在 bash 中更改变量
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Change variables in bash
提问by pkruk
How do I change this var ?
如何更改此 var ？
max=0;
min=20000000;
cat |while read
do
read a
if [[ $a -gt $max ]]
then
max=a`
fi
`if [[ $a -lt $min ]]
then
min=a
fi
done
echo $max
echo $min
My min and max are still the same, 0 and 2000000. Can anybody help me with this ? I have no idea.
我的最小值和最大值仍然相同，0 和 2000000。有人可以帮助我吗？我不知道。
回答by Mat
The (main) problem with your script is that setting minand maxhappens in a subshell, not your main shell. So the changes aren't visible after the pipeline is done.
您的脚本的（主要）问题是设置min和max发生在子 shell 中，而不是您的主 shell。因此，管道完成后更改不可见。
Another one is that you're calling read twice - this might be intended if you want to skip every other line, but that's a bit unusual.
另一个是您调用 read 两次 - 如果您想跳过每隔一行，这可能是有意的，但这有点不寻常。
The last one is that min=asets minto a, literally. You want to set it to $a.
最后一个是min=a设置min为a，字面意思。您想将其设置为$a.
Using process substitutionto get rid of the first problem, removing the (possibly) un-necessary second read, and fixing the assignments, your code should look like:
使用进程替换来解决第一个问题，删除（可能）不必要的第二次读取，并修复分配，您的代码应如下所示：
max=0
min=20000000
while read a
do
if [[ $a -gt $max ]]
then
max=$a
fi
if [[ $a -lt $min ]]
then
min=$a
fi
done < <(cat) # careful with the syntax
echo $max
echo $min
