Question

当我点击“subcribeButton”时，我希望它保存该值并将该值传递给我的PHP文件。我希望该PHP文件然后发布到SQL数据库。

这是我的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("");

});

这是我的PHP代码，我希望我的php代码接受来自我的js文件的数据，然后将数据发布到我的sql数据库：

<?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();
?>

结果是我收到以下错误：

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

那么为什么不在你的ajax请求中添加一个回调，这样你就可以在控制台或警报中调试它，看看它可能对你有帮助

$.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
});

所以在你的php文件中你可以添加这个

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

您遇到语法错误您的代码缺少此行末尾的;：

echo $text

应该是：

echo $text;

Answer 3

解决方案：

在js文件中，网址错误，电子邮件中的引号丢失

$(".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文件中使用了错误的变量

// 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();
?>

百万感谢所有回复此帖子并帮助我解决问题的人。

无法将数据从我的JavaScript传递到我的PHP文件

3 个答案: