Ajax method post and url does not work to change my database

Question

I'm learning to use ajax. I'm facing problems while fixing the issues because there is no error notifications like PHP or MySQL. Till now, I have learn how to manipulate Database using CURD. Everything went well but MySQL itself still do nothing in this part. Please give me some idea, because I don't know what's the problem with this. I think I have not made any mistake.

$("#prodButton").on('click', function(){
    if (jamStock< jamReq || fruitsStock< fruitsReq || sugarStock< sugarReq){
        alert("ingredients is not enough");
    } else {
        $.ajax({
            method: "POST",
            url: "production.php",
            data: {
                jam: jamStock- jamReq,
                fruits: fruitsStock- fruitsReq,
                sugar: sugarStock - sugarReq,
            },
            success: function(data) {
                alert("Data has been changed");
                location.reload();
            }
        });
    }
    return false;
});

oh well about the production.php:

<?php
include 'connect.php';
if (isset($_POST['jam'])) {
    $sql = "UPDATE ingre SET
         jam =" . $_POST['jam'] . "
        , fruits =" . $_POST['fruits'] . "
        , sugar=" . $_POST['sugar'] . "
         WHERE 1";
    mysql_query($sql) or die(mysql_error());
} else {
    die("browser unable to load");
}
?>

"UPDATE `ingre` SET `jam` =" . $_POST['jam'] . " , `fruits` =" . $_POST['fruits'] . " , `sugar`=" . $_POST['sugar'] . " WHERE 1"; — Sudhakar Annadurai, Aug 03 '17 at 05:11
@sudhakar im using PHP 5.3 and mysql is not deprecated, and yes i'm using mysql. it's not an issue because i'm still learning ajax, and mysql is much faster to type than pdo or mysqli. — Obink, Aug 03 '17 at 05:11
You should learn to use secure programming from the very start. Please use prepared statements and *never* directly put your input un-sanitized into an SQL query. — Afaan Bilal, Aug 03 '17 at 05:14
@AfaanBilal i usually using this as my security >>>>>> function array_sanitize(&$item){ $item = htmlentities(strip_tags(mysql_real_escape_string($item))); } <<<<<<< and this >>>>>>function sanitize($data){ return htmlentities(strip_tags(mysql_real_escape_string($data))); } <<<<< — Obink, Aug 03 '17 at 05:17
in production.php which part is reflect in result ? if OR Else ? — maulik, Aug 03 '17 at 05:18
@maulik i'm sorry i'm don't get you. but i guess, i have no result statement in there. The result is in the index.php after url:production.php? — Obink, Aug 03 '17 at 05:20
@JYoThI no no, not at all. that's why i asks in here, i don't know what is my mistake. — Obink, Aug 03 '17 at 05:21
did you noticed that you missed `single` quotes in this line `include connect.php';` — JYoThI, Aug 03 '17 at 05:22
@JYoThI hahaha sorry sorry, i was change my file name, because to connect it is using my name. LoL — Obink, Aug 03 '17 at 05:24
@JYoThI im using PHP 5.3 and in that php version, mysql is not deprecated, and yes i'm using mysql. it's not an issue because i'm still learning ajax, and mysql is much faster to type than pdo or mysqli. — Obink, Aug 03 '17 at 05:28
@JYoThI not yet, but if u want me to try to convince me about mysql. i can doing query in phpmyadmin using mysql. — Obink, Aug 03 '17 at 05:31
For testing purpose change all your code into respectively `GET` method and call the file directly in browser with `query string parameters` and make sure that your `php code` is fine or not — JYoThI, Aug 03 '17 at 05:33
in first place echo "check"; exit(); before & after include 'connect.php'; So you get notify is all ok on your include 'connect.php' file.! — maulik, Aug 03 '17 at 05:36
@maulik i just did that and it's not echoing, nothing to display, it just get back to my index file do nothing like before. — Obink, Aug 03 '17 at 05:41
it means issue is in your url: "production.php" ... your file is yet not call for ajax. set ABSPATH in URL . — maulik, Aug 03 '17 at 05:42
Don't touch those ajax part . simple change your `PHP code` and call the file directly in browser like this `localhost/folder/file.php?jam=valu&fruits=value&sugar=value` — JYoThI, Aug 03 '17 at 05:43
@JYoThI all of the ingredients, with their values on database? — Obink, Aug 03 '17 at 05:46
or add error handling part in ajax like this `success: function(data) { alert("Data has been changed"); location.reload(); }, error : function(xhr,error_string,error_r){ console.log(error_string); console.log(error_r); }` — JYoThI, Aug 03 '17 at 05:48
@JYoThI i'm using chrome and i do right click inspect or f12 to open the console, is that right? — Obink, Aug 03 '17 at 05:52
yeah your right . Then we will debug it add alert in else part `if (jamStock< jamReq || fruitsStock< fruitsReq || sugarStock< sugarReq){ alert("ingredients is not enough"); } else { alert('coming to else part'); ... }` — JYoThI, Aug 03 '17 at 05:54

Zeljka · Answer 1 · 2017-08-03T09:04:33.690

0

-when you write query for insert or update record in table its important to use these signs for name of columns `, and its important for string values ( like varchar, char, dates, text..) to use single quotes When to use single quotes, double quotes, and backticks in MySQL

$sql = "UPDATE `ingre` 
      SET
     `jam` ='{ $_POST['jam']}', `fruits` = '{$_POST['fruits']}', sugar='{$_POST['sugar']}' WHERE 1";

-when using ajax its important to return something (catch in success and chech what happend) to controle success or fail ajax

-and one more thing, I use this "form" for ajax to control response

$.ajax({

    url: ,
    method: '',
    data: {

    }
}).done(function (data) {

}).fail(function () {

}).always(function () {

});
});

edited Aug 03 '17 at 09:04

answered Aug 03 '17 at 05:23

Zeljka

376
1
10

What do u mean? – Obink Aug 03 '17 at 05:25
Notice the quotes around the values. – Afaan Bilal Aug 03 '17 at 05:32
Please add some text to your answer which shows why they should try your answer and what is different which will solve the problem. – Nigel Ren Aug 03 '17 at 07:04

score 0 · Answer 2 · answered Aug 03 '17 at 05:40

Add alert(data); in script & echo in php file & check what you get in response alert message.

$("#prodButton").on('click', function(){
    if (jamStock< jamReq || fruitsStock< fruitsReq || sugarStock< sugarReq){
        alert("ingredients is not enough");
    } else {
        $.ajax({
            method: "POST",
            url: "production.php",
            data: {
                jam: jamStock- jamReq,
                fruits: fruitsStock- fruitsReq,
                sugar: sugarStock - sugarReq,
            },
            success: function(data) {
                alert(data);
                alert("Data has been changed");
                location.reload();
            }
        });
    }
    return false;
});

<?php
include 'connect.php';
echo "connection working<br>";
if (isset($_POST['jam'])) {
  echo "True<br>";
    $sql = "UPDATE ingre SET
         jam =" . $_POST['jam'] . "
        , fruits =" . $_POST['fruits'] . "
        , sugar=" . $_POST['sugar'] . "
         WHERE 1";
    mysql_query($sql) or die(mysql_error());
} else {
  echo "False<br>";
    die("browser unable to load");
}
?>

it didn't work, and i just get back to index.php without any change or error notification. — Obink, Aug 03 '17 at 05:44

Ajax method post and url does not work to change my database

2 Answers2