关于shell中数据替换的问题
2012060100024166311|00000000000000000000|000|00000000|0|00000000000|1338537227|JC0001| |28010000|28120000|0|客家饭店| |0 | |0 |30 |0 |0 |f90c19896a4986faae8c0727537f3f33| |000 |282803406951;282803204738;282801403625| |^A
2012060100024162345|00000000000000000000|000|00000000|0|00000000000|1338537227|JB0001| |28020000|28120000|0|客家饭店| |0 | |0 |30 |0 |0 |f90c19896a4986faae8c0727537f3f33| |000 |282803406951;282803204738;282801403625| |^A
数据格式是这样的，现在的需求如下：
1.以“|”把数据分开
2.取出第十和第十一列，即：28010000和28120000对应的列
3.把取出的数据的后六位全部换成0，前两位保持不变。
新人，求指导！新人没什么分

------解决方案--------------------

PHP code

[User:root Time:00:17:26 Path:/home/liangdong/shell]$ ./test.sh 
28000000,28000000
[User:root Time:00:17:26 Path:/home/liangdong/shell]$ cat test.sh 
#!/bin/bash

echo "2012060100024166311|00000000000000000000|000|00000000|0|00000000000|1338537227|JC0001| |28010000|28120000|0|客家饭店| |0 | |0 |30 |0 |0 |f90c19896a4986faae8c0727537f3f33| |000 |282803406951;282803204738;282801403625| |" | awk -F '|' '{print substr($10, 1, 2)"000000,"substr($11, 1, 2)"000000";}'