查詢不適用於 Mysqli php 類，但適用於 MySQL 工作台

Question

我正在使用在github 上找到的 mysqli 數據庫類

我的查詢看起來像

$db->query('
  SELECT
    memberID,zoneCode,state,zone,countyName,
    CONCAT(state,\'Z\',zone) as fullZoneCode
  FROM members_zonesToWatch
  LEFT JOIN (
    SELECT state,zone,countyName
    FROM countyPublicForcastZoneCorrelation
  ) as zones
    ON zoneCode = CONCAT(state,\'Z\',zone)');

我得到的問題是

Fatal error: Problem preparing query 
(SELECT
  memberID,zoneCode,state,zone,countyName,
   CONCAT(state,'Z',zone) as fullZoneCode
 FROM members_zonesToWatch
 LEFT JOIN (
   SELECT state,zone,countyName
   FROM countyPublicForcastZoneCorrelation
 ) as zones
   ON zoneCode = CONCAT(state,'Z',zone)) 
You have an error in your SQL syntax; check the manual that corresponds to your
MySQL server version for the right syntax to use near
''Z',zone) as fullZoneCode FROM members_zonesToWatch LEFT' at line 1

這個查詢在 mysql workbench 中工作得很好。 使用 mysqli 類有什么問題？

Answer 1

不應該是

SELECT
m.memberID, m.zoneCode, m.state, m.zone, m.countyName,
CONCAT(m.state,'Z',m.zone) as fullZoneCode,
c.state, c.zone, c.countyName
FROM members_zonesToWatch m
LEFT JOIN countyPublicForcastZoneCorrelation c
ON c.zoneCode = CONCAT(m.state, 'Z', m.zone) 

Answer 2

在對您的案例進行測試后，問題是'Z'和$db->query

這是因為MysqliDB內部使用mysqli並且mysqli在prepare語句時拋出異常（從調試器看來，引號'變成了' ）。

相反， MysqliDB為像您這樣的案例提供rawQuery 。 所以使用下面的：

$result = $db->rawQuery('
    SELECT m.memberID, CONCAT(m.state, ? ,m.zone) as fullZoneCode
    FROM members_zonesToWatch m
    LEFT JOIN countyPublicForcastZoneCorrelation c
    ON c.zoneCode = CONCAT(m.state, ? ,m.zone)',
    array('Z','Z'));

參考資料： PHP-MySQLi-Database-Class / MysqliDb.php ；