Form Submission Using Ajax not working

Question

I have a problem getting my form to insert the records into the database. I just can't figure out where I'm wrong...

My form is bellow

 <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script src="script.js"></script>

<form id="form" name="form">
    <div>
        <label>Name :</label>
        <input id="name" type="text">
        <label>Email :</label>
        <input id="submit" onclick="myFunction()" type="button" value="Submit">
    </div>
</form>

a script script.js

function myFunction() {
    var name = document.getElementById("name").value;
    var dataString = 'name1=' + name;
    if (name == '') {
        alert("Please Fill All Fields");
    } else {
        $.ajax({
            type: "POST",
            url: "ajaxjs.php",
            data: dataString,
            cache: false,
            success: function(html) {
                alert(html);
            }
        });
    }
    return false;
}

and a php script ajaxjs.php

<?php
include ('./includes/connection.php');
$name1 = $_POST['name1'];
if (isset($_POST['name1'])) {
    $query = mysql_query("INSERT INTO db VALUES('TEST','name1')");
}
?>

developer tools show message: script.js:5 Uncaught ReferenceError: dataString is not definedmyFunction @ script.js:5onclick @ test.html:9

Turn on developer tools in your browser and see the error your getting update your question with the error — Dan
– Dan, Commented Jul 16, 2016 at 18:08
@Danturnip script.js:5 Uncaught ReferenceError: dataString is not definedmyFunction @ script.js:5onclick @ test.html:9 — Menel
– Menel, Commented Jul 16, 2016 at 18:14
Sidenote: even if it works without errors, you'd be inserting a constant string (name1) instead of the posted value. — FirstOne
– FirstOne, Commented Jul 16, 2016 at 18:16

Quynh Nguyen · Accepted Answer · 2016-07-16 18:16:50Z

2

Your Ajax code working well But your query command look like wrong Please try

INSERT INTO table_name (column1,column2,column3,...)
VALUES (value1,value2,value3,...);

Update: Source code working well

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script type="text/javascript">
    function myFunction() {
        var name = document.getElementById("name").value;
        var dataString = 'name1=' + name;
        if (name == '') {
            alert("Please Fill All Fields");
        } else {
            $.ajax({
                type: "POST",
                url: "ajaxjs.php",
                data: dataString,
                cache: false,
                success: function(html) {
                alert(html);
                }
            });
        }
        return false;
    }
</script>
<form id="form" name="form">
<div>
<label>Name :</label>
<input id="name" type="text">
<label>Email :</label>
<input id="submit" onclick="myFunction()" type="button" value="Submit">
</div>
</form>

And ajaxjs.php

<?php
    // include ('./includes/connection.php');
    $name1 = $_POST['name1'];
    echo $name1;
    // if (isset($_POST['name1'])) {
    // $query = mysql_query("INSERT INTO db VALUES('TEST','name1')");
    // }
?>

edited Jul 16, 2016 at 18:16

answered Jul 16, 2016 at 18:09

Quynh Nguyen

3,0172 gold badges16 silver badges28 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

FirstOne Over a year ago

You can omit the column names if you provide a value for each table column. Reference: If you do not specify a list of column names for INSERT ... VALUES [..], values for every column in the table must be provided by the VALUES [...]

Quynh Nguyen Over a year ago

I think his query not working and I give one example.

Quynh Nguyen Over a year ago

Yes bro, I know it. xD I just give one full example xD

FirstOne Over a year ago

@Menel Hardly... I did nothing xD

Menel Over a year ago

Ah soz, I mean Quynh

|

Collectives™ on Stack Overflow

Form Submission Using Ajax not working

1 Answer 1

6 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Your Answer

Sign up or log in

Post as a guest

Related