我想写一个循环15个字符串的脚本(可能是数组?)这可能吗?
类似于:
for databaseName in listOfNames
then
# Do something
end
我想写一个循环15个字符串的脚本(可能是数组?)这可能吗?
类似于:
for databaseName in listOfNames
then
# Do something
end
当前回答
您可以使用${arrayName[@]}的语法
#!/bin/bash
# declare an array called files, that contains 3 values
files=( "/etc/passwd" "/etc/group" "/etc/hosts" )
for i in "${files[@]}"
do
echo "$i"
done
其他回答
listOfNames="db_one db_two db_three"
for databaseName in $listOfNames
do
echo $databaseName
done
或者只是
for databaseName in db_one db_two db_three
do
echo $databaseName
done
我在GitHub更新中使用了这种方法,我发现它很简单。
## declare an array variable
arr_variable=("kofi" "kwame" "Ama")
## now loop through the above array
for i in "${arr_variable[@]}"
do
echo "$i"
done
您可以使用带有三个表达式(C样式)的计数器遍历bash数组值,以读取循环语法的所有值和索引:
declare -a kofi=("kofi" "kwame" "Ama")
# get the length of the array
length=${#kofi[@]}
for (( j=0; j<${length}; j++ ));
do
print (f "Current index %d with value %s\n" $j "${kofi[$j]}")
done
脚本或函数的隐式数组:
除了anubhava的正确答案:如果循环的基本语法是:
for var in "${arr[@]}" ;do ...$var... ;done
bash中有一个特殊情况:
当运行脚本或函数时,在命令行传递的参数将被分配给$@数组变量,您可以通过$1、$2、$3等进行访问。
可以通过以下方式填充(用于测试)
set -- arg1 arg2 arg3 ...
这个数组上的循环可以简单地写:
for item ;do
echo "This is item: $item."
done
请注意,中的保留工作不存在,也没有数组名称!
示例:
set -- arg1 arg2 arg3 ...
for item ;do
echo "This is item: $item."
done
This is item: arg1.
This is item: arg2.
This is item: arg3.
This is item: ....
注意,这与
for item in "$@";do
echo "This is item: $item."
done
然后进入脚本:
#!/bin/bash
for item ;do
printf "Doing something with '%s'.\n" "$item"
done
将其保存在脚本myscript.sh、chmod+x myscript.shh中,然后
./myscript.sh arg1 arg2 arg3 ...
Doing something with 'arg1'.
Doing something with 'arg2'.
Doing something with 'arg3'.
Doing something with '...'.
在函数中相同:
myfunc() { for item;do cat <<<"Working about '$item'."; done ; }
Then
myfunc item1 tiem2 time3
Working about 'item1'.
Working about 'tiem2'.
Working about 'time3'.
当然,这是可能的。
for databaseName in a b c d e f; do
# do something like: echo $databaseName
done
有关详细信息,请参阅、while和until的Bash循环。
单线循环,
declare -a listOfNames=('db_a' 'db_b' 'db_c')
for databaseName in ${listOfNames[@]}; do echo $databaseName; done;
你会得到这样的输出,
db_a
db_b
db_c