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sql injection security problems(sql 注入)
已完成
#I3BOFQ
suyonghao
创建于
2021-03-16 21:18
Hello, I hava found that there two sql injetions in your program, So I create this issue to help you improve the security of program first place: http://demo.j2eefast.com/fast/sys/role/list and the parameter is deptId second place http://demo.j2eefast.com/fast/sys/role/authUser/list and the parameter is roleId ## 1.how I found this vulnerability? in the mapper file: src/main/resources/mapper/sys/SysUserMapper.xml you use **${}** join the sql statements.  **'${compId}'** is may be vulnerable to sql injection. And then I look into the file src/main/java/com/j2eefast/framework/sys/service/SysUserService.java and found that you use String to make it a string type. this leads to the sql injection.  ## 2.how to reproduce? 1. login with username 00008 and request people management function the request and response is:  the parameter compId is vulnerable to sql injection. use these sql statements I can get the databse info, when the case is true, the server will response the info while the false case will be null. 1. 判断database name's length is 6; 1'+or+length(database())=6+and+'1'='1  2. the first letter of database name is f  3. when the case is false, the info will be null.  4. finally I found that the database name is fastdb. ## 3.how to repair? 1. use #{} but ${} 2. use type cast make compId a int type. ## 4.error info 
Hello, I hava found that there two sql injetions in your program, So I create this issue to help you improve the security of program first place: http://demo.j2eefast.com/fast/sys/role/list and the parameter is deptId second place http://demo.j2eefast.com/fast/sys/role/authUser/list and the parameter is roleId ## 1.how I found this vulnerability? in the mapper file: src/main/resources/mapper/sys/SysUserMapper.xml you use **${}** join the sql statements.  **'${compId}'** is may be vulnerable to sql injection. And then I look into the file src/main/java/com/j2eefast/framework/sys/service/SysUserService.java and found that you use String to make it a string type. this leads to the sql injection.  ## 2.how to reproduce? 1. login with username 00008 and request people management function the request and response is:  the parameter compId is vulnerable to sql injection. use these sql statements I can get the databse info, when the case is true, the server will response the info while the false case will be null. 1. 判断database name's length is 6; 1'+or+length(database())=6+and+'1'='1  2. the first letter of database name is f  3. when the case is false, the info will be null.  4. finally I found that the database name is fastdb. ## 3.how to repair? 1. use #{} but ${} 2. use type cast make compId a int type. ## 4.error info 
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