求一个多文件编辑的shell（已经解决了）

grufield

现在有以下几个文件
1、a.txt
aa.dat 381263
bb.dat 515821

2、b.dat
CLASS 20120622192745 N GRADE COUNT


3、aa.bat
aa 20120107000000000 19950427000000 20120107000000 01001500
aa 20120107000000000 19950427000000 20120107000000 01001500
aa 20120107000000000 19950427000000 20120107000000 01003640
aa 20120107000000000 19950427000000 20120107000000 01003640
aa 20120107000000000 20060124000000 20120107000000 02080011
aa 20120107000000000 20060124000000 20120107000000 02080011
aa 20120107000000000 20051216000000 20120107000000 02080012


4、bb.dat
bb 20120107000000000 19950427000000 20120107000000 01001500
bb 20120107000000000 19950427000000 20120107000000 01001500
bb 20120107000000000 19950427000000 20120107000000 01003640
bb 20120107000000000 19950427000000 20120107000000 01003640
bb 20120107000000000 20060124000000 20120107000000 02080011
bb 20120107000000000 20060124000000 20120107000000 02080011
bb 20120107000000000 20051216000000 20120107000000 02080012


求一个shell，在a.txt文件中取每个dat对应的数字，替换b.dat文件中的COUNT，再把替换后的b.dat文件内容，加到aa.dat,bb.dat文件的最前面。（文件字段间用tab分割）谢谢大家啊~~~~

生成后的文件：
1、aa.bat
CLASS 20120622192745 N GRADE 381263
aa 20120107000000000 19950427000000 20120107000000 01001500
aa 20120107000000000 19950427000000 20120107000000 01001500
aa 20120107000000000 19950427000000 20120107000000 01003640
aa 20120107000000000 19950427000000 20120107000000 01003640
aa 20120107000000000 20060124000000 20120107000000 02080011
aa 20120107000000000 20060124000000 20120107000000 02080011
aa 20120107000000000 20051216000000 20120107000000 02080012


2、bb.dat
CLASS 20120622192745 N GRADE 515821
bb 20120107000000000 19950427000000 20120107000000 01001500
bb 20120107000000000 19950427000000 20120107000000 01001500
bb 20120107000000000 19950427000000 20120107000000 01003640
bb 20120107000000000 19950427000000 20120107000000 01003640
bb 20120107000000000 20060124000000 20120107000000 02080011
bb 20120107000000000 20060124000000 20120107000000 02080011
bb 20120107000000000 20051216000000 20120107000000 02080012


答案：（谢谢大家的帮忙）
awk '{if(NR==FNR){insert=$0}else{to_insert=insert;sub("COUNT", $2, to_insert);cmd="sed -i \"1i"to_insert"\" "$1;system(cmd)}}' b.dat a.txt

alan9101

#!/bin/bash
a1=(`cat a.txt | sed 's/\([a-z].*[a-z]\) .*/\1/g'`)
a2=(`cat a.txt  | sed 's/.* \([0-9]\+\)/\1/g'`)
n=0
for i in ${a1[*]}
do
        m=`cat b.dat`
        sed -i "1i\\$m\\t${a2[$n]}" $i
        n=`expr $n + 1`
done

rdcwayx

alan9101 发表于 2012-08-21 23:13
#!/bin/bash
a1=(`cat a.txt | sed 's/\([a-z].*[a-z]\) .*/\1/g'`)
a2=(`cat a.txt  | sed 's/.* \([0-9 ...

1. while read file count
   
2. do   
   
3.     str=$(sed "s/COUNT/$count/" b.dat)
   
4.     sed -i "1 i $str" $file
   
5. done < a.txt

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