[英]How to display argument in shell script
I have a shell script called displayArg.sh This is how I intend to run it- 我有一个名为displayArg.sh的shell脚本这就是我打算运行它的方式 -
./displayArg hello
and the output is entered arg is hello 并输入输出arg是你好
The following is the script- 以下是剧本 -
if [ $1 == "" ]; then
default="Default"
echo "no value is given. Output is $default"
else
value=$?
echo "entered arg is $value" #I know I am wrong in these 2 lines, but not sure how to fix it
fi
Kindly bear with me. 请耐心等待我。 I'm new to Shell scripting 我是Shell脚本的新手
You want: 你要:
value="$1"
( $?
is the status of the last command, which is 1 because the test command is what was executed last.) ( $?
是最后一个命令的状态,为1,因为测试命令是最后执行的命令。)
Or you can simplify to: 或者您可以简化为:
if [ "$1" == "" ]
then
echo "no value is given. Output is Default"
else
echo "entered arg is $1"
fi
Note the quotes around "$1"
in the test. 请注意测试中"$1"
左右的引号。 If the string is empty, you get a syntax error. 如果字符串为空,则会出现语法错误。 Your alternative with bash
is to use a [[ $1 == "" ]]
test. 使用bash
替代方法是使用[[ $1 == "" ]]
测试。
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