無法將數據從JavaScript傳遞到PHP文件

Question

當我單擊“ subcribeButton”時，我希望它保存值並將該值傳遞給我的PHP文件。 我想要那個PHP文件然后發布到SQL數據庫。

這是我的js：

$(".subscribeButton").on( "click", function() {

    var email = $(".subscribeInput").val();
    var isValid = isEmailValid(email);

    if(isValid){
        $(".errorMsg").css( "display", "none" );
        modal.style.display = "block";

        $.post( "save.php", { email: email });

        $(".subscribeInput").val("");

    } else {
        $(".errorMsg").css( "display", "initial" );
    };

});

$(".subscribeInput").on( "click", function() {

    $(".subscribeInput").val("");

});

這是我的php代碼，我希望我的php代碼接受我js文件中的數據，然后將數據發布到我的sql數據庫中：

<?php
$servername = "localhost";
$username = "user";
$password = "pw";
$dbname = "db_name";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$text = $_POST['email'];

echo $text

$sql = "INSERT INTO MyGuests (email)
 VALUES ($text)";



if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?>

結果是出現以下錯誤：

POST http://flockto.it/home/save.php 500 (Internal Server Error)
send    @   jquery.js:4
ajax    @   jquery.js:4
m.(anonymous function)  @   jquery.js:4
(anonymous function)    @   email.js:13
dispatch    @   jquery.js:3
r.handle    @   jquery.js:3

Answer 1

因此，為什么不在您的ajax請求中添加回調，以便您可以在控制台中對其進行調試，也可以在出現警報時看到它可能會對您有所幫助

$.post( "save.php", { "email" : email } , function(result){
   console.log(result);//here the result will display what you will echo after getting the post in your php file
});

因此，在您的php文件中，您可以添加此內容

if (isset($_POST['email'])){
   $text = $_POST['email'];
   echo $text;
}
//so if you check the console you will get the email
//value from your php file so then you can insert it at the DB as you want

Answer 2

您有語法錯誤。
您的代碼缺少; 在此行的末尾：

echo $text

它應該是：

echo $text;

Answer 3

解決方案：

js文件中的網址錯誤，並且電子郵件周圍的引號丟失了

$(".subscribeButton").on( "click", function() {

    var email = $(".subscribeInput").val();
    var isValid = isEmailValid(email);

    if(isValid){
        $(".errorMsg").css( "display", "none" );
        modal.style.display = "block";

        $.post( "data/save.php", { "email": email });

        $(".subscribeInput").val("");

    } else {
        $(".errorMsg").css( "display", "initial" );
    };

});

$(".subscribeInput").on( "click", function() {

    $(".subscribeInput").val("");

});

在php文件中使用了錯誤的變量

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$text = $_POST['email'];

if (isset($_POST['email'])){
   $text = $_POST['email'];
   echo $text;
}

$sql = "INSERT INTO MyGuests (email)
 VALUES ('$text')";



if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?>

一百萬感謝所有回復此主題並幫助我解決問題的人。