Question

我有一张名为：＆＃39; paper＆＃39; （question_bank）

            id  |  exam_id | question

            1   |  1       | What will be 35*9?
            2   |  1       | What will be 5-9?
            3   |  1       | A + B
            4   |  1       | What is a circle?
            5   |  1       | If we have four corners with equal height and width, then which shape is that?
            6   |  1       | What is Maths?
            7   |  1       | What is a triangle?
            21  |  1       | what is Nikhil surname?
            22  |  2       | Last name of Bhavesh is
            23  |  2       | Last name of Harsh is
            27  |  3       | What is Maths?
            28  |  3       | What is a triangle?
            30  |  3       | Last name of Harsh is

我有一个php页面，我必须在exam_id＆＃39; 3＆＃39;中添加问题。来自表格＆＃39; paper＆＃39;。我插入问题的查询如下：

INSERT INTO paper (question exam_id) SELECT question, '3' FROM paper WHERE id = '2'
                        OR
INSERT INTO paper (question exam_id) SELECT question, '3' FROM paper WHERE id = '1'
                        OR

我还可以添加一个新问题，因此查询为：

INSERT INTO paper (question, exam_id) VALUES ('blah blah blah', '3')

这取决于我选择的问题或我添加的任何新问题。

现在，当我想向exam_id＆＃39; 3＆＃39;添加更多问题时从表格＆＃39; paper＆＃39;中，它显示了所有问题。我的选择查询如下：

SELECT * FROM paper WHERE exam_id != '3'

它显示了所有问题，但是例如我已经添加了id =＆＃39; 23＆＃39;，所以当我添加更多问题时，我不希望显示该问题。请帮我处理我的选择查询。如果我错过了什么，请告诉我！提前谢谢！

Answer 1

我不确定我是否理解你，但你可以列出你不想看到的问题的ids。

例如：排除ids 23,24,25

SELECT * FROM paper WHERE exam_id != 3 AND id NOT IN (23, 24, 25)

Answer 2

好的，我明白了.....

我需要做的就是再次触发查询，以免重复之前选择的问题....

SELECT * FROM paper WHERE question = '$row['question']' AND exam_id = '3'
if($select)
{
    ************************* SHOW RESULT *********************
}

mysql选择查询显示在php中

2 个答案: