AJAX call return success message but post data never insert into db

Question

I have suggestion.php form as follow and call the ajax method. I always got "Thanks you" alert but post data never insert into db. I am very sure that my connection string is correct because it is sometime insert the record but most of the time it is not inserted.

Any help, please. Regards, Alex

suggestion.php

<form class="suggestion" role="form" method="POST" action="suggestion_add.php">
      <div class="form-group"><label name="warnning" id="urlwarnning">Currently we only support YouTube as well as Vimeo videos</label></div>
      <div class="form-group"><input name="url" id="url" type="text" class="form-control" placeholder="URL" required></div>
      <div class="form-group"><input name="title" id="title" type="text" class="form-control" placeholder="Title" required></div>
      <div class="form-group"><textarea id="desc" name="desc" class="form-control" rows="5" placeholder="Description.." required></textarea></div>
      <div class="form-group"><input name="name" id="name" type="text" class="form-control" placeholder="Name" required></div>
      <div class="form-group"><input name="email" id="email" type="text" class="form-control" placeholder="Email ID" required></div>
      <button class="btn btn-lg btn-primary btn-block" type="submit">SUBMIT</button>
    </form>

Function

$(function () {
    $(document).on('submit', ".suggestion", function(e) {
        var name=$("#name").val();
        var email=$("#email").val();
        var url=$("#url").val();
        var title=$("#title").val();
        var desc=$("#desc").val();
        var formURL = $(this).attr("action");
        var youtube_regExp = /^.*(youtu.be\/|v\/|u\/\w\/|embed\/|watch\?v=|\&v=)([^#\&\?]*).*/;
        var vimeo_regExp = /https?:\/\/(?:www\.)?vimeo.com\/(?:channels\/|groups\/([^\/]*)\/videos\/|album\/(\d+)\/video\/|)(\d+)(?:$|\/|\?)/;
        var match = url.match(youtube_regExp);
        if (match&&match[2].length==11){
            var vid = match[2];
            var src = "youtube";
            var check = 1;
        }else{
            var match = url.match(vimeo_regExp);
            if (match){
                var vid = match[3];
                var src = "vimeo";
                var check = 1;
            }else{
                alert("not a valid url");
                var check = 0;

            }
        }
        if(check == 1){
            var data="title="+title+"&desc="+desc+"&src="+src+"&vid="+vid+"&name="+name+"&email="+email;

            $.ajax(
            {
                url : "suggestion_add.php",
                type: "POST",
                data : data,
                success:function(data, textStatus, jqXHR)
                {
                    $("#name").val("");
                    $("#email").val("");
                    $("#url").val("");
                    $("#title").val("");
                    $("#desc").val("");
                    alert("Thank you !");

                },
                error: function(jqXHR, textStatus, errorThrown)
                {
            alert(errorThrown);
                }
            });
        }else{
            alert("Check again!");
        }
        e.preventDefault();
    });

});

suggestion_add.php

<?php
include("config.php");
$con=mysql_connect($db_host,$db_uname,$db_pwd,$db_name);
mysql_select_db($db_name);
$src=$_POST['src'];
$name=$_POST['name'];
$email=$_POST['email'];
$title=$_POST['title'];
$desc=$_POST['desc'];
$vid=$_POST['vid'];
$query="INSERT INTO `suggestion` (`id`,`src`,`vid`,`title`,`desc`,`name`,`email`)VALUES (NULL,'$src','$vid','$title','$desc','$name','$email');";
$result=mysql_query($query);
?>

Answer 1

Thanks you so much for all of your suggestion. It is happened because of the Single quote.

After I have added $dbc->real_escape_string($title) it is working perfectly.

Answer 2

Success will be called as long as the script suggestion_add.php runs.
I recommend that you print something at the end of the script, like:

header("Content-type: application/json");
die(json_encode(array("success" => $result))); 
// Where $result is true if query is OK else false.
// If you want the errror message to be passed to the javascript 
// function too you can add "error" => $error to the array too, 
// and then print it in the 'success' function.

Then in your 'success' function check:

success:function(data){
  if(data.success){
     //Do success
  } else {
     //Do failure
  }
}

Also, do not use the deprecated mysql_* functions, check out mysqli or PDO for better APIs which both have prepared statements, so that you do not open up for sql-injection vulnerabilities.

---

Edit:
You don't have to set up the data as a post request when using the jquery ajax function, you can do:

var data = {
  "title": title,
  "desc": desc,
  "src": src,
  "vid": vid,
  "name": name,
  "email": email
}

And then just pass the object data as the data parameter.

Edit2:
If you do not know how to, to get the mysql error from the old deprecated mysql_* api, you use the mysql_error function.
Something like:

$result = mysql_query($query) or die(array("success" => false, "error" => mysql_error());

Answer 3

Could it be that you have setup a unique index on the table and sometimes mysql cannot insert a duplicate key?

You can do this to see if there are errors with mysql:

$result=mysql_query($query) or die('Query failed: ' . mysql_error());

UPDATE:

$con = mysql_connect($db_host,$db_uname,$db_pwd) or die (mysql_error());
mysql_select_db($db_name, $con);

Answer 4

I thing your data is not passing to php file because you are postting the data twice(inside form tag)

action="suggestion.php"

and in jQuery ajax

 $.ajax(
            {
                url : "suggestion_add.php",
                type: "POST",
                data : data,
                success:function(data, textStatus, jqXHR)
                {
}});

better use anyone to post the value.. change form tag and try

 <form class="suggestion" >

        </form>

i think it will work..