submitting form from jquery to php

Question

I'm going crazy! I'm trying to submit a form from jquery to php and insert a record in my database. I'm getting no error, but no record is being submitted. All works fine when I just go to the php page and put variables in the URL. But it doesn't work when I submit the variables via the jquery page. Can anyone help me out, please?

HTML:

<form id="tellusForm" >
      <div class="formHead">Tell us your story</div>
      <div id="thank">Thank you.</div>
      <table cellpadding="5">
      <tr><td>Name:</td><td><input id="tellName" type="text"/></td></tr>
      <tr><td>Country of origin:</td><td><select id="tellCountry"></select></td></tr>
      <tr><td>Age:</td><td><input type="text" id="tellAge"/></td></tr>
      <tr><td>Occupation:</td><td><input type="text" id="tellOccupation"/></td></tr>
      <tr><td>Email address:</td><td><input type="text" id="tellEmail"/></td></tr>
      <tr><td>Phone number:</td><td><input type="text" id="tellPhone"/></td></tr>
      <tr><td>A bit about you:</td><td><textarea  id="tellAbout" style="width: 100%;"></textarea></td></tr>
      <tr><td></td><td><input type="submit" value="Send" class="submit"/></td></tr>
      </table>
      </form>

jquery:

$('.submit').click(function(){

        $.get('tellus.php', { name: "laura"}, function(data){
                 eval(data);
        });


});

tellus.php:

<?php
require_once ('../constants_test.php'); 

    $name = $_GET['name'];

    $db = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
                if (mysqli_connect_errno()) {
                        printf("Connect failed: %s", mysqli_connect_error());
                        exit;
                }

    $q = "Insert into tellus(`Name`) values ('" . $name . "')";
    if ($db->query($q)) {
        echo "console.log('you got it')";
    };
    $db->close();

?>

Answer 1

Thanks everyone who tried to help me! I ended up having to do it this way:

$.post("./tellus.php", {tellName:name}, function(data){
alert(data)
}); 

The ajax call must have had something that wasn't working.

Answer 2

Use serialize():

$("#tellusForm").on("submit", function() {

    event.preventDefault();

    $.ajax({
      url: 'tellus.php',
      data: $(this).serialize()
    }).done(function(data) {
      $(html).append(data);
    });

});

Also make sure to have names on your form elements:

<input type="text" name="tellAge" id="tellAge" />

This topic is very common here so try searching for other posts - e.g. Submit form using AJAX and jQuery could work too.

Answer 3

The ajax call wasn't working because you are using $.serialize to serialize your form objects, which will return ;

{key:'tellName', value='name'} // <input name="tellName" value="name" />

whereas you are expecting a key:value pair. In order to get the data to send what you expecting, you should modify the object using this jQuery function:

$.fn.serializeObject = function(){
    var o = {};
    var a = this.serializeArray();
    $.each(a, function() {
        if (o[this.name]) {
            if (!o[this.name].push) {
                o[this.name] = [o[this.name]];
            }
                o[this.name].push(this.value || '');
            } else  {
                o[this.name] = this.value || '';
            }
        });
    return o;
}

And call it whenever you're submitting form. It will structure your form to have the key:value pair.

<form id="register">
    <input type="text name="username" />
    <input type="text" name="email" />
    <input type="submit value="Register!" />
</form>

and the javascript (remember to include the serializeObject function)

$('#register').on('submit', function(){
    var data = $(this).serializeObject()
    // data will return {username: value, email:value}
    $.post('process.php', data, function(response){
        try{
            var r = JSON.parse(response) // Check if the response is an object
            console.log(JSON.stringify(r))
        }
        catch(e){
            console.log(response) // The response is a string
        }

    })
})

Cheers!

Answer 4

I suggest you should use POST for actions which include insert/delete/update of data in database.  It keeps the data safe.

If you want to still use GET and test out something.  I suggest do not use jquery in the middle to handle the get.  Submit button on click submits the form anyway, so the jquery $.get is unnecessary.

To your form element add 
<form name='' action='tellus.php'>...</form>

This should fix the problem.

Answer 5

How about this one

jquery

$("#submit").click(function() {
    formData = {
        name: "abc"
    }
    $.ajax({
        type: 'GET',
        contentType: 'application/json',
        url: "http://yourpage/tellus.php",
        dataType: "json",
        data: formData,
        success: function(data) {
            console.log(data.message);
        },
        error: function(data) {
            console.log(data.message);
        }
    });
});

php

  <?php

// array for JSON response
$response = array();

// check for required fields
if (isset($_GET['name'])) {

    $name = $_GET['name'];

    // include db connect class
    require_once ('../constants_test.php'); 

    // connecting to db
    $db = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
                    if (mysqli_connect_errno()) {
                            printf("Connect failed: %s", mysqli_connect_error());
                            exit;
                    }

    // mysqli inserting a new row
    $q = "INSERT INTO tellus('Name') values ('" . $name . "')";

    // check if row inserted or not
    if ($db->query($q)) {
        // successfully inserted into database
        $response["success"] = 1;
        $response["message"] = "you got it";

        // echoing JSON response
        echo json_encode($response);
        $db->close();
    } else {
        // failed to insert row
        $response["success"] = 0;
        $response["message"] = "Oops! An error occurred.";

        // echoing JSON response
        echo json_encode($response);
        $db->close();
    }
} else {
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

    // echoing JSON response
    echo json_encode($response);
}
?>