除非在双引号单卡更换空白 [英] Replacing whitespace with single tab unless in double quotes

问题描述

假设由一个或多个空格分隔的字符串多行文件。进一步假设，串组可以用双引号括起来。

Assume a multi-line file with strings separated by one or more whitespaces. Assume further that groups of strings can be enclosed by double quotes.

> cat file
foo bar "foobar baz qux"
foo "bar foobar baz" qux
"foo   bar foobar" baz   qux   # multiple whitespaces in this line

如果我想使用，以取代单一的制表符双引号外的所有空格的 AWK 的下面列出，我收到以下内容：

If I wish to replace all whitespaces outside the double quotes with single tab characters using awk as listed below, I receive the following:

awk '{OFS="\t"; FPAT="([^, ]+)|(\"[^\"]+\")"; $1=$1; print}' file
# foo   bar "foobar baz qux" # In this line, strings inside the quote are separated by tabs
# foo   "bar foobar baz"    qux
# "foo  bar foobar" baz qux

问题只似乎仅限于以双引号结束行。

The problem only seems to be restricted to the line that ends with a double quote.

的编辑1：的
为了更好地可视化的问题在眼前：

EDIT 1: To better visualize the issue at hand:

awk '{OFS="\t"; FPAT="([^, ]+)|(\"[^\"]+\")"; $1=$1; print}' file | cat -A
# foo^Ibar^I"foobar^Ibaz^Iqux"$
# foo^I"bar foobar baz"^Iqux$
# "foo   bar foobar"^Ibaz^Iqux$

的编辑2：的
看来，这两个命令回答部分做工精细建议，除非非字母字符一定数量或组合在输入present。下面是一个例子：

EDIT 2: It appears that both commands suggested in the answer section work fine unless a certain number or combination of non-letter characters are present in the input. Here is an example:

> cat file
foo_bar_baz foo foo_bar . Name=foo;product="bar baz qux"
foo_bar_baz foo foo_bar . Name=foo;product="bar baz qux"
foo_bar_baz foo foo_bar . Name=foo;product="bar baz qux"

> awk -v FPAT='"[^"]*"|[^[:blank:]]+' -v OFS='\t' '{$1=$1} 1' file | cat -A
foo_bar_baz^Ifoo^Ifoo_bar^I.^IName=foo;product="bar^Ibaz^Iqux"$
foo_bar_baz^Ifoo^Ifoo_bar^I.^IName=foo;product="bar^Ibaz^Iqux"$
foo_bar_baz^Ifoo^Ifoo_bar^I.^IName=foo;product="bar^Ibaz^Iqux"$

> awk '{$1=$1}1' OFS='\t' FPAT='"[^"]+"|[^ ]+' file | cat -A
foo_bar_baz^Ifoo^Ifoo_bar^I.^IName=foo;product="bar^Ibaz^Iqux"$
foo_bar_baz^Ifoo^Ifoo_bar^I.^IName=foo;product="bar^Ibaz^Iqux"$
foo_bar_baz^Ifoo^Ifoo_bar^I.^IName=foo;product="bar^Ibaz^Iqux"$

的编辑3：的
这个问题提出的编辑2 的进一步这里讨论：<一href=\"http://stackoverflow.com/questions/34231514/replacing-whitespace-with-single-tab-unless-in-double-quotes-part-ii#34231597\">Replacing空白单标签，除非在双引号 - 第二部分

EDIT 3: This question posed EDIT 2 is further discussed here: Replacing whitespace with single tab unless in double quotes - Part II

推荐答案

使用的GNU AWK 你可以做到这一点很容易：

Using gnu-awk you can do this easily:

awk -v FPAT='"[^"]*"|[^[:blank:]]+' -v OFS='\t' '{$1=$1} 1' file
foo bar "foobar baz qux"
foo "bar foobar baz"    qux
"foo   bar foobar"  baz qux

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