php - 根据条件更改元素的显示

Question

我希望用户能够区分他们是否已投票给某事(通过加粗)或尚未投票(未加粗)。

例如:

       Voted For Post     Unvoted forVotes:  77 ↑ ↓             12 ↑ ↓

Here is how my database is set up:

intro

Contains messages

    message_id    intro            user_id      up_vote     10            Voted For Post   5            77     11            Unvoted for      5            12

voting

Contains votes

    Voting_id     message_id_fk      user_id     18             10               5        19             10               3     

users

Contains user names

    first_name    user_id     BOB           5

I don't know how to properly query the database then make the proper if statement that will distinguish between voted for and unvoted for posts.

Here's what I have so far:

$sql = mysql_query("SELECT * FROM intro
         INNER JOIN users ON intro.user_id = users.user_id
         ORDER BY `up` DESC ");

echo $row['first_name'] . " " .  $row['intro'];

if( ??? ) {
  echo "<strong>" . $row['up_vote'] . "</strong>";
} else {
  echo $row['up_vote'];
}

有什么想法吗？

Answer 1

左加入投票表，查看是否找到对应项:

SELECT intro.message_id, intro.intro, intro.user_id, intro.up_vote,
    IF(Voting.user_id IS NULL, 0, 1) AS has_voted
FROM intro
INNER JOIN users ON intro.user_id = users.user_id
LEFT JOIN Voting ON Voting.message_id_fk=intro.message_id
    AND Voting.user_id = 5
ORDER BY `up` DESC 

然后在 PHP 中:

if($row['has_voted'){
    echo "<strong>".$row['up_vote']."</strong>";
}else {
    echo $row['up_vote'];
}

一些解释:

• 如果未找到匹配的行，LEFT JOINed 表中的列为NULL
• IF() 是一个函数，如果第一个参数的计算结果为真，则返回第二个参数，否则返回第三个参数。作为一个函数，它可以很容易地在 SELECT 子句中使用
• 我将 SELECT * 替换为显式选择所需的列，这被认为是最佳实践，在这种情况下是必要的，因为列名不明确
• 当然，您必须将文字 5 替换为您当前的用户 ID。使用准备好的语句或像这样连接查询:"...Voting.user_id = "。整数($current_user_id)。 “……”