Bash while 循环，如何读取输入直到条件为假
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Bash while loop, how to read input until a condition is false
提问by Tyler Pfaff
I keep getting a run time error. I'm running this in a terminal on OSX. The error is,
我不断收到运行时错误。我在 OSX 上的终端中运行它。错误是，
test.sh: line 15: while[!false]: command not found
test.sh: line 16: syntax error near unexpected token `do'
test.sh: line 16: `do'
I just can't figure where I've got wrong syntactically as I'm new to writing bash scripts.
因为我是编写 bash 脚本的新手，所以我无法弄清楚我在语法上哪里出错了。
ipa build &
TASK_PID=$!
sleep 5
kill $TASK_PID
finished=false
declare -a schemes
echo "*****************************************************************************************"
echo "| View the list of available build configs above."
echo "| Enter the name of the build you want,one at a time."
echo "| Type \"done\" to finish entering scheme names"
echo "*****************************************************************************************"
while[!${finished}]
do
read input
if[$input == "done"]
then
finished=true
else
schemes=("${schemes[@]}" $input)
echo ${schemes[0]}
fi
done
echo "Do you want a verbose build? (y/n)"
read verbose
echo "Building your selected schemes....."
ipa build -s ${schemes[0]}
回答by chepner
trueand falseare not boolean keywords in bash; they are simply strings (and the names of commands; more on that in a moment). Even if you fix your syntax by supplying whitespace where necessary:
true并且false不是bash; 中的布尔关键字 它们只是字符串（以及命令的名称；稍后会详细介绍）。即使您通过在必要时提供空格来修复语法：
while ! [ "${finished}" ]; do
...
done
this loop will never run. Why? Whether finishedhas the value trueor false, it is simply a non-empty string. This code will run the [command (yes, it's a command, not syntax) and succeed because its argument is a non-empty string. The !negates it, so that the condition for the whileloop then alwaysfails.
这个循环永远不会运行。为什么？无论是否finished具有值trueor false，它都只是一个非空字符串。此代码将运行[命令（是的，它是一个命令，而不是语法）并成功，因为它的参数是一个非空字符串。The!否定它，因此while循环的条件总是失败。
The most direct fix is to explicitly compare $finishedto the string "true".
最直接的解决方法是明确$finished地与字符串“true”进行比较。
while [ "$finished" != "true" ]; do
...
done
I mentioned that trueand falseare also commands: truealways succeeds, and falsealways fails. Usually, you do not want to do what I am about to suggest, but here it's OK because trueand falseare about as simple a pair of commands as you can imagine.
我提到，true和false也命令：true总是成功，false总是失败。通常，您不想执行我将要建议的操作，但是这里可以，因为true和false是您可以想象的简单命令对。
finished=false
while ! $finished; do
...
# At some point
finished=true
done
Here, we are letting $finishedexpand to the name of a command, which then executes and has its exit status negated by the !. As long as finished=false, the negated exit status is always 0 and the while loop will continue to run. Once you change the value of finished, the negated exit status will be 1 and the loop will exit.
在这里，我们让$finished扩展为命令的名称，然后执行该命令并使其退出状态由!. 只要finished=false，否定退出状态始终为 0，while 循环将继续运行。一旦更改 的值finished，否定退出状态将为 1，循环将退出。
回答by Amit Verma
Give space around brackets in test conditions
在测试条件下在括号周围留出空间
while [ ! ${finished} ]
&
&
if [ $input = "done" ]
回答by Emmet
Why not try something like this:
为什么不尝试这样的事情：
#!/bin/bash
list='Foo Bar Baz Quux Xyzzy QUIT'
select item in $list; do
case $item in
QUIT)
break
;;
*)
echo "You picked '$item'!"
;;
esac
done
