bash - bash循环跳过注释行

Question

我正在循环文件中的行。我只需要跳过以“#”开头的行。我怎么做？

 #!/bin/sh 

 while read line; do
    if ["$line doesn't start with #"];then
     echo "line";
    fi
 done < /tmp/myfile

谢谢你的帮助！

Answer 1

while read line; do
  case "$line" in \#*) continue ;; esac
  ...
done < /tmp/my/input

然而，坦率地说，转向grep：

grep -v '^#' < /tmp/myfile | { while read line; ...; done; }

Answer 2

这是一个老问题，但我最近偶然发现了这个问题，所以我也想分享我的解决方案。

如果你不反对使用一些 python 技巧，这里是：

让它成为我们的名为“my_file.txt”的文件：

this line will print
this will also print # but this will not
# this wont print either
      # this may or may not be printed, depending on the script used, see below

让它成为我们的 bash 脚本，名为“my_script.sh”：

#!/bin/sh

line_sanitizer="""import sys
with open(sys.argv[1], 'r') as f:
    for l in f.read().splitlines():
        line = l.split('#')[0].strip()
        if line:
            print(line)
"""
echo $(python -c "$line_sanitizer" ./my_file.txt)

调用脚本将产生类似于：

$ ./my_script.sh
this line will print
this will also print

注意：空白行没有打印

如果您想要空行，您可以将脚本更改为：

#!/bin/sh

line_sanitizer="""import sys
with open(sys.argv[1], 'r') as f:
    for l in f.read().splitlines():
        line = l.split('#')[0]
        if line:
            print(line)
"""
echo $(python -c "$line_sanitizer" ./my_file.txt)

调用此脚本将产生类似于：

$ ./my_script.sh
this line will print
this will also print