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2 回答 2

2

那不应该是

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)
于 2012-07-07T05:05:45.920 回答
2

在对您的案例进行测试后，问题'Z'是$db->query

这是因为当您的语句（从调试器看来，引号变为）时，MysqliDB内部使用mysqli并mysqli引发了异常。prepare''

相反，请MysqliDB提供rawQuery像您这样的案例。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'));

参考：PHP-MySQLi-Database-Class / MysqliDb.php；

于 2012-07-14T02:00:02.177 回答