Question

mysql> SELECT LOCATE("n", "München") COLLATE utf8_general_ci;
ERROR 1253 (42000): COLLATION 'utf8_general_ci' is not valid for CHARACTER SET 'binary'

如何摆脱此错误？

我已经尝试过（复制和粘贴）：

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mysql> SELECT LOCATE("n", "München") COLLATE utf8_general_ci;
ERROR 1253 (42000): COLLATION 'utf8_general_ci' is not valid for CHARACTER SET 'binary'
mysql> SET NAMES utf8;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT LOCATE("n", "München") COLLATE utf8_general_ci;
ERROR 1253 (42000): COLLATION 'utf8_general_ci' is not valid for CHARACTER SET 'binary'
mysql> SELECT LOCATE(_utf8"n", _utf8"München") COLLATE utf8_general_ci;
ERROR 1253 (42000): COLLATION 'utf8_general_ci' is not valid for CHARACTER SET 'binary'
mysql> SHOW VARIABLES LIKE "character_set_database";
+------------------------+-------+
| Variable_name          | Value |
+------------------------+-------+
| character_set_database | utf8  |
+------------------------+-------+
1 row in set (0.00 sec)

Answer 1

可能服务器已使用默认字符集二进制进行编译，因此字符串文字正在被解释为，或者客户端设置为在与服务器通信时使用二进制模式。您可以通过调用SET NAMES utf8来更改客户端和连接字符集（尽管如果您的SQL语句是从PHP发出的，则不建议这样做，因为PHP将有自己的用于设置连接字符集的命令）。请参阅MySQL参考手册中的Connection Character Sets and Collations。

或者，您可以使用“介绍人”明确指定用于 LOCATE 函数中字符串文字的字符集，例如：

LOCATE(_utf8"n", _utf8"München")

有关详细信息，请参阅参考手册页Character String Literal Character Set and Collation。

Answer 2

我的示例sets the collation of the return value中的COLLATE LOCATE，其结果类型为binary。

设置参数的排序规则：

mysql> SELECT LOCATE(_utf8"n" COLLATE utf8_general_ci,
                     _utf8"München" COLLATE utf8_general_ci) AS locate;
+--------+
| locate |
+--------+
|      3 |
+--------+
1 row in set (0.00 sec)

我的动机实际上是找出MySQL是否进行整理在搜索子字符串时考虑到。不幸的是不的。查看第二个命令的结果：

mysql> SELECT LOCATE(_utf8"ü" COLLATE utf8_general_ci,
                     _utf8"München" COLLATE utf8_general_ci) AS locate;
+--------+
| locate |
+--------+
|      2 |
+--------+
1 row in set (0.00 sec)

mysql> SELECT LOCATE(_utf8"u" COLLATE utf8_general_ci,
                     _utf8"München" COLLATE utf8_general_ci) AS locate;
+--------+
| locate |
+--------+
|      0 |
+--------+
1 row in set (0.00 sec)

使用临时表进行测试（在WHERE子句中考虑整理，但不在。{ LOCATE）：

mysql> CREATE TEMPORARY TABLE test
       (text VARCHAR(100) CHARACTER SET utf8 COLLATE utf8_general_ci);
Query OK, 0 rows affected (0.00 sec)

mysql> INSERT INTO test VALUES("München");
Query OK, 1 row affected (0.00 sec)

mysql> SELECT text FROM test WHERE text LIKE "%u%";
+---------+
| text    |
+---------+
| München |
+---------+
1 row in set (0.00 sec)

mysql> SELECT LOCATE("u", text) AS locate FROM test WHERE text LIKE "%u%"; 
+--------+
| locate |
+--------+
|      0 |
+--------+
1 row in set (0.01 sec)

Answer 3

我知道这已经很晚了，但我希望它有所帮助。我一直得到同样的错误，我知道我的字符集和校对很好。

检查＆＃39; @＆＃39;您声明中不属于的符号。我正在测试我的存储过程作为带变量的select语句，然后在创建存储过程时忘记删除＆＃39; @＆＃39;符号。不用说，我觉得很傻。

我也知道这个问题在这个问题上似乎并非如此，但这是我的第一篇SO帖子，而且我没有足够的代表去做其他事情，所以我道歉。

COLLATION'utf8_general_ci'对CHARACTER SET'二进制'无效？

3 个答案: