将文本文件中的变量加载到 bash 脚本中
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原文地址: http://stackoverflow.com/questions/2652753/
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Loading variables from a text file into bash script
提问by S1syphus
Is it possible to load new lines from a text file to variables in bash?
是否可以将文本文件中的新行加载到 bash 中的变量？
Text file looks like?
文本文件长什么样？
EXAMPLEfoo
EXAMPLEbar
EXAMPLE1
EXAMPLE2
EXAMPLE3
EXAMPLE4
Variables become
变量变成
= EXAMPLEfoo
= EXAMPLEbar
ans so on?
等等？
回答by ghostdog74
$ s=$(<file)
$ set -- $s
$ echo
EXAMPLEfoo
$ echo
EXAMPLEbar
$ echo $@
EXAMPLEfoo EXAMPLEbar EXAMPLE1 EXAMPLE2 EXAMPLE3 EXAMPLE4
I would improve the above by getting rid of temporary variable s:
我将通过摆脱临时变量 s 来改进上述内容：
$ set -- $(<file)
And if you have as input a file like this
如果你有这样的输入文件
variable1 = value
variable2 = value
You can use following construct to get named variables.
您可以使用以下构造来获取命名变量。
input=`cat filename|grep -v "^#"|grep "\c"`
set -- $input
while [ ]
do
eval =
shift 3
done
回答by Андрей Костенко
cat somefile.txt| xargs bash_command.sh
bash_command.sh will receive these lines as arguments
bash_command.sh 将接收这些行作为参数
回答by Paused until further notice.
saveIFS="$IFS"
IFS=$'\n'
array=($(<file))
IFS="$saveIFS"
echo ${array[0]} # output: EXAMPLEfoo
echo ${array[1]} # output: EXAMPLEbar
for i in "${array[@]}"; do echo "$i"; done # iterate over the array
Edit:
编辑：
The loop in your pastebin has a few problems. Here it is as you've posted it:
您的 pastebin 中的循环有一些问题。这是您发布的内容：
for i in "${array[@]}"; do echo " "AD"$count = "$i""; $((count=count+1)); done
Here it is as it should be:
这是它应该的样子：
for i in "${array[@]}"; do declare AD$count="$i"; ((count=count+1)); done
or
或者
for i in "${array[@]}"; do declare AD$count="$i"; ((count++)); done
But why not use the array directly? You could call it AD instead of array and instead of accessing a variable called "AD4" you'd access an array element "${AD[4]}".
但是为什么不直接使用数组呢？您可以将其称为 AD 而不是数组，而不是访问名为“AD4”的变量，而是访问数组元素“${AD[4]}”。
echo "${AD[4]}"
if [[ ${AD[9]} == "EXAMPLE value" ]]; then do_something; fi
回答by AnrDaemon
None of the above will work, if your values are quoted with spaces.
如果您的值用空格引用，则上述方法均无效。
However, not everythinf is lost.
然而，并不是所有的东西都丢失了。
Try this:
尝试这个：
eval "$(VBoxManage showvminfo "$VMname" --details --machinereadable | egrep "^(name|UUID|CfgFile|VMState)")"
echo "$name {$UUID} $VMState ($VMStateChangeTime) CfgFile=$CfgFile"
P.S.
Nothing will ever work, if your names are quoted or contain dashes.
If you have something like that, as is the case with VBoxManage output ("IDE-1-0"="emptydrive" and so on), either egreponly specific values, as shown in my example, or silence the errors.
PS 如果您的名字被引用或包含破折号，则什么都不会起作用。如果你有类似的东西，就像 VBoxManage 输出的情况（“IDE-1-0”=“emptydrive”等等），要么egrep只有特定的值，如我的例子所示，要么消除错误。
However, silencing erors is always dangerous. You never know, when the next value will have unquoted "*" in it, thus you must treat values loaded this way very careful, with all due precaution.
然而，沉默错误总是危险的。你永远不知道，下一个值什么时候会有不带引号的“*”，因此你必须非常小心地对待以这种方式加载的值，并采取一切应有的预防措施。
回答by Andy
This can be done be with an array if you don't require these variables as inputs to a script. push()function lifted from the Advanced Scripting Guide
如果您不需要这些变量作为脚本的输入，则可以使用数组来完成此操作。push()从高级脚本指南中提取的功能
push() # Push item on stack.
{
if [ -z "" ] # Nothing to push?
then
return
fi
let "SP += 1" # Bump stack pointer.
stack[$SP]=
return
}
The contents of /tmp/test
的内容 /tmp/test
[root@x~]# cat /tmp/test
EXAMPLEfoo
EXAMPLEbar
EXAMPLE1
EXAMPLE2
EXAMPLE3
EXAMPLE4
SP=0; for i in `cat /tmp/test`; do push $i ; done
Then
然后
[root@x~]# echo ${stack[3]}
EXAMPLE1
