Question

这是我第一次使用jQuery AJAX所以请耐心等待。我正在尝试为我的网站创建一个类似于Facebook按钮的类似按钮。用户必须通过Facebook登录，这将从我的数据库中提取用户ID。当用户单击like按钮时，它会使用用户信息和页面信息将一行插入数据库。如果用户和页面已有条目，则会更新该行。

我的代码在我的localhost上运行得很好。我可以整天点击相似的按钮，图像会不断地从不同的方向来回变换。数据库每次都会更新，没有任何问题。

当我将其上传到我的实时服务器时，脚本运行完全正常。一旦我的页面完成加载，数据库就不再从点击中更新。图像将照常更改，我将收到警报（“成功”），但不会对实际数据库进行任何更改。

<?php
$tempstat = '1';
if ($stmt = $mysqli -> prepare("SELECT count(id)
    FROM votes
    WHERE businessid=? and status=? ")) {
    $stmt -> bind_param('ss', $businessid, $tempstat);
    $stmt -> execute();
    $stmt->bind_result($likecount);
    $stmt->fetch();
    $stmt->close();
}
if ($likecount == '') { $likecount = 0; }
?>
<style>
#karmabar { padding-left: 4px; }
#counter { margin: 0 auto; text-align: center; width: 50px; height: 25px; background-image: url('/images/karma-counter.png'); }
p#counter { font-size: 12px; padding-top: 5px; }
.karma { margin: 0 auto; font-size: 16px; color: blue; text-align: center; padding: 5px; }
</style>
<?php
if(login_check($mysqli) == true) { 
$userid = $_SESSION['userurl']; 
if ($stmt = $mysqli -> prepare("SELECT status
    FROM votes
    WHERE userid=? and businessid=? ")) {
    $stmt -> bind_param('ss', $userid, $businessid);
    $stmt -> execute();
    $stmt -> bind_result($like_status);
    $stmt->fetch();
    $stmt->close();
}
if ($like_status == '' or $like_status == NULL) { $like_status = '0'; }
?>

<script type="text/javascript">
  $(document).ready(function() {
    $('#like_post').click(function() {
      $.ajax({
        url: "/includes/like.php",
        type: "GET",
        data: 'userid=<?php echo $userid; ?>&businessid=<?php echo $businessid; ?>&like_status=1',
        success: function() {
            alert("Success");
        },
        error: function() {
           alert("Something went wrong");
        }
      });
      $('#counter').html(function(i, val) { return val*1+1 });
      $('#like_post').hide();
      $('#unlike_post').show();
    });

  $('#unlike_post').click(function() {
     $.ajax({
        url: "/includes/like.php",
        type: "GET",
        data: 'userid=<?php echo $userid; ?>&businessid=<?php echo $businessid; ?>&like_status=0',
        success: function() {
            alert("Success");
        },
        error: function() {
            alert("Something went wrong");
        }
        });
      $('#counter').html(function(i, val) { return val*1-1 });
      $('#unlike_post').hide();
      $('#like_post').show();
  });
});
</script>
<div id="karmabar">
    <table cellpadding="0px">
        <?php if ($like_status == '0') { ?>
        <tr><td><a href="javascript:;" id="unlike_post" class="hide"><img src="/images/karma-active.png" title="Undo Karma" /></a><a href="javascript:;" id="like_post"><img src="/images/karma-inactive.png" title="Spread Karma" /></a><td><p id="counter"><?php echo $likecount; ?></p></td></tr>
        <?php } else { ?>
        <tr><td><a href="javascript:;" id="unlike_post"><img src="/images/karma-active.png" title="Undo Karma" /></a><a href="javascript:;" id="like_post" class="hide"><img src="/images/karma-inactive.png" title="Spread Karma" /></a></td><td><p id="counter"><?php echo $likecount; ?></p></td></tr>
        <?php } ?>
    </table>
</div>
<?php } else { ?>
<div id="karmabar">
    <table cellpadding="0px">
        <tr><td><a href="#" onclick="alert('Login with Facebook to Spread Karma');" ><img src="/images/karma-inactive.png" title="Spread Karma" /></a><td><p id="counter"><?php echo $likecount; ?></p></td></tr>
    </table>
</div>

