如果数据库中不存在行，则显示链接，否则显示文本

Question

我的数据库中有一个名为flagged_posts 的表，它包含以下列：

id
thought_id
flagged_by_id

我想要做的是，如果登录用户已经标记了帖子，那么不要让他们再次标记帖子，我试图通过删除锚链接并将其替换为一条消息。

这是我的代码的 sn-p：

<?php

$query = mysqli_query($connect, "SELECT * FROM user_thoughts WHERE added_by='$user' AND shared ='yes' "."ORDER BY id DESC LIMIT {$start}, {$limit}");

while ($row = mysqli_fetch_array($query)) {
    $thought_id      = $row['id'];
    $message_content = $row['message'];
    $date_of_msg     = $row['post_details'];
    $thoughts_by     = $row['added_by'];
    $attachent       = $row['attachment'];
    $shared          = $row['shared'];

    // getting the id of the user who is logged in.
    $see_if_flagged_q = mysqli_query($connect, "SELECT id FROM users WHERE username = '$username'");
        $getting_deets = mysqli_fetch_assoc ($see_if_flagged_q);
            $logged_in_user_id = $getting_deets ['id'];

    echo "
        <div class='more_options' style='float: right;'>"; 
            $see_if_flagged_q2 = mysqli_query($connect, "SELECT * FROM flagged_posts WHERE flagged_by_id ='$logged_in_user_id' ");
                while ($getting_deets2 = mysqli_fetch_assoc ($see_if_flagged_q2)){
                        $flagged_post_by_id = $getting_deets2 ['flagged_by_id']; 

                    // If the user logged in has not flagged the post, i.e. there is no data in the database ..
                    // .. which says their user id has flagged this thought_id.. then display the link...       
                    if ($logged_in_user_id == $flagged_post_by_id){
                            echo "<a href='/inc/flagged_post.php?id=$thought_id'> Flag </a>";
                    } 
                    // if there is data stating this user has flagged this thought_id, then echo a message
                    if ($logged_in_user_id != $flagged_post_by_id) {
                            echo "Flagged";
                    }

                }
    echo "  </div>";                
}
?>

所以假设我以Conor 登录。 Conor 的 id 为 8（从 users 表获得的 ID）。 Conor 标记一个 id 为 209 的帖子（thought_id 来自user_thoughts 表）。所以在我的flagged posts 表中，我将看到以下行：

id: 1
thought_id: 209
flagged_by_id: 8

目前，链接和消息都没有出现。如果我更改我的查询，即$see_if_flagged_q2 = mysqli_query($connect, "SELECT * FROM flagged_posts ");（删除了 WHERE 子句），那么我会收到四次回显消息Flagged（因为flagged_posts 表中有四行，它们在每个帖子上都是回显，甚至那些没有被登录用户标记的。

更新：

首先是更新后的代码：

            $see_if_flagged_q2 = mysqli_query($connect, "SELECT * FROM flagged_posts WHERE flagged_by_id = '$logged_in_user_id'");
            $test_num = mysqli_num_rows ($see_if_flagged_q2);
                $getting_deets2 = mysqli_fetch_assoc ($see_if_flagged_q2);
                        $flagged_post_by_id = $getting_deets2['flagged_by_id'];

                        if ($flagged_post_by_id == $logged_in_user_id){
                            echo "<a href='/inc/flagged_post.php?id=$thought_id'> Flag </a>";
                            echo $test_num;
                        } 
                        if ($flagged_post_by_id != $logged_in_user_id) {
                            echo "Flagged";
                        }

有了以上内容，现在所有帖子的链接都会出现，即使它们被标记。我已经回显了 $flagged_post_by_id 和 '$logged_in_user_id'，它们都回显了 12 的值（users 表中的 Conor 的 ID）。值是正确的，$test_num 返回的行数也是正确的。

Answer 1

好的，这是对原始代码的修改。我将数据收集部分移到了前面，因此在对想法运行 while 循环之前，我们有一个设置部分。我在这里和那里更改了一个变量名。基本上，我们建立一个标记条目列表，然后在 while 循环中，工作更简单。如果当前行 id 在 flagged_posts 数组中，它被标记，否则显示链接。

// get the id of the current user
$user_id_q = mysqli_query($connect, "SELECT id FROM users WHERE username = '$username'");
$getting_deets = mysqli_fetch_assoc($user_id_q);
$logged_in_user_id = $getting_deets['id'];

// build array of posts flagged by current user
$flagged_posts_q = mysqli_query($connect, "SELECT thought_id FROM flagged_posts WHERE flagged_by_id = '$logged_in_user_id'");
$flagged_posts = array();
while ($row = mysqli_fetch_array($flagged_posts_q)) {
    $flagged_posts[] = $row['thought_id'];
}

$query = mysqli_query($connect, "SELECT * FROM user_thoughts WHERE added_by='$user' AND shared ='yes' "."ORDER BY id DESC LIMIT {$start}, {$limit}");

while ($row = mysqli_fetch_array($query)) {
    //You could just use $row['foo'] down below, and skip all this
    /* 
    $thought_id      = $row['id'];
    $message_content = $row['message'];
    $date_of_msg     = $row['post_details'];
    $thoughts_by     = $row['added_by'];
    $attachent       = $row['attachment'];
    $shared          = $row['shared'];
    */

    echo "<div class='more_options' style='float: right;'>"; 
        if (in_array($row['id'], $flagged_posts)){ 
            echo "Flagged";
        } else {
            echo "<a href='/inc/flagged_post.php?id=".$row['id']."'> Flag </a>";
        }
    echo "</div>";                
}