Question

我正在开发一个Xamarin.Forms社交媒体应用。我试图抓住与用户联系的帖子，这样我就可以显示海报的用户名。

我使用以下方式抓住帖子：

$sql = "SELECT * FROM posts AS P INNER JOIN users AS U ON U.id = P.user_id WHERE P.user_id=:userid";
$results = DB::query($sql, array(':userid' => $target_id));

使用print_r将其显示为：

阵（ [0] =＆gt;排列（ [id] =＆gt; 剪断 [0] =＆gt; 剪断 [user_id] =＆gt; 剪断 [1] =＆gt; 剪断 [body] =＆gt; 56565 [2] =＆gt; 56565 [data] =＆gt; 0 [3] =＆gt; 0 [timestamp] =＆gt; 2018-03-22 00:00:00 [4] =＆gt; 2018-03-22 00:00:00 [type] =＆gt; 1 [5] =＆gt; 0 [评论] =＆gt; 0 [6] =＆gt; 0 [users] =＆gt; [7] =＆gt; [8] =＆gt; 剪断 [username] =＆gt; username_test [9] =＆gt; username_test [密码] =＆gt; 剪断 [10] =＆gt; 剪断 [user_password] =＆gt; [11] =＆gt; [email] =＆gt; 剪断 [12] =＆gt; 剪断 [name] =＆gt;乔 [13] =＆gt;乔 [bio] =＆gt; Hello World Test Print [14] =＆gt; Hello World Test Print [birthday] =＆gt; 2001-05-30 00:00:00 [15] =＆gt; 2001-05-30 00:00:00 [性别] =＆gt; 0 [16] =＆gt; 0 [created] =＆gt; 2018-03-25 20:29:44 [17] =＆gt; 2018-03-25 20:29:44））

在添加连接之前，我使用类似下面的内容循环结果：

foreach($results as $p) { 
    $username = $p[username']
}

但是现在我不明白如何从用户或帖子中获取内容，例如如果我需要获取用户ID，我如何指定用户的“id”而不是来自帖子的“id”并且不小心获得帖子ID

我试过像$ p [users =＆gt;用户名]，但不起作用

Answer 1

不确定这是否适合您，但由于您只对userId感兴趣，为什么不选择您想要的？然后$ result数组每个记录集只有一个条目。

<!DOCTYPE html>
<html>
    <head>
        <title> My Life</title>
        <link rel="stylesheet" type="text/css" href="MyLife.css">
        <link href="https://fonts.googleapis.com/css?family=Contrail+One" rel="stylesheet">

        <style type="text/css">
         
            ul {
                list-style-type: none;
                margin: 0;
                padding: 0;
                overflow: hidden;
                background-color: #333;
            }

            li {
                float: left;
            }

            li a {
                display: block;
                color: white;
                text-align: center;
                padding: 14px 16px;
                text-decoration: none;
            }

            /* Change the link color to #111 (black) on hover */
            li a:hover {
                background-color: #111;
            }

            h1{
                width:20%
                text-align:center;
            }

            h2{
                text-align: center;
            }

            body{
                background-color: white;
            }
            img{
                width: 30%;
                float: left;


                margin: 1.66%;
            }

            p{
                margin-left: 1.66%;
                font-family: 'Contrail One', cursive;
                font-size: 35px;
                text-transform: uppercase;
                border-bottom: 2px solid black;
                width: 30%;
                padding-bottom: 20px;

            } 

            div.a{
                text-align: center;
            }
            
        </style>
    </head>
    <body>

        <!-- 
        This is the code for the Home Page -->

        <div class ="a">
            <h1> My Name is John </h1>
        </div>
        <h2> Welcome to my Website!</h2>


        <ul>
            <li><a href="">Home</a></li>
            <li><a href="MyHistory.html">My History</a></li>
            <li><a href="MyEducation.html">My Education</a></li>
            <li><a href="MyTravels.html">My Travels</a></li>
            <li><a href="">My Form Page</a></li>
            <li><a href="">What I Like to Do</a></li>
        </ul>
<ul>                <img src = "http://c1.staticflickr.com/9/8450/8026519634_f33f3724ea_b.jpg">
            <img src = "http://c2.staticflickr.com/8/7218/7209301894_c99d3a33c2_h.jpg ">
            <img src = "http://c2.staticflickr.com/8/7231/6947093326_df216540ff_b.jpg ">
            <img src = "http://c1.staticflickr.com/9/8788/17367410309_78abb9e5b6_b.jpg ">
            <img src = "http://c2.staticflickr.com/6/5814/20700286354_762c19bd3b_b.jpg ">
            <img src = "http://c2.staticflickr.com/6/5647/21137202535_404bf25729_b.jpg ">
            <img src = "http://c2.staticflickr.com/6/5588/14991687545_5c8e1a2e86_b.jpg">
            <img src = "http://c2.staticflickr.com/4/3888/14878097108_5997041006_b.jpg ">
            <img src = "http://c2.staticflickr.com/8/7579/15482110477_0b0e9e5421_b.jpg">
</ul>
    </body>
</html>

Answer 2

试试这个：

SELECT * FROM posts
INNER JOIN users
ON users.id = posts.user_id
WHERE posts.user_id=:userid

您的foreach语句缺少引号并以分号结束：

foreach($results as $p) { 
    $username = $p['username'];
}

获取明确的ID：

SELECT *, users.id AS uid, posts.id AS postId
FROM posts
INNER JOIN users ON users.id = posts.user_id
WHERE posts.user_id=:userid

要在同一查询中获取相似的数量：

SELECT *, users.id AS uid, posts.id AS postId,
COUNT(L.likes) AS likeCount 
FROM posts
INNER JOIN users ON users.id = posts.user_id
LEFT JOIN like AS L ON L.post_id = P.id
WHERE posts.user_id=:userid

从MySQL查询中获取值

2 个答案: