Question

我已经使用了2天，现在转向stackoverflow寻求帮助。我试图为用户的每一篇文章返回display_name以查找“故事”，其中故事（SID）可以有多个用户（也就是多个写作ID，这里称为WID）。

除了每一篇文章的显示名称始终相同外，一切都有效。当我为$ getauthor做一个var_dump（）时，我简单地得到wp_users中所有display_names的数组。但我需要的是只返回在写作和故事表中写入该故事（SID）的写作表中写一篇文章（WID）的作者的显示名称。

更清晰一点：当有人启动博客帖子（称为故事）时，会在该故事中为具有新SID的用户ID创建记录。然后，当人们添加到该SID时，每个添加都在写入表中给出具有相同SID的WID记录。我的代码正在做的是返回所有结果（写作文件）的SID的ORIGINAL作者，实际上应该发生的是每个WID应该有添加该部分的作者。

应该看起来像：

Author: Kate
This is my blog
Author: Mike
This is my addition.

但它看起来像是：

Author: Kate 
This is my blog
Author: Kate
This is my addition.

这是数据库架构：

wp_users表：

ID
display_name

故事表：

SID
ID
story_name

写作表：

WID
ID 
SID
approved
text

代码：

<?php
global $wpdb;
$user_ID = get_current_user_id();
// get the SID
$the_SID = ( isset( $_GET['writing'] ) ) ? $_GET['writing'] : false;

$results = $wpdb->get_results("
SELECT wp_users.ID, wp_users.display_name,
stories.ID, stories.SID, stories.story_name, stories.category,
writing.ID, writing.text, writing.approved
FROM stories
JOIN wp_users ON stories.ID = wp_users.ID
JOIN writing ON stories.SID = writing.SID
WHERE (stories.SID = $the_SID) AND (writing.approved = 'Y')");

$sql = "SELECT display_name FROM wp_users where ID = ". $results->writing.ID .";";  
$getauthor = $wpdb->get_results($sql);
$author = $getauthor[0]->display_name;
?>

<?php 
foreach ($results as $result) {
echo "<br>" . "Author: " . $author . "<br>" . $results[0]->text . "<br>";
}
?>

Answer 1

这是您的查询：

SELECT wp_users.ID, wp_users.display_name,
----------------^
       stories.ID, stories.SID, stories.story_name, stories.category,
---------------^
       writing.ID, writing.text, writing.approved
---------------^
FROM stories JOIN
     wp_users
     ON stories.ID = wp_users.ID JOIN
     writing
     ON stories.SID = writing.SID
WHERE (stories.SID = $the_SID) AND (writing.approved = 'Y');

标识的列都具有相同的名称ID。您应该为它们定义单独的别名：

SELECT wp_users.ID as userID, wp_users.display_name,
       stories.ID as storiesID, stories.SID, stories.story_name, stories.category,
       writing.ID as writingID, writing.text, writing.approved
FROM stories JOIN
     wp_users
     ON stories.ID = wp_users.ID JOIN
     writing
     ON stories.SID = writing.SID
WHERE (stories.SID = $the_SID) AND (writing.approved = 'Y');

那么也许这会奏效：

$sql = "SELECT display_name FROM wp_users where ID = ". $results->writingID .";";

但是，只需再次加入wp_users，即可在同一查询中获得所需内容：

SELECT u.ID as userID, u.display_name,
       s.ID as storiesID, s.SID, s.story_name, s.category,
       w.ID as writingID, w.text, w.approved,
       uw.display_name as writer_name
FROM stories s JOIN
     wp_users u
     ON s.ID = u.ID JOIN
     writing w
     ON s.SID = w.SID JOIN
     wp_users uw
     ON w.ID = uw.ID
WHERE (s.SID = $the_SID) AND (w.approved = 'Y');

无法获取查询以引用第三个表中的用户名

1 个答案: