Replace ", with ",

Question

我在以下格式中具有以下文件中的数据

给出

"253539","","Company1 Name  
","0","1"
"255229","","Ramu ","0","1"
"253548","","Mater Name/Id 
","0","1"
"255229","","Ram Lakh","0","1"
"253619","","CHild Name/Ids 
","0","1"

每行都附加了 n。

在shell脚本中如何替换 n"，"，以获取输出为

"253539","","Company1 Name/Id  ","0","1"
"255229","","Ramu ","0","1"
"253548","","Mater Name/Id ","0","1"
"255229","","Ram Lakh","0","1"
"253619","","CHild Name/Id ","0","1"

需要自动化一个过程，因此必须在shell脚本中写入。

请帮助。

Answer 1

不确定要在外壳中轻松执行此操作，但是我做了一个抛出的perl脚本来完成：

perl -e 'foreach (<>) { chop; print; print "n" if /"$/ }' < mydata.csv

它仅在 "中的线结束时输出新线来起作用。

Answer 2

使用awk您可以做：

$ awk '/"$/ {print a$0; a=""; next} {a=$0}' file
"253539","","Company1 Name  ","0","1"
"255229","","Ramu ","0","1"
"253548","","Mater Name/Id ","0","1"
"255229","","Ram Lakh","0","1"
"253619","","CHild Name/Ids ","0","1"

它的作用是将其存储在a中，以防该行不会以"结束。如果这样做，它将打印存储的线加上当前的线。

Answer 3

可以做

之类的事情

#!/bin/bash
while read -r line; do
 [[ $line =~ '"'$ ]] && echo "$line" || echo -n "$line"
done < file

例如。在您的输入文件

上

> ./abovescript
"253539","","Company1 Name","0","1"
"255229","","Ramu ","0","1"
"253548","","Mater Name/Id","0","1"
"255229","","Ram Lakh","0","1"
"253619","","CHild Name/Ids","0","1"

Answer 4

如果可以接受 @fedorqui的答案，则是较短的代码。

awk '{printf $0 (/"$/?RS:X)}' file