grep文件中的特定字段也打印該行的文件名

Question

我有一個文件。它包含這樣

領域的I/P文件：

a.txt 
9842,5003,a,a,100 
9942,5003,a,a,100 
9852,5003,a,a,100 
98456,5003,a,a,100 
9742,5003,a,a,100 

b.txt 
7842,5003,a,a,100 
7942,5003,a,a,100 
7852,5003,a,a,100 
98456,5003,a,a,100 
7742,5003,a,a,100 

c.txt 
8842,5003,a,a,100 
9842,5003,a,a,100 
8852,5003,a,a,100 
88456,5003,a,a,100 
8742,5003,a,a,100 

工作要做： 我需要到grep在上面輸入文件（A.TXT特定行，b.txt & c.txt ） 輸出線應包含所有字段和也是作爲FILNAME最後場

O/p格式： 說，我的grep線與「9」開始指低於格式將需要在輸出

9842,5003,a,a,100,a.txt 
9942,5003,a,a,100,a.txt 
9852,5003,a,a,100,a.txt 
98456,5003,a,a,100,a.txt 
9742,5003,a,a,100,a.txt 
98456,5003,a,a,100,b.txt 
9842,5003,a,a,100,c.txt 

Answer 1

grep -H '^9' a.txt b.txt c.txt | awk -F : '{print $2","$1;}' 

grep命令以9開始，通過使用-h參數來打印文件名，然後用awk把它變成你想要的格式線。 awk的-F標誌將字段分隔符更改爲'：'

Answer 2

您可以只使用一個awk命令，不需要grep。 grep和awk是互補的。

awk '/^9/{print $2","$1","FILENAME}' [abc].txt