我正在尝试删除前10个引号，然后再删除最后4个引号。
例如：
"one","two","three","four","ten","eleven","thirteen","fourteen"
"one","two","three","four","ten","eleven","thirteen","fourteen"
"one","two","three","four","ten","eleven","thirteen","fourteen"
应变成：
one,two,three,four,five,"eleven",thirteen,fourteen
one,two,three,four,five,"eleven",thirteen,fourteen
one,two,three,four,five,"eleven",thirteen,fourteen
我想重复此sed过程，直到文件结尾（csv）并将其应用于每一行。 一件事是，在一个元素中有换行符，双引号，逗号字符，并留下了引号。我相信这不会改变计数过程，但我担心它可能会与sed混淆，因为它会在看到第一个换行符后停止大小写，并在下一个换行符大小写之后继续，即使它可能位于同一元素中。这是一个例子
"eleven,"quotationmarks"
newline endofeleven"
在这种情况下，应该将整个内容算作1个元素， 这样的过程就像
看着行->删除前10个标记->如果换行符不在其中，则忽略换行符->删除最后4个标记->在每一行重复
我相信sed / awk可以做到这一点，它不必是一个班轮，并且可以有多行代码来实现这一目标。
正在寻找类似https://unix.stackexchange.com/questions/155805/sed-replace-first-k-instances-of-a-word-in-the-file
的内容