PHP SQLSTATE [42000]語法錯誤或訪問衝突：在查詢

我運行一個查詢，但是當我運行它，它會返回一個PHP SQLSTATE [42000]語法錯誤或訪問衝突：在查詢

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 ' count(distinct exams.user_id) as count_user_mhs from `users` where `users.type`' at line 1 (SQL: select `universities`.`name`, count(distinct exams.user_id) as count_user_dsn from `users` where `users.type` = 0, count(distinct exams.user_id) as count_user_mhs from `users` where `users.type` = 1, count(exams.subject_id) as count_subject, sum(exams.score) as count_score, TIME_FORMAT(SUM(TIMEDIFF(exams.exam_end_date, exams.exam_start_date)), "%H:%i:%s") as count_time from `exams` inner join `universities` on `universities`.`id` = `exams`.`university_id` inner join `users` on `users`.`id` = `exams`.`user_id` where exists (select * from `universities` where `exams`.`university_id` = `universities`.`id`) group by `user_id` order by `count_score` desc, `count_time` asc)

$exams = Exam::join('universities', 'universities.id', '=', 'exams.university_id') 
       ->join('users', 'users.id', '=', 'exams.user_id') 
       ->select('universities.name',DB::raw('count(distinct exams.user_id) as count_user_dsn from `users` where `users.type` = 0'),DB::raw('count(distinct exams.user_id) as count_user_mhs from `users` where `users.type` = 1'),DB::raw('count(exams.subject_id) as count_subject'),DB::raw('sum(exams.score) as count_score'),DB::raw('TIME_FORMAT(SUM(TIMEDIFF(exams.exam_end_date, exams.exam_start_date)), "%H:%i:%s") as count_time'));

誰能幫助1064？我猜測，這與DB::raw表達式有關。但我不知道到底出了什麼問題。

來源

2017-06-05 goateee25

1）這不是一個警告，這是一個錯誤。 2）你沒有複製整個錯誤信息，所以我們甚至不知道可能發生了什麼問題。 3）你還沒有共享從你的代碼創建的sql查詢laravel，這也有助於識別錯誤。但是那些原始sql部分中的計數子查詢看起來是錯誤的。 4）如果你描述了你想要達到的目標，這也會有所幫助。因爲我們可以告訴你犯了什麼錯誤是一回事，但另一個想法是幫助你創建一個工作解決方案。 – Shadow

[進一步閱讀]（https://stackoverflow.com/questions/23515347/how-cani-i-fix-mysql-error-1064）。 – tadman

粘貼普通錯誤消息，它會幫助調試。 – BDS

你的生成的查詢看起來像這樣：

select 
    `universities`.`name`, 
    count(distinct exams.user_id) as count_user_dsn from `users` where `users.type` = 0, 
    count(distinct exams.user_id) as count_user_mhs from `users` where `users.type` = 1, 
    count(exams.subject_id) as count_subject, 
    sum(exams.score) as count_score, 
    TIME_FORMAT(SUM(TIMEDIFF(exams.exam_end_date, exams.exam_start_date)), "%H:%i:%s") as count_time 
from 
    `exams` 
     inner join 
    `universities` on `universities`.`id` = `exams`.`university_id` 
     inner join 
    `users` on `users`.`id` = `exams`.`user_id` 
where 
    exists (select * from `universities` where `exams`.`university_id` = `universities`.`id`) 
group by 
    `user_id` 
order by 
    `count_score` desc, 
    `count_time` asc

的問題是在其上由所述DB::raw代碼生成線3和4。

要解決這個問題，你需要更新你這樣的雄辯查詢：

<?php 
$exams = Exam::join('universities', 'universities.id', '=', 'exams.university_id') 
    ->join('users', 'users.id', '=', 'exams.user_id') 
    ->select(
     'universities.name', 
     DB::raw('(select count(distinct exams.user_id) as count_user_dsn from `users` where `users.type` = 0)'), 
     DB::raw('(select count(distinct exams.user_id) as count_user_mhs from `users` where `users.type` = 1)'), 
     DB::raw('count(exams.subject_id) as count_subject'),DB::raw('sum(exams.score) as count_score'), 
     DB::raw('TIME_FORMAT(SUM(TIMEDIFF(exams.exam_end_date, exams.exam_start_date)), "%H:%i:%s") as count_time') 
    );

也就是說......在主SELECT語句執行的SELECT COUNT子查詢是不是一個好主意，因爲他們會爲每個被執行由主查詢選擇的行。

來源

2017-06-07 17:57:22 Jan

PHP SQLSTATE [42000]語法錯誤或訪問衝突：在查詢

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