ID: 12829
Updated by: sniper
Reported By: [EMAIL PROTECTED]
Old Status: Open
Status: Bogus
Bug Type: Scripting Engine problem
Operating System: Linux
PHP Version: 4.0.6
New Comment:
This works: ($origins+1)
Not a bug.
Previous Comments:
------------------------------------------------------------------------
[2001-08-18 18:32:56] [EMAIL PROTECTED]
1 $sq = "SELECT * FROM TLinks WHERE LinkID = 1";
2 $result = mysql_query($sq)
3 or die ("Error- Could not run Query: $sq!\nError (" .
mysql_errno() . ") - " . mysql_error());
4
5 if ($row = mysql_fetch_array($result))
6 {
7 $link = $row['Link'];
8 $origins = $row["Origins"];
9
10 $q = "UPDATE TLinks SET Origins = " . $origins + 1 . " WHERE LinkID =
1";
11 $r3 = mysql_query($q)
12 or die ("Error- Could not run Query: $q!\nError (" .
mysql_errno() . ") - " . mysql_error());
13
14 }
The above lines cause the following error:
Error- Could not run Query: 1 WHERE LinkID = '01a_gennote'! Error (1064) - You have an
error in your SQL syntax near '1 WHERE LinkID = '01a_gennote'' at line 1
Please note that the error happends with reference to the SELECT query on line 2.
SOLUTION
To solve the problem, I had to remove '+ 1' following the '$origins' on line 10. The
following code executes without any errors.
1 $sq = "SELECT * FROM TLinks WHERE LinkID = 1";
2 $result = mysql_query($sq)
3 or die ("Error- Could not run Query: $sq!\nError (" .
mysql_errno() . ") - " . mysql_error());
4
5 if ($row = mysql_fetch_array($result))
6 {
7 $link = $row['Link'];
8 $origins = $row["Origins"];
9
10 $q = "UPDATE TLinks SET Origins = " . $origins . " WHERE LinkID = 1";
11 $r3 = mysql_query($q)
12 or die ("Error- Could not run Query: $q!\nError (" .
mysql_errno() . ") - " . mysql_error());
13
14 }
------------------------------------------------------------------------
Edit this bug report at http://bugs.php.net/?id=12829&edit=1
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