$ businessid和$ userid都在页面上定义得更高。

like.php

<?php
include_once $_SERVER['DOCUMENT_ROOT'].'/includes/db_connect.php';
include_once $_SERVER['DOCUMENT_ROOT'].'/includes/functions.php';
$userid = $_GET['userid'];
$businessid = $_GET['businessid'];
$IP = $mysqli->real_escape_string(getClientIP());
$like_status = $_GET['like_status'];
if ($stmt = $mysqli -> prepare("SELECT count(id)
    FROM votes
    WHERE userid=? and businessid=? ")) {
        $stmt -> bind_param('ss', $userid, $businessid);
        $stmt -> execute();
        $stmt ->bind_result($count);
        $stmt ->fetch();
        $stmt ->close();
}
if ($count == '1') {
    if ($stmt = $mysqli -> prepare("UPDATE votes 
    SET ip=?, status=?
    WHERE userid=? and businessid=? ")) {
        $stmt -> bind_param('ssss', $IP, $like_status, $userid, $businessid);
        $stmt -> execute();
        $stmt -> close();
        $mysqli -> close();
    }
}
if ($count == '0') {
    if ($stmt = $mysqli -> prepare("INSERT INTO votes (userid, businessid, ip, status) 
    VALUES ( ?, ?, ?, ? ) ")) {
        $stmt -> bind_param('ssss', $userid, $businessid, $IP, $like_status );
        $stmt -> execute();
        $stmt -> close();
        $mysqli -> close();
    }
}
?>

我感谢你们给我的任何帮助。就像我说的，我是这种编码的新手，正在寻找你可能有的任何指针。

Answer 1

这听起来像连接错误。您是否已将连接参数从localhost更改为实际站点。即您要连接的数据库名称和密码？

Answer 2

迈克，我不确定发生了什么，但也许你可以尝试使用这种类型的ajax请求/ PHP返回来解决问题：

用以下代码替换整个AJAX函数：

$(document).ready(function() {
    $('#like_post').click(function() {
        $.post("/includes/like.php", {userid:<?php echo $userid; ?>, businessid:<?php echo $businessid; ?>, like_status:"1"}, function(data){
            alert(data); //Check to see what is being returned
            if(data == 1){ //Later, you can echo "1" in your PHP file on success
                //This is success
                alert("Success");
                $('#counter').html(function(i, val) { return val*1+1 });
                $('#like_post').hide();
                $('#unlike_post').show();
            } else {
                //This is failure
                alert("Failure");
            }
        });
    });

    $('#unlike_post').click(function() {
        $.post("/includes/like.php", {userid:<?php echo $userid; ?>, businessid:<?php echo $businessid; ?>, like_status:"0"}, function(data){
            alert(data); //Check to see what is being returned
            if(data == 1){ //Later, you can echo "1" in your PHP file on success
                //This is success
                alert("Success");
                $('#counter').html(function(i, val) { return val*1-1 });
                $('#unlike_post').hide();
                $('#like_post').show();
            } else {
                //This is failure
                alert("Failure");
            }
        });
    });
});

这将是PHP - 注意，我将$_GET更改为$_POST并在最后添加了回显。我解释了为什么之后。

<?php
    include_once $_SERVER['DOCUMENT_ROOT'].'/includes/db_connect.php';
    include_once $_SERVER['DOCUMENT_ROOT'].'/includes/functions.php';
    $userid = $_POST['userid'];
    $businessid = $_POST['businessid'];
    $like_status = $_POST['like_status'];
    $IP = $mysqli->real_escape_string(getClientIP());
    if ($stmt = $mysqli -> prepare("SELECT count(id) FROM votes WHERE userid=? and businessid=? ")) {
        $stmt -> bind_param('ss', $userid, $businessid);
        $stmt -> execute();
        $stmt ->bind_result($count);
        $stmt ->fetch();
        $stmt ->close();
        echo "Got the count: " . $count;
    }
    if ($count == '1') {
        echo "The Count is 1";
        if ($stmt = $mysqli -> prepare("UPDATE votes SET ip=?, status=? WHERE userid=? and businessid=? ")) {
            $stmt -> bind_param('ssss', $IP, $like_status, $userid, $businessid);
            $stmt -> execute();
            $stmt -> close();
            $mysqli -> close();
            echo "And the post is successful (1)";
        }
    }
    if ($count == '0') {
        echo "The Count is 0";
        if ($stmt = $mysqli -> prepare("INSERT INTO votes (userid, businessid, ip, status) VALUES ( ?, ?, ?, ? ) ")) {
            $stmt -> bind_param('ssss', $userid, $businessid, $IP, $like_status );
            $stmt -> execute();
            $stmt -> close();
            $mysqli -> close();
            echo "And the post is successful (0)";
        }
    }
    echo "finished";
?>

基本上 - 我已经添加了一堆回声，你将能够倾听＆＃39;在返回数据时使用$.post ajax调用。如果一切都成功，你应该得到一个句子，回复说＆＃34;得到计数：1TheCount是1并且帖子成功（1）完成＆＃34;。当帖子成功时，所有这些回声都应该开启。如果你没有得到其中一个句子，那就出问题了。

尝试此操作来解决问题，让我们知道会发生什么。

完整页面加载完成后，jQuery AJAX无法更新数据库

2 个答案: