CODEWITHSUNDEEP
javascriptphpjqueryhtmlajax
user2166164
user2166164

Reputation:

Can not add data into database using jquery ajax with php

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";

    }

}

Upvotes: 2

Views: 609

Answers (2)

Er. Sajal Tiwari
Er. Sajal Tiwari

Reputation: 118

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";

    }

}

Upvotes: 1

Xlander
Xlander

Reputation: 1331

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
}

Upvotes: 0

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