我一直致力于使用jQuery，AJAX和PHP的Like and Different功能。我从间接页面获取jQuery post请求。例如，我有2个PHP页面，import tkinter as tk
root = tk.Tk()
tk.Label(root, text = "This is the main window").pack()
sub_window = tk.Toplevel(root)
tk.Label(sub_window, text = "This is the other window").pack()
root.mainloop()
和viewProfile.php。 LikeMail.php中的AJAX函数正在调用LikeMail.php。
这是viewProfile.php页面描述的部分
viewProfile.php以下是这个AJAX函数实际来自 -----------------
| Like/Unlike |
-----------------
的按钮：
LikeMail.phpHTML：
function like()
{
var req = new XMLHttpRequest();
req.onreadystatechange = function()
{
if(req.readyState==4 && req.status==200)
{
document.getElementById('like1').innerHTML=req.responseText;
}
}
req.open('POST','LikeMail.php','true');
req.send();
}
setInterval(function(){like()},1000);
此<div id="like1"></div>
中显示的是输出。 以上按钮可以是div或Like取决于Unlike中的条件，这将在LikeMail.php说明部分中进行说明。
当其中一个（按钮）LikeMail.php或Like被点击时。然后它调用相应的jQuery点击功能，将发布请求发送到Unlike。我在标题中提到了LikeMail.php页面，因为Indirect或Like按钮实际存在于Unlike中页。但由于AJAX调用，这些按钮显示在LikeMail.php页面中。然后，我通过viewProfile.php将帖子请求发送到实际页面viewProfile.php
这是LikeMail.php按钮的
Unlike 这是jQuery帖子或$(document).ready(function(){
$('#Unlike').unbind().click(function(){
$.post("LikeMail.php",
{Unlike: this.id},
function(data){
$('#response').html(data);
}
);
});
});
按钮
Like
$(document).ready(function(){
$('#Like').unbind().click(function(){
$.post("LikeMail.php",
{Like: this.id},
function(data){
$('#response').html(data);
}
);
});
});
页面的描述部分
以下是LikeMail.php页面描述的部分
viewProfile.php页面中显示的{p> Like或Unlike按钮取决于此代码：
viewProfile.php按钮取决于上述两个条件
现在，当点击$check_for_likes = mysqli_query($conn, "SELECT * FROM liked WHERE user1='$user1' AND user2='$user2'");
$numrows_likes = mysqli_num_rows($check_for_likes);
if (false == $numrows_likes) {
echo mysqli_error($conn);
}
if ($numrows_likes >= 1) {
echo '<input type="submit" name="Unlike" value="Unlike" id="Unlike" class="btn btn-lg btn-info edit">';
}
elseif ($numrows_likes == 0) {
echo '<input type="submit" name="Like" value="Like" id="Like" class="btn btn-lg btn-info edit">';
}
按钮时，来自Like的帖子请求就会显示在此处。
viewProfile.php同样点击if(isset($_POST['Like'])) //When Like button in viewProfile.php is clicked then this peace of code inside if condition should run and insert some record in database
{
$total_likes = $total_likes+1;
$like = mysqli_query($conn, "UPDATE user SET user_Likes = '$total_likes' WHERE user_id = '$user2'");
$user_likes = mysqli_query($conn, "INSERT INTO liked (user1,user2) VALUES ('$user1','$user2')");
$query3 = "INSERT INTO notification (user1, user2, alert, notificationType) VALUE ('$user1','$user2','unchecked','like')";
if (mysqli_query($conn, $query3)) {
echo "Like";
} else {
echo mysqli_error($conn);
}
}
按钮时。这种代码的和平应该运行。
Unlike问题：
主要问题是jQuery帖子请求未从if(isset($_POST['Unlike'])) //This is the condition for Unlike button. It should delete record from databse
{
$total_likes = $total_likes-2;
$like = mysqli_query($conn, "UPDATE user SET user_Likes='$total_likes' WHERE user_id='$user2'");
$remove_user = mysqli_query($conn, "DELETE FROM liked WHERE user1='$user1' AND user2='$user2'");
$query3 = "DELETE FROM notification WHERE user1='$user1' AND user2='$user2' AND notificationType='like'";
$check = mysqli_query($conn, $query3);
if ($check) {
echo "Unlike";
} else {
echo mysqli_error($conn);
}
}
发送到viewProfile.php。有没有办法从间接页面发送jQuery post请求？