Question

我有一个在命令行上运行的sql查询（在我的情况下，使用GUI运行查询），但是在PHP上，通过mysqli_query（）函数，它不起作用。该错误是重复的键错误。

奇怪的是：错误不存在。它在命令行上运行的事实证明了这一点。此外，在进一步调查后，记录显示唯一密钥从未被破坏。不过，mysqli_query（）会抛出唯一的键错误。

我的查询是：

UPDATE
   frequencia 
   INNER JOIN
      (
         SELECT
            *,
            @row_number_aux : = 
            CASE
               WHEN
                  @atual_aux = disciplina_id 
               THEN
                  @row_number_aux + 1 
               ELSE
                  1 
            END
            AS "row_number", @atual_aux : = disciplina_id AS "disciplina_atual" 
         FROM
            frequencia 
         WHERE
            frequencia.data > "2019-06-17" 
            AND frequencia.aluno_id IN 
            (
               8835, 5553, 3488, 7769, 143, 4406, 1472, 698, 8731, 5221, 4240, 4934, 8426, 8107, 5180, 2146, 5589, 4565
            )
      )
      frequencia_aux 
      ON frequencia.id = frequencia_aux.id 
   INNER JOIN
      (
         SELECT
            *,
            @row_number : = 
            CASE
               WHEN
                  @atual = disciplina_id 
               THEN
                  @row_number + 1 
               ELSE
                  1 
            END
            AS "row_number", @atual : = disciplina_id AS "disciplina_atual" 
         FROM
            (
               SELECT
                  disciplina_id,
                  data,
                  tempo,
                  turma_id 
               FROM
                  frequencia 
               WHERE
                  data > "2019-06-17" 
                  AND turma_id = "87" 
               GROUP BY
                  data,
                  disciplina_id,
                  tempo 
            )
            frequencia 
      )
      aux 
      ON frequencia_aux.disciplina_id = aux.disciplina_id 
      AND frequencia_aux.data = aux.data 
      AND frequencia_aux.row_number = aux.row_number 
SET
   frequencia.turma_id = aux.turma_id,
   frequencia.tempo = aux.tempo;

我以前已经使用@变量，没有问题，并且使用过所有类型的复杂语句，但从未遇到过这样的错误。有人曾经经历过类似的事情吗？

Answer 1

已解决。

问题是：在mysql cli上直接运行时@变量似乎是自动初始化的，而不是通过mysqli_query进行的。一个简单的子查询联接就完成了工作。

感谢@ spencer7593的见识。

最终的sql：

UPDATE 
   frequencia 
   INNER JOIN
      (
         SELECT
            *,
            @row_number_aux := 
            CASE
               WHEN
                  @atual_aux = disciplina_id 
               THEN
                  @row_number_aux + 1 
               ELSE
                  1 
            END
            AS "row_number", @atual_aux := disciplina_id AS "disciplina_atual" 
         FROM
            frequencia                     
            join (select @row_number_aux := 0, @atual_aux := 0) xxx

         WHERE
            frequencia.data > "2019-06-17" 
            AND frequencia.aluno_id IN 
            (
               8835, 5553, 3488, 7769, 143, 4406, 1472, 698, 8731, 5221, 4240, 4934, 8426, 8107, 5180, 2146, 5589, 4565
            )
      )
      frequencia_aux 
      ON frequencia.id = frequencia_aux.id 
   INNER JOIN
      (
         SELECT
            *,
            @row_number := 
            CASE
               WHEN
                  @atual = disciplina_id 
               THEN
                  @row_number + 1 
               ELSE
                  1 
            END
            AS "row_number", @atual := disciplina_id AS "disciplina_atual" 
         FROM
            (
               SELECT
                  disciplina_id,
                  data,
                  tempo,
                  turma_id 
               FROM
                  frequencia 
               WHERE
                  data > "2019-06-17" 
                  AND turma_id = "87" 
               GROUP BY
                  data,
                  disciplina_id,
                  tempo 
            )
            frequencia 
           join (select @row_number := 0, @atual := 0) xxx
      )
      aux 
      ON frequencia_aux.disciplina_id = aux.disciplina_id 
      AND frequencia_aux.data = aux.data 
      AND frequencia_aux.row_number = aux.row_number
SET
   frequencia.turma_id = aux.turma_id,
   frequencia.tempo = aux.tempo

为什么我的sql语句可以在命令行上运行，但不能在PHP上与mysqli_query一起使用？

1 个答案: