我有两个表问题和答案，我想显示所有问题及其答案的详细信息。 例如。 23（意见）12（答案）

This doesn't show questions without answers

SELECT 
  M.id, M.j_surname, 
  Q.id as qid, Q.qus_view_count, Q.qus_owner, Q.qus_title, Q.qus_description,
  Q.qus_createddate, 
  COUNT(A.qus_id) as ans_count, 
  A.qus_id FROM jt_questions Q 
LEFT JOIN jt_members M ON M.id = Q.qus_owner 
LEFT JOIN jt_answers A ON Q.id = A.qus_id 
GROUP BY A.qus_id 
ORDER BY Q.qus_createddate DESC 
LIMIT $offset, $recordsPerPage;

Try GROUP BY Q.id not GROUP BY A.qus_id

For unanswered questions A.qus_id will all have a value of NULL

$offset = "1; drop table jt_answers; --";

EDIT

if what you did in your code looked like this:

$SQL = "SELECT 
  M.id, M.j_surname, 
  Q.id as qid, Q.qus_view_count, Q.qus_owner, Q.qus_title, Q.qus_description,
  Q.qus_createddate, 
  COUNT(A.qus_id) as ans_count, 
  A.qus_id FROM jt_questions Q 
LEFT JOIN jt_members M ON M.id = Q.qus_owner 
LEFT JOIN jt_answers A ON Q.id = A.qus_id 
GROUP BY A.qus_id 
ORDER BY Q.qus_createddate DESC 
LIMIT $offset, $recordsPerPage;";

All I would need to do is to get $offset to look like my original answer above. That would drop your jt_answers table.

You could validate these values, such as making sure $offset is a positive integer.

But really you should use MySQLi's parameter binding instead. That would replace the variables with ? and let you pass the variables into it so the database can decide how to handle them.

http://www.php.net/manual/en/mysqli-stmt.bind-param.php