php mysql left join syntax error

Question

hello i have some error , i can't see it but members who visit site , tell me there ir an error

$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());

and error

 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

What are the values of `$forum_prefix` and `$id` that give rise to this error? — eggyal, Sep 03 '12 at 13:56
What does your string concatenated evaluates to ? Can't you write that in a debug file or something ? I highly suspect you have `WHERE m.member_id = ` or a similar problem... — Eregrith, Sep 03 '12 at 13:58
$forum_prefix and $id work's fine , 17 line are at m.msg_count_new as msg_count_new, — fees, Sep 03 '12 at 13:59
Please put [the last query received by your mysql](http://stackoverflow.com/questions/650238/how-to-show-the-last-queries-executed-on-mysql) — Eregrith, Sep 03 '12 at 14:02
try single quotes for ".$forum_prefix." , ".$forum_prefix." and ".$id." — Wearybands, Sep 03 '12 at 14:02
"near '' at line 17": The 17 line is the last line: `WHERE m.member_id = ".$id.""`. So I guess $id is empty. — valex, Sep 03 '12 at 14:09

open source guy · Answer 1 · 2012-09-03T14:17:50.400

0

$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());

edited Sep 03 '12 at 14:17

answered Sep 03 '12 at 14:00

open source guy

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don,t think so , that is not a problem – M Khalid Junaid Sep 03 '12 at 14:03
You forgot a `.` after table_name. This will evaluate to `FROM table_namemembers m` and `LEFT JOIN table_nameprofile_portal p` – Eregrith Sep 03 '12 at 14:03

score 0 · Answer 2 · answered Sep 03 '12 at 14:15

The line 17 is the last line of your query.
The only thing in this line is
```
WHERE m.member_id = ".$id."
```
The message error is ““ (empty string). This is a symptom of a "truncated" query.

So, the error is in your php code because the $id var is not set.

php mysql left join syntax error

2 Answers2