, Sep 29 '15 at 05:14

Nimmy Alice Mathew · Answer 3 · 2015-09-29T04:44:37.020

0

"my current folder is localhost/abc/bsd.php while the one that i am going is localhost/test/testForm.php".

From the above comment, you have to change the ajax url to url: "/test/testForm.php",

Apart from this , your php code shows

$num  = mysqli_fetch_assoc($list);

if($num >0)
    echo "OK";

This is incorrect as mysqli_fetch_assoc returns an associative array of strings. So, the comparison $num >0 is illogical.

edited Sep 29 '15 at 04:44

answered Sep 25 '15 at 10:40

Nimmy Alice Mathew

95
9

Tried this before. Not working. Any other possible way? – yan Sep 28 '15 at 00:40
Are you sure about the path?like slashes at correct places? – Nimmy Alice Mathew Sep 28 '15 at 09:14
Yes. I had copy the path to the url to test out. I am sure it is correct – yan Sep 29 '15 at 00:42

score 0 · Answer 4 · answered Sep 25 '15 at 11:06

Try this also both are running on my localhost:

<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
    <script>
  $(function () {
   $('form').on('submit', function (e) {
   e.preventDefault();
   $.ajax({
     type: 'post',
     url: 'post.php',
     data: $('form').serialize(),
     success: function (d) {
      alert(d);
     }
    });
   });
  });
    </script>

<form method="post" id="myform">
      <input type="text" name="time" /><br>
      <input type="text" name="date" /><br>
      <input name="submit" type="submit" value="Submit">
    </form>

make new file of post.php and pest this code. This code will alert your input field value:

<?php print_R($_POST); ?>

Pass data from ajax to php

4 Answers4