bash 如何在bash脚本中正确循环？
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How to while loop correctly in bash script?
提问by frodo
I have wrote this script which takes a stop and start number and counts out the numbers in between - I am yting to get it to keep increasing no matter whether the "stop" number is reaced or not so it keeps counting until ctrl+z is pressed but it is not recognising the while condition for me - could anyone correct the syntax for me please ?
我写了这个脚本，它需要一个停止和开始的数字，并计算出它们之间的数字 - 我很想让它继续增加，无论“停止”数字是否被重新接收，所以它会一直计数直到 ctrl+z 是按下但它不能识别我的 while 条件 - 任何人都可以为我纠正语法吗？
#!/bin/sh
stopvalue=
startvalue=
if [ $# -ne 2 ]
then
echo "Error - You must enter two numbers exactly - using default start value of 1"
#exit 0
fi
echo ${startvalue:=1}
while (test "$startvalue" -le "$stopvalue" || "$startvalue" -ge "$stopvalue")
do
startvalue=$((startvalue+1))
echo $startvalue
done
回答by Michael Krelin - hacker
Now that you have two answers about the while loop, I'll suggest using forloop instead:
既然您对 while 循环有了两个答案，我建议改用for循环：
for((i=$startvalue;i<=$stopvalue;++i)) do
echo $i
done
回答by Tarun Sharma
I use the simple while loop program. Which works for me. as per my understanding before passing any variable in while loop we have to declare it:
我使用简单的 while 循环程序。这对我有用。根据我的理解，在 while 循环中传递任何变量之前，我们必须声明它：
sum=0
temp=1
while [ $temp -ne 0 ]
do
echo "Enter number: OR 0 to Quit :"
read temp
sum=`expr $sum + $temp`
done
echo $sum
回答by Kevin
Your main problem is in your condition:
您的主要问题在于您的状况：
while (test "$startvalue" -le "$stopvalue" || "$startvalue" -ge "$stopvalue")
You are telling it to continue if $startvalueis (less than or equal to) or (greater than or equal to) $stopvalue. For ANY combination of $startvalueand $stopvalue, the one will be less than, equal to, or greater than the other. Take out the second half so it only continues if $startvalueis less than or equal to $stopvalue.
如果$startvalue是 (小于或等于) 或 (大于或等于)，您告诉它继续$stopvalue。对于任意组合$startvalue和$stopvalue，一个将小于，等于或大于另一个。取出后半部分，以便它仅在$startvalue小于或等于 时继续$stopvalue。
And you should write it:
你应该写它：
while [ "$startvalue" -le "$stopvalue” ]
回答by Andrew Clark
while [ "$startvalue" -le "$stopvalue" -o "$startvalue" -ge "$stopvalue" ]
do
startvalue=$((startvalue+1))
echo $startvalue
done
