我有一行csv包含很多元素。現在我想在bash/shell腳本中的每個第n個元素之後插入一個換行符。Shell命令插入一個換行符的每一個大數行中的第n個元素分隔字符串
獎勵:我想在描述符的前面添加一行,並使用描述符的計數作爲'n'。
例子:
"4908041eee3d4bf98e606140b21ebc89.16","7.38974601030349731","45.31298584267982221","94ff11ce7eb54642b0768dde313e8b25.16","7.38845318555831909","45.31425320325949713", (...)
到
"id","lon","lat"
"4908041eee3d4bf98e606140b21ebc89.16","7.38974601030349731","45.31298584267982221"
"94ff11ce7eb54642b0768dde313e8b25.16","7.38845318555831909","45.31425320325949713"
(...)
編輯:我做了第一次嘗試,但逗號分隔符丟失,則:
(...) | xargs --delimiter=',' -n3
"4908041eee3d4bf98e606140b21ebc89.16" "7.38974601030349731" "45.31298584267982221"
"94ff11ce7eb54642b0768dde313e8b25.16" "7.38845318555831909" "45.31425320325949713"
試圖取代 「」 with「,」
(...) | xargs --delimiter=',' -n3 -i echo ${{}//" "/","}
-bash: ${{}//\": bad substitution
向我們展示你嘗試的東西。 – Fazlin
| xargs --delimiter =','-n3但刪除逗號 – mstra001
編輯您的文章並添加您獲得的命令和輸出,以便您可以獲得幫助調試/修復 – Fazlin