我收到並顯示所有帖子的功能,並且每個帖子都有upvote/downvote按鈕。我不希望我的php頁面在執行功能後刷新
點擊按鈕,我調用upvotePost和downvotePost函數。
它一切正常,但它刷新頁面,我想了解如何使它不刷新頁面。
我知道它是由ajax/jquery完成的,但不知道如何製作它。
我的按鈕,例如:
<a href="fun.php?upvote-btn=true?action=select&image_id=<?php echo $post['id'];?>">
函數調用:
if(isset($_GET['upvote-btn'])){
$fun->upvotePost();
}
我的功能:
public function upvotePost(){
try
{
if(isset($_SESSION['user_session'])){
$user_id = $_SESSION['user_session'];
$stmt = $this->runQuery("SELECT * FROM users WHERE id=:id");
$stmt->execute(array(":id"=>$user_id));
$myRow=$stmt->fetch(PDO::FETCH_ASSOC);
}else{
$_SESSION["error"]='Sorry, You have to login in you account!';
}
$id = $_GET['image_id'];
$user_id = $myRow['id'];
$stmt2 = $this->conn->prepare("SELECT count(*) FROM fun_post_upvotes WHERE image_id=('$id') AND user_id=('$user_id')");
$stmt2->execute();
$result2 = $stmt2->fetchColumn();
if($result2 == 0){
$stmt3 = $this->conn->prepare("INSERT INTO fun_post_upvotes (image_id,user_id) VALUES(:image_id,:user_id)");
$stmt3->bindparam(":image_id", $id);
$stmt3->bindparam(":user_id", $user_id);
$stmt3->execute();
$stmt4 = $this->conn->prepare("SELECT * FROM fun_posts WHERE id=('$id')");
$stmt4->execute();
$result4 = $stmt4->fetchAll();
foreach($result4 as $post){
$newUpvotes = $post['upvotes']+1;
$stmt5 = $this->conn->prepare("UPDATE fun_posts SET upvotes=$newUpvotes WHERE id=('$id')");
$stmt5->execute();
$_SESSION["result"]='You have succesfully liked this post!';
}
}else{
$_SESSION["error"]='You have already liked this post!';
}
$stmt6 = $this->conn->prepare("SELECT count(*) FROM fun_post_downvotes WHERE image_id=('$id') AND user_id=('$user_id')");
$stmt6->execute();
$result6 = $stmt6->fetchColumn();
if($result6 > 0){
$stmt7 = $this->conn->prepare("DELETE FROM fun_post_downvotes WHERE image_id=('$id') AND user_id=('$user_id')");
$stmt7->execute();
$stmt8 = $this->conn->prepare("SELECT * FROM fun_posts WHERE id=('$id')");
$stmt8->execute();
$result8 = $stmt8->fetchAll();
foreach($result8 as $post){
$newDownvotes = $post['downvotes'] - 1;
$stmt9 = $this->conn->prepare("UPDATE fun_posts SET downvotes=$newDownvotes WHERE id=('$id')");
$stmt9->execute();
}
}
}
catch(PDOException $e)
{
echo $e->getMessage();
}
}
你有什麼試圖讓它在AJAX中工作?有很多關於SO的解釋和通過谷歌來引導你完成AJAX請求。 –
我從來沒有學過AJAX,但最近幾周試圖弄清楚它是如何工作的,但無法理解。 – Crelix