
這是我的文字檔:
0019121002313002316003135
0057936000814000814003023
0081638001519001523001176
0090531001841001842002633
0111210001515001518000912
0115400001807001828001593
我希望處理後的輸出為:
0019121,002313,002316,003135
0057936,000814,000814,003023
0081638,001519,001523,001176
0090531,001841,001842,002633
0111210,001515,001518,000912
0115400,001807,001828,001593
我該怎麼做呢?
答案1
$ sed -E 's/(.{7})(.{6})(.{6})(.{6})/\1,\2,\3,\4/' file
0019121,002313,002316,003135
0057936,000814,000814,003023
0081638,001519,001523,001176
0090531,001841,001842,002633
0111210,001515,001518,000912
0115400,001807,001828,001593
也就是說,匹配組成新欄位的每一行的位元並在它們之間插入逗號。字段的匹配是使用.{7}
或.{6}
取決於所需的字段長度來完成的。
答案2
使用 GNU awk ( gawk
),您可以設定明確欄位寬度:
$ gawk '{$1=$1} 1' FIELDWIDTHS='7 6 6 6' OFS=, file
0019121,002313,002316,003135
0057936,000814,000814,003023
0081638,001519,001523,001176
0090531,001841,001842,002633
0111210,001515,001518,000912
0115400,001807,001828,001593
答案3
假設第一個逗號在7個字元之後,接下來的逗號是6的倍數,這不是一個錯誤,我建議
sed 's/\(.......\)\(......\)\(......\)/\1,\2,\3,/'