【发布时间】:2017-07-01 19:19:35
【问题描述】:
这是我目前的 QB。
$institutes = count($request->input("selected_institutes"));
$courses = count($request->input("selected_courses"));
$institutes_data = $request->input("selected_institutes");
$courses_data = $request->input("selected_courses");
$search=DB::table('users as u');
if($institutes > 0 && $courses = 0){
$search=$search->select('u.id','u.first_name','u.profile_picture','inst.full_name')
->leftJoin('user_institutes as ins','ins.user_id','=','u.id')
->leftJoin('institutes as inst','inst.id','=','ins.institute_id');
}
else if($courses > 0 && $institutes = 0){
$search=$search->select('u.id','u.first_name','u.profile_picture','crse.full_name')
->leftJoin('user_courses as crs','crs.user_id','=','u.id')
->leftJoin('courses as crse','crse.id','=','crs.course_id');
}
else if($institutes > 0 && $courses > 0){
$search=$search->select('u.id','u.first_name','u.profile_picture','inst.full_name','crse.full_name')
->leftJoin('user_institutes as ins','ins.user_id','=','u.id')
->leftJoin('institutes as inst','inst.id','=','ins.institute_id')
->leftJoin('user_courses as crs','crs.user_id','=','u.id')
->leftJoin('courses as crse','crse.id','=','crs.course_id');
}
$search=$search->WhereNull('u.deleted_at');
for($i=0;$i<$institutes;$i++){
$search=$search->orWhere('inst.full_name','=',"`$institutes_data[$i]`");
}
for($i=0;$i<$courses;$i++){
$search=$search->orWhere('crse.full_name','=',"`$courses_data[$i]`");
}
$search=$search->get();
结果基于输入,给定场景是 $institutes = 2 和 $courses = 2, 我打印出来的选择查询 如果如下所示,我希望它落在第二个其他位置
'select * from `users` as `u` where `u`.`deleted_at` is null or `inst`.`full_name` = ?', array('`institutes 1`') or `inst`.`full_name` = ?', array('`institutes 2`')
基于这种情况,我想执行如下查询
select('u.id','u.first_name','u.profile_picture','inst.full_name','crse.full_name')
leftJoin 'user_institutes as ins' on 'ins.user_id''=''u.id'
leftJoin 'institutes as inst' on 'inst.id''=''ins.institute_id'
leftJoin 'user_courses as crs' on 'crs.user_id''=''u.id'
leftJoin 'courses as crse' on 'crse.id''=''crs.course_id'
where 'u.deleted_at' is null
and ('inst.full_name' = 'institutes 1'
or 'inst.full_name' = 'institutes 2')
and ('crse.full_name' = 'courses 1'
or 'crse.full_name' = 'courses 2'}
谁能指导我用这种 if else 条件构建我的 QB,并为 where 子句变量输入运行 for 循环。 (为我的语法道歉)
【问题讨论】:
-
什么会被覆盖?
-
如果特定数量的 $institutes = 2 和 $courses = 2,也许你应该再添加一个。
-
@jedrzej.kurylo,select 语句被覆盖,但我试图修改 QB,这是最新的,输出不是我想要的
-
@Grynets,以下陈述是否满足条件?否则 if($institutes > 0 && $courses > 0){ ..... }
-
“select 语句被覆盖”是什么意思?