我正在按照 youtube 上的教程为我的状态系统制作一个简单的喜欢/不喜欢按钮,我完成了大部分工作,但它不会更新我的喜欢,也不会将喜欢插入数据库,请帮我说怎么了,我已经尝试了很多了..
获取状态的函数:
function getStatus($conn) {
$sql = "SELECT * FROM status ORDER BY sid DESC";
$query = mysqli_query($conn, $sql);
while ($row = $query->fetch_assoc()) {
echo "<div class='post'>".$row['message']."<br>";
$result = mysqli_query($conn, "SELECT * FROM status_like WHERE uid=1 and sid=".$row['sid']."");
if (mysqli_num_rows($result) == 1) {
echo "<span><a href='' class='unlike' id='".$row['sid']."'>unlike</a></span>";
} else {
echo "<span><a href='' class='like' id='".$row['sid']."'>like</a></span></div>";
}
}
}
jquery代码
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('.like').click(function(){
var sid = $(this).attr('id');
$.ajax({
url: 'test.php',
type: 'post',
async: false,
data: {
'liked': 1,
'sid': sid
},
success:function(){
}
});
});
});
</script>
我认为问题所在的最后一个 php 代码是:
if (isset($_POST['liked'])) {
$sid = $_POST['sid'];
$sql = "SELECT * FROM status WHERE sid=$sid";
$query = mysqli_query($conn, $sql);
$row = mysqli_fetch_array($query);
$n = $row['likes'];
$uid = 1;
$sql2 = "UPDATE status SET likes=$n+1 WHERE sid=$sid";
$sql3 = "INSERT INTO status_like (uid, sid, username) VALUES (1, '$sid', '$uid')";
mysqli_query($conn, $sql2);
mysqli_query($conn, $sql3);
exit();
}
最佳答案
if (isset($_POST['liked'])) {
$sid = $_POST['sid'];
$sql = "SELECT * FROM status WHERE sid=$sid";
$query = mysqli_query($conn, $sql);
$row = mysqli_fetch_array($query);
//$n = $row['likes']; // your code
$n = (int) $row['likes']; // try like this.. might be likes in string so convert to int
$uid = 1;
//$sql2 = "UPDATE status SET likes=$n+1 WHERE sid=$sid"; // Your code
// Do like this `status` in query because status is reserved keyword of MySql for more details you could visit this link https://dev.mysql.com/doc/refman/5.7/en/keywords.html
$sql2 = "UPDATE `status` SET likes=$n+1 WHERE sid=$sid";
$sql3 = "INSERT INTO status_like (uid, sid, username) VALUES (1, '$sid', '$uid')";
mysqli_query($conn, $sql2);
mysqli_query($conn, $sql3);
exit();
}
关于PHP Like/Unlike 按钮与 jquery,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42545919/