Question

查询以显示用户发布的帖子。

$sql = "SELECT B.title, B.article, B.article_id, B.posted_by, B.users_id,DATE_FORMAT(updated, '%m-%d-%Y %l:%i:%s') AS updated , U.user_pic_path, U.user_id, U.username, I.filename
        FROM users U 
        INNER JOIN blog B
        ON B.posted_by = U.user_id
        LEFT OUTER JOIN images AS I 
        ON B.image_id = I.image_id
        WHERE B.users_id = ?
        AND B.posted_by = ?
        ORDER BY updated  DESC"

查询选择我朋友的用户。

$sql = "SELECT F.status,U.user_id, U.username, U.email, U.user_pic_path
      FROM users U, friends F
      WHERE
      CASE
      WHEN F.friend_one =? 
      THEN F.friend_two = U.user_id  
      WHEN F.friend_two = ? 
      THEN F.friend_one= U.user_id 
      END 
      AND F.status = 2";

所以我想展示作为我的朋友的用户所做的所有博客文章。要做到这一点，我必须加入两个加入两个查询。我被卡住了，不知道我怎么能这样做。

Answer 1

您可以使用POST https://www.googleapis.com/gmail/v1/users/arpan%40intricare.net/messages/send?fields=raw&key={YOUR_API_KEY} { "raw": "UkNQVCBUTzogYWN0dWFsX3RvIDx0b0ByZWNpcGllbnQClRsuY29tPgpGcm9tOiBTZW5kZXIgPHNl bmRlckBzZW5kZXIuY29tPiAKVG86IER1bW15X3RvIDxkdW1teUBkdW1teS5jb20-IApTdWJqZWN0 OiBTYXlpbmcgSGVsbG8gCiAgICAKVGhpcyBpcyBhIG1lc3NhZ2UganVzdCB0byBzYXkgaGVsbG8u IFNvLCAiSGVsbG8iLg==" }：

执行此操作

帖子查询WHERE id_users IN（朋友查询）。

这样的事情：

conn.request("GET",path,headers=headers)
res = conn.getresponse()
if res.status == 500:
    print "Shell Shock Exploitable "

MYSQL和PHP加入2个查询

1 个答案: