如何使用 PHP 从多个表中选择数据并显示它?

问题描述：

我正在尝试制作一个私人消息应用程序.我需要收件箱同时显示主要消息"和消息答复".

I'm trying to make a Private Message Application. And I need the Inbox to show both "main messages" and "answers on messages".

我的表如下:

发布

id - 标题 - 内容 - from_user - to_user -receiver_opened - user_bywho - 时间戳.

id - header - content - from_user - to_user - receiver_opened - user_bywho - timestamp.

post_answer

id - answer_from_user - answer_to_user - answer_user_bywho - answer_header - answer_content - answer_id - timestamp_svar.

现在我遇到了麻烦.因为我不知道如何使这成为可能.

And now I'm having troubles. Because I don't know how to make this possible.

我的 mysql_query 现在看起来像这样:

My mysql_query looks like this right now:

$user = $_SESSION['username'];    

$sql = mysql_query("SELECT * FROM post,post_answer WHERE ".
                  "answer_to_user='$user' AND to_user='$user'")
       or die(mysql_error());

提前致谢.

答

该查询无效，因为它对于 id 字段会产生歧义.

The query is not valid as it will be ambiguous about the id field.

为此，您可以使用 JOIN

SELECT * 
    FROM post 
    INNER JOIN post_answer
    WHERE post.to_user = post_answer.answer_to_user

愚蠢的方式(易于理解)

SELECT post.id, 
       header, 
       content, 
       from_user, 
       to_user, 
       receiver_opened, 
       user_bywho, timestamp,
       post_content.id, 
       answer_from_user, 
       answer_to_user, 
       answer_user_bywho, 
       answer_header, 
       answer_content, 
       answer_id, 
       timestamp_svar
    FROM post, 
         post_content
    WHERE answer_to_user=$user AND 
          to_user=$user

希望能帮到你

假设您的表格包含以下字段:

Lets suppose your table contains fields like-

名字
姓氏
数量

看看你是否在数组中得到结果，比如 $results.

See if you get the result in an array say $results.

foreach ($results as $result)
{
    echo $result->firstname;
    echo $result->lastname;
    echo $result->number;
}

如何使用 PHP 从多个表中选择数据并显示它?

愚蠢的方式(易于理解)

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