MySQL的選擇查詢顯示在PHP

Question

我有一個名爲表：「紙」（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 

我有一個PHP頁面，我從表「紙」在exam_id「3」添加的問題。我的查詢插入的問題是如下：

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 

我還可以添加一個新的問題，因此該查詢：

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

這取決於問題我選擇或任何新的問題，我補充。

現在，當我想從table'paper'向exam_id'3'添加更多問題時，它會顯示所有問題。我的選擇查詢如下：

SELECT * FROM paper WHERE exam_id != '3' 

它顯示了所有的問題，但比如我已經添加ID =「23」，所以我不想這個問題，當我加入了更多的問題來顯示。請幫我選擇查詢。讓我知道如果我錯過任何東西！提前致謝！

Answer 1

好吧我知道了.....

所有我需要做的是，爲不重複再次觸發查詢以前選擇的問題....

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

Answer 2

我不確定我是否正確理解了你，但是你可以用你不想看到的問題的ID列出一個列表。

例如：排除ids 23，24，25

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