[英]Run Linux shell script with arguments
Can somone help me with this: so i have this script 有人可以帮我吗:所以我有这个脚本
#!/bin/bash
echo -n "Enter a value for X:(999 to exit): "
read x
until [[ $x == 999 ]]
do
echo -n "Enter a value for Y: "
read y
echo "X="$x
echo "Y="$y
((a=y+x))
echo "X+Y="$a
((s=y-x))
echo "X-Y="$s
((m=y*x))
echo "X*Y="$m
((d=y/x))
echo "X/Y="$d
((m=y%x))
echo "X%Y="$m
echo -n "Enter a value for X:(999 to exit): "
read x
if [[ $x == 999 ]];
then
exit 0
fi
done
exit 0
but i didnt know how to write the rest of it, the missing thing is: Use the two command line arguments when the script starts if the user supplied them, and then prompt for more numbers to continue in the loop. 但我不知道如何编写其余内容,所缺少的是:如果用户提供了脚本,则在脚本启动时使用两个命令行参数,然后提示更多数字继续循环。
Am guessing the arguments you are looking from the user are x
and y
values. 我猜您从用户那里看到的参数是
x
和y
值。 The easiest way to check if user provided arguments is to use $#
which gets you the number of arguments given by the user. 检查用户提供的参数是否最简单的方法是使用
$#
,它使您获得用户提供的参数数量。
So use it like this: 所以像这样使用它:
if [ "$#" -eq 2 ]; #2 arguments provided by user
then
x=$1
...
fi
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