MySQL:utf8字符串上的奇怪LENGTH()行为

Question

我正在对请求生成器进行单元测试,而且我遇到了LENGTH函数问题.

我有2个请求相互跟随:

SHOW VARIABLES LIKE '%character%'

返回以下结果:

array(8) {
  [0] =>
  array(2) {
    'Variable_name' =>
    string(20) "character_set_client"
    'Value' =>
    string(4) "utf8"
  }
  [1] =>
  array(2) {
    'Variable_name' =>
    string(24) "character_set_connection"
    'Value' =>
    string(4) "utf8"
  }
  [2] =>
  array(2) {
    'Variable_name' =>
    string(22) "character_set_database"
    'Value' =>
    string(6) "latin1"
  }
  [3] =>
  array(2) {
    'Variable_name' =>
    string(24) "character_set_filesystem"
    'Value' =>
    string(6) "binary"
  }
  [4] =>
  array(2) {
    'Variable_name' =>
    string(21) "character_set_results"
    'Value' =>
    string(4) "utf8"
  }
  [5] =>
  array(2) {
    'Variable_name' =>
    string(20) "character_set_server"
    'Value' =>
    string(4) "utf8"
  }
  [6] =>
  array(2) {
    'Variable_name' =>
    string(20) "character_set_system"
    'Value' =>
    string(4) "utf8"
  }
  [7] =>
  array(2) {
    'Variable_name' =>
    string(18) "character_sets_dir"
    'Value' =>
    string(26) "/usr/share/mysql/charsets/"
  }
}

我的第二个要求是:

SELECT LENGTH('??') as len

它返回6而不是2.

这有什么不对？我的charset参数看起来不错.

Answer 1

我在MySQL文档中找到了答案:

该LENGTH函数计算字节数:

mysql> SELECT LENGTH('??') ;
+------------------+
| LENGTH('??')   |
+------------------+
|                6 |
+------------------+
1 row in set (0.00 sec)

该CHAR_LENGTH函数计算字符数:

mysql> SELECT CHAR_LENGTH('??') ;
+-----------------------+
| CHAR_LENGTH('??')   |
+-----------------------+
|                     2 |
+-----------------------+
1 row in set (0.00 sec)