什么$？和$＃立场在Shell脚本？ [英] What do $? and $# stand for in Shell Script?

问题描述

我有code以下两个大块，我不能肯定什么$？和$＃立场。请帮助。

I have the following two chunks of code, I am not so sure what $? and $# stand for. Please help.

code ONE

 #!/bin/bashUSERID="$1" 
 /bin/id $USERID 2>/dev/null
[ $? -eq 0 ] && echo "User found" || echo "User not found"

/bin/id -g $USERID 2>/dev/null
[ $? -eq 0 ] && echo "Group found" || echo "Group not found"    `


$ cat > mtable

code两个

#!/bin/sh
#
#Script to test for loop
#
#
if [ $# -eq 0 ]
then
echo "Error - Number missing form command line argument"
echo "Syntax : $0 number"
echo "Use to print multiplication table for given number"
exit 1
fi
n=$1
for i in 1 2 3 4 5 6 7 8 9 10
do
echo "$n * $i = `expr $i \* $n`"
done

感谢您

推荐答案

$？为您提供了previous命令的返回code执行。

$? gives you the return code of the previous command executed.

$＃给你提供给脚本的命令行参数的个数。

$# give you the number of command line arguments given to the script.

所以，基本上，如果条件检查，如果给定的参数数分别为合适。

So, basically that if condition checks if the number of arguments given were proper or not.

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