Question

我正在使用php Phalcon框架。

使用phalcon执行查询：

$rawQuery = 
    'SELECT users.ID, users.Name, users.ProfilePictureUrl, users.Birthday, GetAge(users.Birthday) as Age, ' .
    '    users.Sex, users.LookingFor, commons.Name as CommonName, provinces.Name as ProvinceName, users.LastOnline FROM favorites ' .
    'INNER JOIN users ON favorites.FavoriteUserID = users.ID ' .
    'INNER JOIN commons ON commons.ID = users.CommonID ' .
    'INNER JOIN provinces ON provinces.ID = commons.ProvinceID ' .
    'WHERE favorites.OriginUserID = ' . $this->ID;
return $this->getReadConnection()->query($rawQuery)->fetchAll();

此查询正确返回所有字段，但CommonName除外，它应为字符串'Aglié'，但为null。样本输出：

[{"ID":"2","0":"2","Name":"Olga","1":"Olga","ProfilePictureUrl":"asd","2":"asd","Birthday":"2014-07-09","3":"2014-07-09","Sex":"f","4":"f","LookingFor":"girls","5":"girls","CommonName":null,"6":null,"ProvinceName":"Torino","7":"Torino","LastOnline":null,"8":null}]

另一方面，通过phpMyAdmin执行相同的查询：

SELECT users.ID, users.Name, users.ProfilePictureUrl, users.Birthday, GetAge(users.Birthday) as Age,
    users.Sex, users.LookingFor, commons.Name as CommonName, provinces.Name as ProvinceName, users.LastOnline FROM favorites
INNER JOIN users ON favorites.FavoriteUserID = users.ID 
INNER JOIN commons ON commons.ID = users.CommonID
INNER JOIN provinces ON provinces.ID = commons.ProvinceID
WHERE favorites.OriginUserID = 1

它无限期地挂起。但是，如果我取出Age字段，它将返回所有正确的字段，包括CommonName（有效字符串）。样本输出：

ID, Name, ProfilePictureUrl, Birthday, Sex, LookingFor, CommonName, ProvinceName, LastOnline
2, Olga, asd, 2014-07-09, f, girls, Agliè, Torino, NULL

GetAge（生日）定义如下：

return DATE_FORMAT(FROM_DAYS(DATEDIFF(curdate(),birthday)), '%Y')

您可以简单地假设只有4个表用户，收藏夹，公共区和省，只有查询中使用的那些字段。其他任何东西都没用。

这对我来说完全没有意义......请帮助！

Apache版本2.4.4

PHP版本5.4.12

MySql版本5.6.12

Phalcon版本1.3.2

Windows 7 64位操作系统

Answer 1

这是因为普通名字末尾的特殊字符'é'。显然json_encode会跳过该字段，如果它包含这样的字符。我通过直接在公共表中转义特殊字符来解决。

相同的连接查询：通过Phalcon和phpMyAdmin获得两个不同的结果

1 个答案: