blocks|key|844302|text|是的，这可以在awk中很好地完成。很容易获得所有的字段，而不需要任何严重的黑客攻击。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|844303|(此示例在The+One+True+Awk和gawk中都有效。)|offset|length|844304|{
++split($0,+a,+"\"")
++$2+=+a[2]
++$3+=+$(NF+-+1)
++$4+=+$NF
++print+"and+the+fields+are+",+$1,+"%2B",+$2,+"%2B",+$3,+"%2B",+$4
}|code-block|syntax|javascript|844305|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/AWK#Versions_and_implementations^0|0|5|G|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@$D|U|E|V|1|W]]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Yes, this can be done nicely in awk. It's easy to get all the fields without any serious hacks.

(This example works in both <a href="http://en.wikipedia.org/wiki/AWK#Versions_and_implementations" rel="noreferrer">The One True Awk</a> and in gawk.)

<pre><code>{
 split($0, a, "\"")
 $2 = a[2]
 $3 = $(NF - 1)
 $4 = $NF
 print "and the fields are ", $1, "+", $2, "+", $3, "+", $4
}
</code></pre>

blocks|key|629066|text|另一种选择是使用FPAT变量，该变量定义一个正则表达式来描述每个字段的内容。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|629067|将此AWK脚本另存为parse.awk|629068|#!/bin/awk+-f

BEGIN+{
++FPAT+=+"([%5E+]%2B)%7C(\"[%5E\"]%2B\")"
}
{
++print+$2
}|code-block|syntax|javascript|629069|使其可用chmod+%2Bx+./parse.awk执行，并将数据文件解析为./parse.awk+data.txt|629070|"I+am+ABC"
"I+am+not+ABC"|629071|entityMap|0|LINK|mutability|MUTABLE|url|https://www.gnu.org/software/gawk/manual/html_node/Splitting-By-Content.html^0|8|4|8|4|0|0|A|9|0|0|4|K|10|K|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|17|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|1D|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

Another alternative would be to use the <a href="https://www.gnu.org/software/gawk/manual/html_node/Splitting-By-Content.html" rel="noreferrer"><code>FPAT</code></a> variable, that defines a regular expression describing the contents of each field.

Save this AWK script as <code>parse.awk</code>:

<pre><code>#!/bin/awk -f

BEGIN {
 FPAT = "([^ ]+)|(\"[^\"]+\")"
}
{
 print $2
}
</code></pre>

Make it executable with <code>chmod +x ./parse.awk</code> and parse your data file as <code>./parse.awk data.txt</code>:

<pre><code>"I am ABC"
"I am not ABC"
</code></pre>

blocks|key|844221|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|844222|$+cat+data.txt+%7C+awk+-F\"+'{print+$2}'
I+am+ABC
I+am+not+ABC|code-block|syntax|javascript|844223|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this:

<pre><code>$ cat data.txt | awk -F\" '{print $2}'
I am ABC
I am not ABC
</code></pre>

blocks|key|844614|text|此问题的最高答案仅适用于具有单引号字段的行。当我发现这个问题时，我需要一些可以处理任意数量的引用字段的东西。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|844615|最终我遇到了an+answer+by+Wintermute+in+another+thread，他为这个问题提供了一个很好的通用解决方案。我刚刚修改了它，删除了引号。请注意，在运行以下程序时，您需要使用-F\"调用awk。|offset|length|style|CODE|844616|BEGIN+{+OFS+=+""+}+{
++++for+(i+=+1;+i+<=+NF;+i+%2B=+2)+{
++++++++gsub(/[+\t]%2B/,+",",+$i)
++++}
++++print
}|code-block|syntax|javascript|844617|这是通过观察数组中的每个其他元素都位于引号内的方式来实现的，当您使用"-character“分隔时，它会用逗号替换分隔不在引号中的元素的空格。|844618|然后，您可以轻松地链接awk的另一个实例来执行所需的任何处理(只需再次使用字段分隔符开关-F,)。|844619|请注意，如果第一个字段被引用，这可能会中断-我还没有测试它。如果是这样的话，如果行的第一个字符是“，那么通过添加一个If语句从2开始，而不是从1开始，应该很容易修复。|844620|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/28063726/3756197^0|0|2T|4|6|15|0|0|0|0|18|3|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@$D|12|E|13|F|G]]|9|@$D|14|E|15|1|16]]|A|$]]|$1|H|3|I|5|J|7|17|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|19|8|@$D|1A|E|1B|F|G]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1C|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

The top answer for this question only works for lines with a single quoted field. When I found this question I needed something that could work for an arbitrary number of quoted fields.

Eventually I came upon <a href="https://stackoverflow.com/a/28063726/3756197">an answer by Wintermute in another thread</a>, and he provided a good generalized solution to this problem. I've just modified it to remove the quotes. Note that you need to invoke awk with <code>-F\"</code> when running the below program.

<pre><code>BEGIN { OFS = "" } {
 for (i = 1; i &lt;= NF; i += 2) {
 gsub(/[ \t]+/, ",", $i)
 }
 print
}
</code></pre>

This works by observing that every other element in the array will be inside of the quotes when you separate by the "-character, and so it replaces the whitespace dividing the ones not in quotes with a comma.

You can then easily chain another instance of awk to do whatever processing you need (just use the field separator switch again, <code>-F,</code>).

