CODEWITHSUNDEEP
phpjquerymysqlajax
pocockn
pocockn

Reputation: 2063

Row not being deleted in database when using AJAX and PHP

I have created a button on my application that is supposed to remove a record in my database. At the moment I get a message saying the record has been successfully removed but it isn't actually deleting it.

Here is my button HTML

<button id="deleteButton" class="btn btn btn-danger btn-lg active" role="button" value="<?php echo $rows['recipe_id']; ?>">Delete Recipe</button>

I first send the ID of the record via ajax with this code

$(document).ready(function() {
    $('#deleteButton').click(function(){
        var clickBtnValue = $(this).val();
        var recipeID = JSON.stringify(clickBtnValue);
        var data = {
            "action" : recipeID
        }
        var ajaxurl = '/recipe-project/ajax.php';
        $.ajax({
            url: ajaxurl,
            type: "POST",
            data: data,
            success:function(data) {
                alert(data);
            }
        });
    });
});

And here is my PHP

include_once('classes/class-database-functions.php');
include_once ('classes/User.php');

if (isset($_POST['action'])) {
    $recipe_id = $_POST['action'];
    $test = new User();
    if (!empty($test)) {
        $test->delete_recipe($recipe_id);
    }
}

and here is the function in the class that actual removes the record

public function delete_recipe($recipeID) {

        $query = "DELETE FROM recipes WHERE recipe_id= '$recipeID'";

        if (mysqli_query($this->connection, $query)) {
            echo "Record deleted successfully";
        } else {
            echo "Error deleting record: " . mysqli_error($this->connection);
        }
    }

I am not getting errors in my logs and i've checked the ajax is sending the correct ID to the function, can anyone see why this problem is occurring?

Upvotes: 1

Views: 39

Answers (2)

Lord-Y
Lord-Y

Reputation: 157

As it's a POST, this part of your code is good:

if (isset($_POST['action'])) {
    $recipe_id = $_POST['action'];
    $test = new User();
    if (!empty($test)) {
        $test->delete_recipe($recipe_id);
    }
}

But for your debug, you can have before this code:

var_dump($_POST);

You can also installed firebug on Firefox in order to debug which datas you are sending.

Upvotes: 1

Technoh
Technoh

Reputation: 1600

I suggest you change your code to this

if (isset($_GET['action'])) {
    $recipe_id = $_GET['action'];
    $test = new User();
    if (!empty($test)) {
        $test->delete_recipe($recipe_id);
    }
}

then test it by accessing the URL directly in your browser and including the $_GET variable, e.g. ?action=the recipe's id. Then you can use var_dump to debug if the ID is correct, if your $test object is created correctly and if the query is executed properly.

Upvotes: 0

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