php mysql左连接语法错误答案

【问题标题】：php mysql left join syntax errorphp mysql左连接语法错误
【发布时间】：2012-09-03 13:55:28
【问题描述】：

你好，我有一些错误，我看不到它，但是访问网站的成员，告诉我那里有错误

$query = mysql_query("
SELECT

m.member_id as member_id,
m.member_group_id as member_group_id,
m.members_display_name as members_display_name,
m.email as email,
m.joined as joined, 
m.member_login_key as member_login_key,
m.msg_count_new as msg_count_new,

p.pp_thumb_photo as pp_thumb_photo,
p.pp_photo_type as pp_photo_type

FROM        ".$forum_prefix."members m
LEFT JOIN   ".$forum_prefix."profile_portal p ON p.pp_member_id = m.member_id

WHERE m.member_id = ".$id."
") or die(mysql_error());

和错误

 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 17

【问题讨论】：

$forum_prefix 和 $id 的值是什么导致此错误？
你的字符串连接的结果是什么？你不能把它写在调试文件或其他东西中吗？我高度怀疑你有WHERE m.member_id = 或类似的问题...
$forum_prefix 的可能值是多少？
$forum_prefix 和 $id 工作正常，第 17 行在 m.msg_count_new 作为 msg_count_new，
请输入the last query received by your mysql

标签： php mysql syntax left-join

【解决方案1】：

line 17 是查询的最后一行。
这一行中唯一的东西是
```
WHERE m.member_id = ".$id."
```
消息错误是““（空字符串）。这是“截断”查询的症状。

因此，错误出现在您的 php 代码中，因为未设置 $id var。

【讨论】：

【解决方案2】：

$forum_prefix = "table_name";// put table name here.its empty in your code
$id = 2;//put your id here and check
$query = mysql_query("
SELECT

m.member_id as member_id,
m.member_group_id as member_group_id,
m.members_display_name as members_display_name,
m.email as email,
m.joined as joined, 
m.member_login_key as member_login_key,
m.msg_count_new as msg_count_new,

p.pp_thumb_photo as pp_thumb_photo,
p.pp_photo_type as pp_photo_type

FROM        ".$forum_prefix."members m
LEFT JOIN   ".$forum_prefix."profile_portal p ON p.pp_member_id = m.member_id

WHERE m.member_id = ".$id."
") or die(mysql_error());

【讨论】：

别这么想，这不是问题
您在 table_name 后忘记了 .。这将评估为 FROM table_namemembers m 和 LEFT JOIN table_nameprofile_portal p