Note that this might break if the first field is quoted - I haven't tested it. If it does, though, it should be easy to fix by adding an if statement to start at 2 rather than 1 if the first character of the line is a ".

blocks|key|628819|text|我整理了一个函数，它将$0重新拆分到一个名为B的数组中。双引号之间的空格不起到字段分隔符的作用。适用于任意数量的字段，可以是带引号的和不带引号的字段的混合。如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|628820|#!/usr/bin/gawk+-f

#+Resplit+$0+into+array+B.+Spaces+between+double+quotes+are+not+separators.
#+Single+quotes+not+handled.+No+escaping+of+double+quotes.
function+resplit(+++++++a,+l,+i,+j,+b,+k,+BNF)+#+all+are+local+variables
{
++l=split($0,+a,+"\"")
++BNF=0
++delete+B
++for+(i=1;i<=l;%2B%2Bi)
++{
++++if+(i+%25+2)
++++{
++++++k=split(a[i],+b)
++++++for+(j=1;j<=k;%2B%2Bj)
++++++++B[%2B%2BBNF]+=+b[j]
++++}
++++else
++++{
++++++B[%2B%2BBNF]+=+"\""a[i]"\""
++++}
++}
}

{
++resplit()

++for+(i=1;i<=length(B);%2B%2Bi)
++++print+i+":+"+B[i]
}|code-block|syntax|javascript|628821|希望能有所帮助。|628822|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I've scrunched up together a function that re-splits $0 into an array called B. Spaces between double quotes are not acting as field separators. Works with any number of fields, a mix of quoted and unquoted ones. Here goes:

<pre><code>#!/usr/bin/gawk -f

# Resplit $0 into array B. Spaces between double quotes are not separators.
# Single quotes not handled. No escaping of double quotes.
function resplit( a, l, i, j, b, k, BNF) # all are local variables
{
 l=split($0, a, "\"")
 BNF=0
 delete B
 for (i=1;i&lt;=l;++i)
 {
 if (i % 2)
 {
 k=split(a[i], b)
 for (j=1;j&lt;=k;++j)
 B[++BNF] = b[j]
 }
 else
 {
 B[++BNF] = "\""a[i]"\""
 }
 }
}

{
 resplit()

 for (i=1;i&lt;=length(B);++i)
 print i ": " B[i]
}
</code></pre>

Hope it helps.

blocks|key|844374|text|好的，如果你真的想要这三个字段，你可以得到它们，但这需要大量的管道：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|844375|$+cat+data.txt+%7C+awk+-F\"+'{print+$1+","+$2+","+$3}'+%7C+awk+-F'+,'+'{print+$1+","+$2}'+%7C+awk+-F',+'+'{print+$1+","+$2}'+%7C+awk+-F,+'{print+$1+","+$2+","+$3}'
ABC,I+am+ABC,35
DEF,I+am+not+ABC,42|code-block|syntax|javascript|844376|在最后一条管道中，您已经获得了所有三个字段，可以随心所欲地处理。|844377|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Okay, if you really want all three fields, you can get them, but it takes a lot of piping:

<pre><code>$ cat data.txt | awk -F\" '{print $1 "," $2 "," $3}' | awk -F' ,' '{print $1 "," $2}' | awk -F', ' '{print $1 "," $2}' | awk -F, '{print $1 "," $2 "," $3}'
ABC,I am ABC,35
DEF,I am not ABC,42
</code></pre>

By the last pipe you've got all three fields to do whatever you'd like with.

blocks|key|628976|text|这是我最终得到的一些东西，它对我的项目更通用。注意，它不使用awk。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|628977|someText="ABC+\"I+am+ABC\"+35+DESC+'1+23'+testing+456"
putItemsInLines()+{
++++local+items=""
++++local+firstItem="true"
++++while+test+$#+-gt+0;+do
++++++++if+[+"$firstItem"+==+"true"+];+then
++++++++++++items="$1"
++++++++++++firstItem="false"
++++++++else
++++++++++++items="$items
$1"
++++++++fi
++++++++shift
++++done
++++echo+"$items"
}

count=0
while+read+-r+valueLine;+do
++++echo+"$count:+$valueLine"
++++count=$((+$count+%2B+1+))
done+<<<+"$(eval+putItemsInLines+$someText)"|code-block|syntax|javascript|628978|以下哪项输出：|628979|0:+ABC
1:+I+am+ABC
2:+35
3:+DESC
4:+1+23
5:+testing
6:+456|628980|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here is something like what I finally got working that is more generic for my project.
Note it doesn't use awk.

<pre><code>someText="ABC \"I am ABC\" 35 DESC '1 23' testing 456"
putItemsInLines() {
 local items=""
 local firstItem="true"
 while test $# -gt 0; do
 if [ "$firstItem" == "true" ]; then
 items="$1"
 firstItem="false"
 else
 items="$items
$1"
 fi
 shift
 done
 echo "$items"
}

count=0
while read -r valueLine; do
 echo "$count: $valueLine"
 count=$(( $count + 1 ))
done &lt;&lt;&lt; "$(eval putItemsInLines $someText)"
</code></pre>

Which outputs:

<pre><code>0: ABC
1: I am ABC
2: 35
3: DESC
4: 1 23
5: testing
6: 456
</code></pre>

Data file - data.txt:

<pre><code>ABC "I am ABC" 35 DESC
DEF "I am not ABC" 42 DESC
</code></pre>

<code>cat data.txt | awk '{print $2}'</code> 

will result the "I" instead of the string being quoted

How to make awk so that it ignore the space within the quote and think that it is one single token?

Awk consider double quoted string as one token and ignore space in between

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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数据文件- data.txt：ABC "I am ABC" 35 DESCDEF "I am not ABC" 42 DESCcat data.txt | awk '{print $2}'将产生"I“而不是被引号的字符串。如何使awk忽略引号中的空格，并认为它是一个单独的令牌？

问Awk将双引号字符串视为一个标记，并忽略其间的空格
EN

回答 7

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问Awk将双引号字符串视为一个标记，并忽略其间的空格EN

回答 7

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问Awk将双引号字符串视为一个标记，并忽略其间的空格
EN