error in syntax

Question

$sql="SELECT * FROM 'image_upload' where uid='$uid' ";

I have written this query and it is showing me 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 ''image_upload' where uid=''' at line 1

Can you please rectify it..

score 9 · Accepted Answer · edited Apr 15 '12 at 13:03

9

This will work:

$sql = "SELECT * FROM `image_upload` where uid='$uid' ";

edited Apr 15 '12 at 13:03

Tim Cooper

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answered Jun 16 '11 at 06:35

Manish Das

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yes, my uid is not char or varchar, it is bigint and when i put the query as suggested by you, it shows me uid 0 in my database – user799100 Jun 16 '11 at 07:24
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are you sure that $uid store the number you want??? or may be the previous insertions into uid column have size less than the length of your number. – Manish Das Jun 16 '11 at 07:54

score 4 · Answer 2 · answered Jun 16 '11 at 06:24

4

Use backticks for table names:

SELECT * FROM `image_upload` ...

answered Jun 16 '11 at 06:24

Pekka

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score 1 · Answer 3 · answered Jun 16 '11 at 06:26

You should be using backticks (`) rather than single quotes ('). In fact, you shouldn't be using either in this case since it's not required:

$sql = "SELECT * FROM image_upload where uid='$uid'";

The backticks are only required if your table name has funny characters in it that would otherwise annoy the SQL parser (like a space for example).

And make sure that your uid column is a textual one (like char or varchar) - otherwise you should not be surrounding $uid with the single quotes.

score 0 · Answer 4 · answered Jun 16 '11 at 06:24

0

$sql="SELECT * FROM image_upload where uid='$uid' ";

answered Jun 16 '11 at 06:24

gmhk

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score 0 · Answer 5 · answered Jun 16 '11 at 06:25

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Can you remove the single quotes, and try again?

SELECT * FROM image_upload where uid='$uid'

answered Jun 16 '11 at 06:25

Liangliang Zheng

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score 0 · Answer 6 · answered Jun 16 '11 at 06:25

0

try this:

$sql="SELECT * FROM image_upload where uid='".$uid."'";

answered Jun 16 '11 at 06:25

Sujit Agarwal

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score 0 · Answer 7 · edited May 23 '17 at 12:04

0

$sql="SELECT * FROM `image_upload` where uid='$uid' ";

You've been rectified ;)

You need to protect against SQL injections. Please see this thread.

edited May 23 '17 at 12:04

Community

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answered Jun 16 '11 at 06:26

Znarkus

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score -1 · Answer 8 · answered Jun 16 '11 at 06:26

-1

Remove single quotes in image_upload Before Query echo $uid; then u ll know the answer

answered Jun 16 '11 at 06:26

K6t

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error in syntax

8 Answers8