我正在编写我的第一个bash脚本,并且不断出错,而且我不确定自己哪里出错了。下面是我要执行的脚本:
#!/bin/ksh
#Script Name: printnum.sh
# Verify the number of arguments and exit if not equal to 1`enter code here`
$mynum = "5"
echo $mynum
if [$mynum -gt 1]
then
printf "error: program must be executed with 1 argument\n"
printf "usage: $0 value (where value >= 1)\n"
exit 1
fi
# Verify argument is a positive number
if [$mynum -lt 1]
then
printf "error: argument must be a positive number\n"
printf "usage: $0 value (where value >= 1)\n"
fi
# Store command line argument in variable i
$mynum="$i"
# Loop and print $i while decrementing variable to =1 (with comma)
while [$i -gt 1]
do
printf "$i, "
done
以下是我遇到的错误:
./printnum.sh[3]: =: not found [No such file or directory]
./printnum.sh[5]: [: ']' missing
./printnum.sh[11]: [: ']' missing
./printnum.sh[16]: =: not found [No such file or directory]`enter code here`
./printnum.sh[17]: [: ']' missing
/export/home/hanko01/HOME/itec400/homework>
这里的任何帮助将不胜感激!
答案 0 :(得分:2)
首先要注意的是:
您可以从更正的代码中了解其余信息:
#!/bin/bash
#Script Name: printnum.sh
# Verify the number of arguments and exit if not equal to 1 `enter code here`
if [ $# -ne 1 ]
then
printf "error: program must be executed with 1 argument\n"
printf "usage: $0 value (where value >= 1)\n"
exit 1
fi
# Verify argument is a positive number
if [ $1 -lt 1 ]
then
printf "error: argument must be a positive number\n"
printf "usage: $0 value (where value >= 1)\n"
fi
# Store command line argument in variable i
i=$1
# Loop and print $i while decrementing variable to =1 (with comma)
while [ $i -gt 1 ]
do
printf "$i, "
i=$((i-1))
done