使用PHP和MySQL的ajaxpost

我试图通过使用PHP和Ajax将数据发布到mysql，唯一的问题是数据没有进入数据库，所以问题似乎出现在javascript中，我认为Ajax请帮助我

请有人可以修复它，如果有任何错误的代码

表格：

AJAX.php

$con=mysqli_connect("localhost","admin","admin","test");
$name = mysqli_real_escape_string($con, $_POST['name']);
$email = mysqli_real_escape_string($con, $_POST['email']);
$sql="INSERT INTO test (name, email) VALUES ('$name', '$email')";
if (!mysqli_query($con,$sql)) {
  die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);

---

提前谢谢。

您必须在AJAX调用中包含要发布到的URL

$.ajax({
    url: "ajax.php",
    type:"POST",
    data: dataString,
    success: success()
});

---

以下是我的建议：

HTML:

PHP:

---

我也相信Mattmo的另一个答案。他纠正的事情肯定是个错误。我想在MARGELANI的剧本中可能还有其他错误，所以我也选择发布我的答案。如果这个脚本不工作，让我知道，我会重新编码。祝你好运D

在将变量插入数据库之前，应始终转义变量。去掉html中的表单标记

$con=mysqli_connect("localhost","admin","admin","test");
$name = mysqli_real_escape_string($con, $_POST['name']);
$email = mysqli_real_escape_string($con, $_POST['email']);
$sql="INSERT INTO test (name, email) VALUES ('$name', '$email')";
if (!mysqli_query($con,$sql)) {
  die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);

$.ajax({
    url: "ajax.php",
    type:"POST",
    data: dataString,
    success: success()
});

<input name="name" id="name" type="text"/>
<input name="email" id="email" type="text"/>
<input type="button" value="Send" onclick="validate();" id="submit" /> 
<span id="error" class="warning">Message</span></p>
<p id="sent-form-msg" class="success">Thanks for your comments.We will update you within 24 hours. </p>

jQuery(document).ready(function($){
    // hide messages 
    $("#error").hide();
    $("#sent-form-msg").hide();
}
    // on submit...
    function validate()
{
        $("#error").hide();

        var name = $("#name").val();
        if(name == ""){
            $("#error").fadeIn().text("Name required.");
            $("input#name").focus();
        }

        var email = $("#email").val();
        if(email == ""){
            $("#error").fadeIn().text("Email required");
            $("input#email").focus();
            return false;
        }

       // var dataString = 'name=' + name + '&email=' + email;

        $.ajax({
            url: "phpScript.php"
            type:"POST",
            data: {name:name, email:email},
            success: success()
        });
    });  

    // on success...
     function success(){
        $("#sent-form-msg").fadeIn();
        $("#contactForm").fadeOut();
    }

}

    $con = mysqli_connect("localhost","admin","admin","test");
    $name = $_POST['name'];
    $email = $_POST['email'];
    $sql = "INSERT INTO test (name, email) VALUES ('$name', '$email')";
//I like to do it like this:

$result = mysql_query($query, $connect);

    /*if (!mysqli_query($con,$sql)) {
      die('Error: ' . mysqli_error($con));
    }*/
    echo "1 record added";
    mysqli_close($con);