bash bash循环跳过注释行
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bash loop skip commented lines
提问by exvance
I'm looping over lines in a file. I just need to skip lines that start with "#". How do I do that?
我正在遍历文件中的行。我只需要跳过以“#”开头的行。我怎么做？
#!/bin/sh
while read line; do
if ["$line doesn't start with #"];then
echo "line";
fi
done < /tmp/myfile
Thanks for any help!
谢谢你的帮助！
回答by pilcrow
while read line; do
case "$line" in \#*) continue ;; esac
...
done < /tmp/my/input
Frankly, however, it is often clearer to turn to grep:
然而，坦率地说，转向grep：
grep -v '^#' < /tmp/myfile | { while read line; ...; done; }
回答by Slav
This is an old question but I stumbled upon this problem recently, so I wanted to share my solution as well.
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。
If you are not against using some python trickery, here it is:
如果你不反对使用一些 python 技巧，这里是：
Let this be our file called "my_file.txt":
让它成为我们名为“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
Let this be our bash script called "my_script.sh":
让这成为我们的 bash 脚本“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)
Calling the script will produce something similar to:
调用脚本将产生类似于：
$ ./my_script.sh
this line will print
this will also print
Note:the blank line was not printed
注意：未打印空行
If you want blank lines you can change the script to:
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)
Calling this script will produce something similar to:
调用此脚本将产生类似于：
$ ./my_script.sh
this line will print
this will also print
