从一个位置开始将长字符串包装成多行

Question

我有一个文件，看起来像这样:

FirstSentences1  bfjkjhdfhizhfzibfkjezfzfiuzehfizdjfldfsdfsljfklj
SecondSentences2 fjlskdjfjoijrgeojrgijgoejrgrjgiorjofgjeirjgoergd
.
.
.
NthhhSentencesN  klkdlffjsldfsljflsfjlskfjldkjflsfjlfkdjfdfjojjij

我必须得到以下输出:

FirstSentences1  bfjkjhdfhizhfzibfkje
FirstSentences1  zfzfiuzehfizdjfldfsd
FirstSentences1  fsljfklj
SecondSentences2 fjlskdjfjoijrgeojrgi
SecondSentences2 jgoejrgrjgiorjofgjei
SecondSentences2 rjgoergd
.
.
.
NthhhSentencesN  klkdlffjsldfsljflsfj
NthhhSentencesN  lskfjldkjflsfjlfkdjf
NthhhSentencesN  dfjojjij

说明:

，例如第一行:

FirstSentences1  bfjkjhdfhizhfzibfkjezfzfiuzehfizdjfldfsdfsljfklj

我们取字符串"bfjkjhdfhizhfzibfkjezfzfiuzehfizdjfldfsdfsljfklj"当长度等于20时将其包装起来

你知道怎么得到这个吗?

Answer 1

您可以使用一个短脚本利用字符串索引和嵌套循环:

#!/bin/bash
declare -i len=${2:-20}     ## take length as 2nd arg (filename is 1st)
while read -r line; do      ## read each line
    while [ ${#line} -gt 0 ]; do            ## if characters remain
        printf "%sn" "${line:0:$((len))}"  ## print len chars
        line="${line:$((len))}"             ## strip len chars from line
    done
done < "$1"

示例输入文件

$ cat dat/longsent.txt
bfjkjhdfhizhfzibfkjezfzfiuzehfizdjfldfsdfsljfklj
fjlskdjfjoijrgeojrgijgoejrgrjgiorjofgjeirjgoergd

使用/输出示例

default 20-chars per-line:

$ bash wrap.sh dat/longsent.txt
bfjkjhdfhizhfzibfkje
zfzfiuzehfizdjfldfsd
fsljfklj
fjlskdjfjoijrgeojrgi
jgoejrgrjgiorjofgjei
rjgoergd

按每行10字符换行:

$ bash wrap.sh dat/longsent.txt 10
bfjkjhdfhi
zhfzibfkje
zfzfiuzehf
izdjfldfsd
fsljfklj
fjlskdjfjo
ijrgeojrgi
jgoejrgrjg
iorjofgjei
rjgoergd

注意:您应该验证len大于0，并且您可以将|| test -n "$line"添加到第一个while子句中，以容纳在最后一行结束的非posix行(为简洁而省略)。

---

包括行前缀

如果您的数据文件包含前缀，(例如FirstSentence1, ...)，并且您需要在输出中包含这些前缀，则只需在line之前添加prefix的读取，并在每个换行行之前输出prefix(具有相同的字段宽度，左对齐)。例如:

#!/bin/bash
declare -i len=${2:-20}     ## take length as 2nd arg (filename is 1st)
declare -i wdth=22          ## set min field width for prefix (so cols align)
while read -r prefix line; do      ## read each line
    while [ ${#line} -gt 0 ]; do   ## if characters remain
        ## print len chars w/prefix width set to wdth, left-justified
        printf "%-*s %sn" $wdth "$prefix" "${line:0:$((len))}"
        line="${line:$((len))}"    ## strip len chars from line
    done
done < "$1"

示例输入文件w/前缀

$ cat dat/longsentpfx.txt
FirstSentence1   bfjkjhdfhizhfzibfkjezfzfiuzehfizdjfldfsdfsljfklj
SecondSentences2 fjlskdjfjoijrgeojrgijgoejrgrjgiorjofgjeirjgoergd

使用/输出示例

$ bash wrap.sh dat/longsentpfx.txt
FirstSentence1         bfjkjhdfhizhfzibfkje
FirstSentence1         zfzfiuzehfizdjfldfsd
FirstSentence1         fsljfklj
SecondSentences2       fjlskdjfjoijrgeojrgi
SecondSentences2       jgoejrgrjgiorjofgjei
SecondSentences2       rjgoergd
$ bash wrap.sh dat/longsentpfx.txt 10
FirstSentence1         bfjkjhdfhi
FirstSentence1         zhfzibfkje
FirstSentence1         zfzfiuzehf
FirstSentence1         izdjfldfsd
FirstSentence1         fsljfklj
SecondSentences2       fjlskdjfjo
SecondSentences2       ijrgeojrgi
SecondSentences2       jgoejrgrjg
SecondSentences2       iorjofgjei
SecondSentences2       rjgoergd

如果你有其他问题，请告诉我。

注意:将宽度设置为比最长的prefix晚一个字符，您需要在实际编写换行行之前读取所有prefix值以找到最长的宽度，然后添加+1。如果您的数据文件很短，您可以将前缀和行读入一对索引数组并首先扫描前缀数组的长度，如果数据文件很大，您可以扫描文件两次(不是最佳的)，或者您可以像上面那样设置一些预定的宽度。

Answer 2

使用substr:

awk '{ for(i=0;i<length($2);i=i+20) print $1,substr($2,i,20) }' file

Answer 3

对于您的示例，您可以:

awk '{n=patsplit($2, a, /.{1,20}/); for(i=1;i<=n;i++) print $1, a[i] }' file