为什么我的变量在我的 bash while 循环中似乎没有增加？
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Why doesn't my variable seem to increment in my bash while loop?
提问by wwwilliam
I am fairly new to bash scripting. I can't seem to get the correct value of my counting variables to display at the end of of a whileloop in my bash script.
我对 bash 脚本很陌生。我似乎无法while在我的 bash 脚本中的循环结束时获得要显示的计数变量的正确值。
Background: I have a fairly simple task: I would like to pass a text file containing a list of file paths to a bash script, have it check for the existence of those files, and count the number of existing/missing files. I got most of the script to work, except for the counting part.
背景：我有一个相当简单的任务：我想将一个包含文件路径列表的文本文件传递给 bash 脚本，让它检查这些文件是否存在，并计算现有/丢失文件的数量。除了计数部分，我让大部分脚本都可以工作。
N=0
correct=0
incorrect=0
cat | while read filename ; do
N=$((N+1))
echo "$N"
if ! [ -f $filename ]; then
incorrect=$((incorrect+1))
else
correct=$((correct+1))
fi
done
echo "# of Correct Paths: $correct"
echo "# of Incorrect Paths: $incorrect"
echo "Total # of Files: $N"
If I have a list of 5 files, 4 of which exist, I expect to get the following output (note the echocommand within the whileloop):
如果我有 5 个文件的列表，其中 4 个存在，我希望得到以下输出（注意循环中的echo命令while）：
1
2
3
4
5
# of Correct Paths: 4
# of Incorrect Paths: 1
Total # of Files: 5
Instead, I get:
相反，我得到：
1
2
3
4
5
# of Correct Paths: 0
# of Incorrect Paths: 0
Total # of Files: 0
What happened to the values of these variables? Google had many suggestions of questionable quality and I think I could get it to work with a little more searching, but a brief explanation of what I'm doing wrong would be very helpful.
这些变量的值发生了什么变化？谷歌有很多质量有问题的建议，我想我可以通过更多的搜索来让它工作，但简要说明我做错了什么会非常有帮助。
回答by kurumi
This is because you are using the useless catcommand with a pipe, causing a subshell to be created. Try it without the cat:
这是因为您使用了cat带有管道的无用命令，导致创建了一个子 shell。尝试没有cat：
while read filename ; do
N=$((N+1))
....
done < file
回答by jlliagre
Alternatively, if you want to keep the catfor some reason, you can fix your script simply by adding this line before the catinstruction:
或者，如果您cat出于某种原因想保留，您可以通过在cat指令前添加以下行来修复您的脚本：
shopt -s lastpipe
回答by Stephen Niedzielski
More generally, sometimes you want to pipe the output of a command. Here's an example that uses process substitutionto lint JavaScript files about to be committed by Git and counts the number of files that failed:
更一般地，有时您希望通过管道传输命令的输出。下面是一个示例，它使用进程替换来 lint 即将由 Git 提交的 JavaScript 文件并计算失败的文件数：
# $@ glob
git-staged-files() {
git diff --cached -C -C -z --name-only --relative --diff-filter=ACMRTUXB "$@"
}
# $@ name
map() { IFS= read -rd $'##代码##' "$@"; }
declare -i errs=0
while map file; do
echo "Checking $file..."
git show ":$file"|
eslint --stdin --stdin-filename "$file" || ((++errs))
done < <(git-staged-files \*.js)
((errs)) && echo -en "\e[31m$errs files with errors.\e[00m " >&2 || :
