bash bash脚本将文件变量读入局部变量
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bash script read file variables into local variables
提问by thegreat078
I have a config file with the following content:
我有一个包含以下内容的配置文件：
msgs.config:
msgs.config:
tmsg:This is Title Message!
t1msg:This is T1Message.
t2msg:This is T2Message.
pmsg:This is personal Message!
I am writing a bash script that reads the msgs.config file variables and stores them into local variables. I will use these throughout the script. Due to permission I do not want to use the .method (source).
我正在编写一个 bash 脚本，它读取 msgs.config 文件变量并将它们存储到局部变量中。我将在整个脚本中使用这些。由于许可，我不想使用该.方法（来源）。
tmsg
t1msg
t2msg
pmsg
Any help would be greatly appreciated.
任何帮助将不胜感激。
回答by Jonathan Leffler
You can use:
您可以使用：
oldIFS="$IFS"
IFS=":"
while read name value
do
# Check value for sanity? Name too?
eval $name="$value"
done < $config_file
IFS="$oldIFS"
Alternatively, you can use an associative array:
或者，您可以使用关联数组：
declare -A keys
oldIFS="$IFS"
IFS=":"
while read name value
do
keys[$name]="$value"
done < $config_file
IFS="$oldIFS"
Now you can refer to ${keys[tmsg]}etc to access the variables. Or, if the list of variables is fixed, you can map the values to variables:
现在您可以参考${keys[tmsg]}etc 来访问变量。或者，如果变量列表是固定的，您可以将值映射到变量：
tmsg="${keys[tmsg]}"
回答by AnBisw
Read the file and store the values-
读取文件并存储值-
i=0
config_file="/path/to/msgs.config"
while read line
do
if [ ! -z "$line" ] #check if the line is not blank
then
key[i]=`echo $line|cut -d':' -f1` #will extract tmsg from 1st line and so on
val[i]=`echo $line|cut -d':' -f2` #will extract "This is Title Message!" from line 1 and so on
((i++))
fi
done < $config_file
Access the array variables as ${key[0]},${key[1]},.... and ${val[0]},${val[1]}...
访问数组变量为${key[0]}, ${key[1]},.... 和${val[0]}, ${val[1]}...
回答by chepner
In case you change your mind about source:
如果您改变主意source：
source <( sed 's/:\(.*\)/=""/' msgs.config )
This does not work if any of your values have double quotes.
如果您的任何值有双引号，这将不起作用。
