用bash删除空白部分

问题描述：

我正在寻找一种方法来删除markdown文档中的空白部分，更具体地说，更新日志。用bash删除空白部分

举例来说，如果我有：

## Version 

### Added 
- something 

### Removed 

### Changed 

- something  

### Fixed

我想直到结束：

## Version 

### Added 
- something 

### Changed 

- something

注意Fixed节走了（空，直到文件的结尾）和Version节仍然存在，因为我只应在###部分采取行动。

为什么downvoting？ –

答

在平原AWK：

#!/usr/bin/awk 

BEGIN { 
    inside_empty_section = "false" 
    buffer = "" 
} 

/^$/ { # Add the empty line to the current section buffer or print it 
    if (inside_empty_section == "true") { buffer = buffer $0 "\n" } else { print $0 }; next } 

/^###/ { 
    # This is the beginning of a new section. 
    # Either the previous one was empty: just forget its buffer. 
    # either it was not empty and has already been printed. 
    # In any case, just start buffering a new empty section. 
    inside_empty_section = "true" 
    buffer = $0 
    next 
} 

{ 
    # Found a non-empty line: the current section is NOT empty. 
    # If it was supposed to be empty, print its buffer. 
    if (inside_empty_section == "true") { print buffer } 
    inside_empty_section = "false" 
    print $0 
    next 
}

与启动它：

$ awk -f ./test.awk < plop 
## Version 

### Added 
- something 

### Changed 

- something

当然，你可以把它更简洁：

awk '/^$/ {if (i) {b=b $0 "\n"} else {print $0 }; next} \ 
/^###/ {i=1; b=$0; next} {if (i) {print b}; i=0; print $0; next}' < plop

所有解决方案都非常好，但我最喜欢这个。我将'/^### /'修改为'/^### /'，以确保只考虑h3标题。 –

答

awk中溶液：

awk -v RS="###" 'NF>1{ printf "%s%s",(NR==1? "" : RS),$0 }' file

RS="###" - 考虑###如记录分隔符
NF>1 - 如果记录包含至少2个字段（包括部名）

输出：

## Version 

### Added 
- something 

### Changed 

- something

不错，简单而优雅。但是，对于出现在其他地方而不是-3级开始部分的###，绝对不是很健壮（例如'####小节或' - 某些###'） –

我同意：好，简单和优雅，但如果一个部分名称可能包含多个单词，则不起作用。 – duthen

答

在普通的bash：

$ cat foo.sh 
#!/usr/bin/env bash 
declare -i p=1 # to-print or not-to-print flag 
declare w="" # buffer 
declare l="" # line 
while read l; do 
    if [[ $l =~ ^###[^#] ]]; then # if new section 
     ((p)) && printf "%s" "$w" # if to-print, print buffer 
     w="$l"$'\n'     # re-initialize buffer 
     p=0       # reset to-print flag 
    else 
     w="$w$l"$'\n'    # update buffer 
     [[ -n $l ]] && p=1  # if non-empty line, set to-print flag 
    fi 
done < $1 
((p)) && printf "%s" "$w" # if to-print, print buffer 

$ ./foo.sh foo.md 
## Version 

### Added 
- something 

### Changed 

- something

你可以，如果你的部分和空部分的定义是不完全的读线的两个测试适应我所假设的。

用bash删除空白部分

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