Can not add data into database using jquery ajax with php

Question

I want to store data into database using jquery ajax using php but when i click on submit button a empty alert shows up and a message below submit button submitted succesfully but no data is added on my databse.i am at very beginer level in jquery and ajax...Any Help will be appreciated.

Here is my HTML

<form method="POST" >
        <pre>
            <label for="">Enter Post Topic</label><input type="text" id="txt_topic_name" name="txt_topic_name"><br>
            <label for="">Enter Detail</label><textarea id="txt_detail" name="txt_detail"></textarea><br/>
            <label for=""></label><input type="button" id="btn_submit" name="btn_submit" value="submit"><br>
        </pre>
    </form>
    <div id="results"></div>

And here is my javascript

$(document).ready(function(){
        $('#btn_submit').on('click',function(){
            var topic_name = $('#txt_topic_name').val();
            var detail     = $('#txt_detail').val();
            $.ajax({
                url   : "ajax/add_topic.php",
                type  : "POST",
                data  : {'txt_topic_name' : topic_name ,'txt_detail' : detail},
                success : function(data){
                    alert(data);
                    console.log(data);
                    $('#results').html("submitted succesfully");
                },
                error  : function(data){
                    // alert(data);
                    // console.log(data);
                }

            });
             // return false;

        });

     });

And PHP

if (isset($_POST['btn_submit'])) {
    mysql_connect("localhost","root","") or die("Could not coonnect");
    mysql_select_db("forum") or die("could not select db");

    $topic_name  = mysql_real_escape_string($_POST['txt_topic_name']);
    $detail = mysql_real_escape_string($_POST['txt_detail']);

    $sql    = "INSERT INTO Topics(name,detail)  VALUES('$topic_name','$detail')";
    $query  = mysql_query($sql);
    if ($query) {
        echo "Sucess";
    }
    else{
        echo "Failed";

    }

}

You can't able to use Form Submit and Ajax both at same time... Both are different from each other — Anto king
– Anto king, Commented Jul 18, 2014 at 7:00
No error just a alert pops up from success function and displays nothing — user2166164
– user2166164, Commented Jul 18, 2014 at 7:01

Sudhanshu Saxena · Accepted Answer · 2014-07-18 07:19:54Z

1

if (isset($_POST['txt_topic_name'])) {
    mysql_connect("localhost","root","") or die("Could not coonnect");
    mysql_select_db("forum") or die("could not select db");

    $topic_name  = mysql_real_escape_string($_POST['txt_topic_name']);
    $detail = mysql_real_escape_string($_POST['txt_detail']);

    $sql    = "INSERT INTO Topics(name,detail)  VALUES('$topic_name','$detail')";
    $query  = mysql_query($sql);
    if ($query) {
        echo "Sucess";
    }
    else{
        echo "Failed";

    }

}

edited Jul 18, 2014 at 7:19

Sudhanshu Saxena

1,19912 silver badges32 bronze badges

answered Jul 18, 2014 at 7:11

Er. Sajal Tiwari

1187 bronze badges

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4 Comments

Wladimir Palant Over a year ago

Feel free to add some text to explain what you did and why.

user2166164 Over a year ago

Thnx Man it's added in the database.Can you plz explain to me what was i doing wrong.

user2166164 Over a year ago

@Er. Sajal Tiwari can you please explain it what i did wrong.?

Er. Sajal Tiwari Over a year ago

actually you are posting the following data 'txt_topic_name' : topic_name ,'txt_detail' : detail ,and in your php file yor are catching button which has not been posted.

Xlander · Accepted Answer · 2014-07-18 07:05:30Z

0

The data object should be like this

data: {txt_topic_name : topic_name ,txt_detail : detail}

instead of this

data: {'txt_topic_name' : topic_name ,'txt_detail' : detail}

And for a better check for a post, use a hidden input

<input type='hidden' value='true' name='submission' />

Then in php check for submission

if(isset($_POST['submission']) && $_POST['submission'] == 'true'){
  //the rest of your code
}

answered Jul 18, 2014 at 7:05

Xlander

1,33112 silver badges24 bronze badges

5 Comments

user2166164 Over a year ago

it still displays nothing on alert

hindmost Over a year ago

@Xlander The data object should be like this... Both are the same

Xlander Over a year ago

Have you include submission: 'true' in the data object?

Xlander Over a year ago

@hindmost: Hmmm... Well, I used to write it this way :p... nevermind.

Xlander Over a year ago

data: {txt_topic_name : topic_name ,txt_detail : detail, submission: $('input[name="submission"]').val()} Then in your success function, check for the value data: if(data == 'Sucess') $('#results').html("submitted succesfully"); else $('#results').html("error");

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