Question

所以我试图使用PHP执行一个Insert语句，但是当我调用应该执行它的代码部分时，没有任何事情发生......

已经检查过，看看出了什么问题，但无法找到答案。

这是调用PHP代码的JavaScript函数。

function comparaSenhas(){
    var pass = document.getElementById("pwd1").value;
    var pass2 = document.getElementById("pwd2").value;
    if(pass !== pass2){
        return false;
    }else{
        return true;
    }
}

function postData(){
    var hr = new XMLHttpRequest();
    var url = "../mysql.php";
    var fstnm = document.getElementById("fn").value;
    var lstnm = document.getElementById("ln").value;
    var dtnasc = document.getElementById("dn").value;
    var email = document.getElementById("em").value;
    var senha = document.getElementById("pwd1").value;
    var vars = "fname="+fstnm+"&lname="+lstnm+"&dt_nasc="+dtnasc+"&email="+email+"&senha="+senha;

    if(comparaSenhas()){

        hr.open("POST", url, true);
        hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        hr.onreadystatechange = function() {
            if(hr.readyState == 4 && hr.status == 200) {
                var return_data = hr.responseText;
                alert(return_data);
            }
        }

        hr.send(vars);

    }

}

这是我正在使用的PHP代码。

<html>
<body>
    <?php
        $servername = "localhost";
        $username = "root";
        $password = "godienski";
        $dbname = "web";

        //create connection
         $conn = new mysqli($servername, $username, $password, $dbname);

        if($conn->connect_error){
            die("Connection failed: " . $conn->connect_error);
        }
        $stmt = $conn->prepare("insert into usuarios(first_name, last_name, data_nascimento, email, senha) values (?,?,?,?,?);");
        $stmt->bind_param("sssss",$firstname,$lastname,$dtnascimento,$mail,$password);

        $firstname = $_POST['fname'];
        $lastname = $_POST['lname'];
        $dtnascimento = $_POST['dt_nasc'];
        $mail = $_POST['email'];
        $password = $_POST['senha'];
        $stmt->execute();

    $stmt->close();
    $conn->close();
    ?>
</body>
</html>

有人可以帮忙吗？

PHP使用多个参数发布到mysql

0 个答案: