AJAX Failed to load resource: the server responded with a status of 403 (Forbidden)

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Make sure you have remote access enabled on your mysql database.

How to allow remote connection to mysql

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Updated on June 04, 2022

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  • Admin
    Admin almost 2 years

    I tried everything available Online but I am still getting a error on AJAX Request. Tried with XMLHttpRequest() Request too but couldn't solve the problem.

    Here is my Ajax Request

    $.ajax({
    type: "POST",
    url: "ajaxsubmit.php",
    data: {
      content: post_data
    },
    cache: false,
    success: function(result){
    alert("The Data have been Entered.");
    }
    
    });
    

    This is My ajaxsubmit.php

    <?php
    session_start();
    
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "dataentry";
    
    $content = $_POST['content'];
    
    $tb_name = "pdf".$content[10];
    
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
    }
    
    $sql = "CREATE TABLE IF NOT EXISTS $tb_name (
    id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
    APNR TEXT, APDA TEXT, TMNM TEXT, TMTY TEXT,OWNN TEXT,OWNA TEXT,
    OWNC TEXT,CORN TEXT,CLAS TEXT,DES TEXT, ENTRY_TYPE TEXT, LOGO_SRC TEXT
    )";
    
    if ($conn->query($sql) === TRUE) {
    echo "Table MyGuests created successfully";
    } else {
    echo "Error creating table: " . $conn->error;
    }
    
    $sql="insert into $tb_name (APNR, APDA, TMNM, TMTY,
    OWNN,OWNA,OWNC,CORN,CLAS,DES,ENTRY_TYPE,LOGO_SRC) values ('$content
    [0]', '$content[1]', '$content[2]','$content[3]','$content[4]','$content
    [5]','$content[6]','$content[7]','$content[8]','$content[9]','$content
    [11]','$content[12]')";
    
    if ($conn->query($sql) === TRUE) {
    echo "Table MyGuests created successfully";
    } else {
    echo "Error creating table: " . $conn->error;
    }
    
    $conn->close();
    ?>
    

    I tried XMLHttpRequest() This way,

    JS PART var http = new XMLHttpRequest();

    var url = "ajaxsubmit.php";
    var params = "lorem=ipsum&name=binny";
    http.open("POST", url, true);
    
    //Send the proper header information along with the request
    http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    http.setRequestHeader("Content-length", params.length);
    http.setRequestHeader("Connection", "close");
    
    http.onreadystatechange = function() {//Call a function when the state
    changes.
    if(http.readyState == 4 && http.status == 200) {
    alert(http.responseText);
    }
    }
    http.send(params);
    

    Please give me your best suggestion as solution.