我正在将我的 zf1 应用程序移动到 zf2 并且我在移动我的映射器时遇到了很多问题。显然,我不能只是复制/过去我的查询(太糟糕了!)
我正在尝试CONCAT
在像这样的连接中使用
$resultSet = $this->select(function (Select $select) use($userId, $filter, $status){
$select->columns(array("id_project", "name", "current_to", "avancement"));
$prj = "PRJ_";
$select->join(array("ru" => "role_users"), "CONCAT( $prj , projects.id_project ) = ru.element", array())
->where("ru.user_id = $userId");
如果我打印$select->getSqlString()
,我得到:
SELECT "projects"."id_project" AS "id_project", "projects"."name" AS "name",
"projects"."current_to" AS "current_to", "projects"."avancement" AS "avancement" FROM
"projects" INNER JOIN "role_users" AS "ru" ON "CONCAT"( "PRJ_" "," "projects"."id_project"
) = "ru"."element" WHERE ru.user_id = 3
批准所有“”,这是正确的(如果我删除所有“”并在phpmyadmin中执行它,它可以工作。但是,如果zf2正在执行它,我会收到一个错误:
SQLSTATE[42000]: Syntax error or access violation: 1064 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 '`projects`.`id_project` ) = `ru`.`element`
WHERE ru.user_id = 3' at line 1
看起来有问题,CONCAT()
但我找不到问题所在。