我尝试创建类似12DEC2013bhav.csv.zip
以下脚本的输出。
但它给了我类似的东西12DEC.csv.zip
:
for (( i = 2013; i <= 2014; i++ ))
do
for m in JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC
do
for (( d = 1; d <= 31; d++))
do
echo "$d$m$ibhav.csv.zip"
done
done
done
我该如何改正它?
答案1
问题是您正在尝试引用一个名为$ibhav
(not $i
) 的变量。
变量可以包含多个字符,在您的示例中,shell 无法判断您的意思是$i
或$ibhav
(或$ibha
或$ibh
或$ib
)。
解决方法是将变量名括起来:
echo "${d}${m}${i}bhav.csv.zip"
这样就明确了您引用的是哪个变量。
从man bash
:
Parameter Expansion
The `$' character introduces parameter expansion, command substitution, or
arithmetic expansion. The parameter name or symbol to be expanded may be
enclosed in braces, which are optional but serve to protect the variable
to be expanded from characters immediately following it which could be
interpreted as part of the name.
When braces are used, the matching ending brace is the first `}' not
escaped by a backslash or within a quoted string, and not within an embed‐
ded arithmetic expansion, command substitution, or parameter expansion.
${parameter}
The value of parameter is substituted. The braces are required
when parameter is a positional parameter with more than one digit,
or when parameter is followed by a character which is not to be
interpreted as part of its name. The parameter is a shell parame‐
ter as described above PARAMETERS) or an array reference (Arrays).