| Bug #6317 | string function CHAR, parameter is NULL, wrong result | ||
|---|---|---|---|
| Submitted: | 29 Oct 2004 9:44 | Modified: | 23 Feb 2005 18:17 |
| Reporter: | Matthias Leich | Email Updates: | |
| Status: | Closed | Impact on me: | |
| Category: | MySQL Server | Severity: | S3 (Non-critical) |
| Version: | 4.1 | OS: | |
| Assigned to: | Sergei Glukhov | CPU Architecture: | Any |
[29 Oct 2004 9:49]
Matthias Leich
The position (first,middle,last) of the function parameter set to NULL does not affect the bug.
[1 Nov 2004 9:28]
Sergei Glukhov
ChangeSet 1.2070 04/11/01 Fix for bug #6317: string function CHAR, parameter is NULL, wrong result
[23 Feb 2005 15:33]
Sergei Glukhov
Fixed in 4.1.11
[23 Feb 2005 18:17]
Paul DuBois
Noted in 4.1.11, 5.0.3 changelogs.
Description: The MySQL manual states : CHAR(N,...) CHAR() interprets the arguments as integers and returns a string consisting of the characters given by the code values of those integers. NULL values are skipped. In my example the NULL values are not skipped and a '^@' is inserted. # Example from the manual for comparison SELECT CHAR(77,121,83,81,'76') as my_column; my_column MySQL # now the NULL must be skipped and we must get 'ySQL' SELECT CHAR(NULL,121,83,81,'76') as my_column; my_column ^@ySQL # skipping one character means we must get a length of 4 SELECT CHAR_LENGTH(CHAR(NULL,121,83,81,'76')) as my_column; my_column 5 My environment: - Intel PC with Linux(SuSE 9.1) - MySQL 4.1 compiled from source last ChangeSet@1.2061, 2004-10-28 How to repeat: Please execute the statements above