选择像A％一样没有给出结果但％A可以做到的地方

Question

My table is filled with a total of 40.000 rows with at least over 2.000 rows starting with the letter A .

Now my SELECT query does return some results when I use

$selectGames = mysql_query("SELECT * FROM Games WHERE GameName LIKE '%A'");

But when I use :

$selectGames = mysql_query("SELECT * FROM Games WHERE GameName LIKE 'A%'");

a total of 0 rows are being returned. Why is that?

Answer 1

%A means "where the end of the string is the letter A". A% is the opposite "a string that STARTS with the letter A"

       %A   A%
APPLE  N    Y
ABBA   Y    Y
GOUDA  Y    N

They're totally different matches, so your two queries are NOT equivalent.

As well, make sure that your table's collation is set to a case-insensitive match. If it's case-senstive, then a and A will be treated differently.

edit: as well, remember whitespace matters:

mysql> select 'gouda ' like '%a';
                    ^---note the space here
+--------------------+
| 'gouda ' like '%a' |
+--------------------+
|                  0 |
+--------------------+
1 row in set (0.00 sec)