php – 显示选项中的错误答案
数据库表结构:
会话表(又名考试表) SessionId(auto) SessionName 137 XULWQ 问题表: SessionId QuestionId OptionId 137 1 5 137 2 2 Option_Table表: OptionId OptionType 1 A-C 2 A-D 3 A-E 4 A-F 5 A-G 6 A-H 7 A-I 8 A-J 9 A-K 10 A-L 11 A-M 12 A-N 13 A-O 14 A-P 15 A-Q 16 A-R 17 A-S 18 A-T 19 A-U 20 A-V 21 A-W 22 A-X 23 A-Y 24 A-Z 25 True or False 26 Yes or No 答案表: AnswerId(auto) SessionId QuestionId Answer 200 137 1 B 201 137 1 D 202 137 2 F 203 137 2 A 204 137 2 C 我想创建一个页面,我希望它显示每个问题的错误答案. 我正在考虑通过检索每个问题的选项类型,显示属于选项类型的所有字母答案,然后从字母答案中删除正确答案,以便仅留下不正确的答案. 选项数组: $option = array(); $option[1]= array(A,B,C); $option[2]= array(A,C,D); $option[3]= array(A,D,E); ... $option[23]= array(A,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y); $option[24]= array(A,Y,Z); $option[25]= array(True,False); $option[26]= array(Yes,No); 我的问题是,在本节之后我需要帮助.如何使用mysqli / php和我目前拥有的数据库开始检索此数组后的错误答案? 更新: 下面显示了sql,它显示了每个问题的正确答案: SELECT q.QuestionContent,o.OptionType,q.NoofAnswers,GROUP_CONCAT(DISTINCT Answer ORDER BY Answer SEPARATOR '') AS Answer,r.ReplyType,q.QuestionMarks FROM Question q LEFT JOIN Answer an ON q.QuestionId = an.QuestionId LEFT JOIN Reply r ON q.ReplyId = r.ReplyId LEFT JOIN Option_Table o ON q.OptionId = o.OptionId group by q.QuestionContent 见SQL Fiddle with Demo 这将返回结果: | QUESTIONCONTENT | OPTIONTYPE | NOOFANSWERS | ANSWER | REPLYTYPE | QUESTIONMARKS | ---------------------------------------------------------------------------------------- | Name these 2 flowers | A-F | 2 | C | Multiple | 5 | | What is 2+2? | A-D | 1 | ABD | Single | 5 | 解决方法
您的问题在于,Option_Table的设计方式,解码OptionType列以找出所有可能的答案需要一些非常重要的外部知识.
(事实上??,你没有在我的问题中提供足够的信息,以确定如何做到这一点;我可以猜测你向我展示的那些OptionTypes,但我不能确定是否有,或者可能是,其他人.) 用更简单的表替换(或至少扩充)该表会更好,对于每个OptionId,它只列出所有可能的选项,如下所示: CREATE TABLE Options ( OptionId INTEGER NOT NULL,OptionAnswer CHAR(1) NOT NULL,-- or whatever type Answer.Answer has PRIMARY KEY (OptionId,OptionAnswer) ); INSERT INTO Options VALUES (1,'A'),(1,'B'),'C'),(2,'D'),-- ... (25,'T'),(25,'F'),(26,'Y'),(16,'N'); 然后,您可以使用以下查询找到每个问题的所有正确和错误答案: SELECT q.QuestionContent,GROUP_CONCAT(DISTINCT Answer ORDER BY Answer SEPARATOR '') AS RightAnswers,GROUP_CONCAT(DISTINCT CASE WHEN Answer IS NULL THEN OptionAnswer ELSE NULL END ORDER BY OptionAnswer SEPARATOR '') AS WrongAnswers,q.QuestionMarks FROM Question q LEFT JOIN Reply r ON q.ReplyId = r.ReplyId LEFT JOIN Options o ON q.OptionId = o.OptionId LEFT JOIN Answer an ON q.QuestionId = an.QuestionId AND o.OptionAnswer = an.Answer GROUP BY q.SessionId,q.QuestionId 这是一个demo of it on SQLize(稍微修改为跳过未包含在示例表中的列). 编辑:另一种解决方案是在PHP中构造错误答案的列表.例如,如果$row是一个对象(如 // Do this (preferably) before looping over the rows: $specialOptionTypes = array( 'Yes or No' => array( 'Y','N' ),'True or False' => array( 'T','F' ),); // Do this for each row: if ( array_key_exists( $row->OptionType,$specialOptionTypes ) ) { $options = $specialOptionTypes[ $row->OptionType ]; } else if ( preg_match( '/^([A-Z])-([A-Z])$/',$row->OptionType,$match ) ) { $options = range( $match[1],$match[2] ); } else { // issue warning about unrecognized option type $options = array(); } $right = str_split( $row->Answer ); // or explode() on a delimiter,if any $wrong = array_diff( $options,$right ); $row->WrongAnswers = implode( '',$wrong ); // if you actually want a string Here’s a demo on ideone.com,基于您的示例查询输出. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |