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
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
其他回答
我真正需要的是这样的东西:
for i in $(the_array); do something; done
例如:
for i in $(ps -aux | grep vlc | awk '{ print $2 }'); do kill -9 $i; done
(将终止所有以vlc命名的进程)
脚本或函数的隐式数组:
除了anubhava的正确答案:如果循环的基本语法是:
for var in "${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 '$item'."; done ; }
Then
myfunc item1 tiem2 time3
Working about 'item1'.
Working about 'tiem2'.
Working about 'time3'.
这也很容易阅读:
FilePath=(
"/tmp/path1/" #FilePath[0]
"/tmp/path2/" #FilePath[1]
)
#Loop
for Path in "${FilePath[@]}"
do
echo "$Path"
done
本着与4ndrew的回答相同的精神:
listOfNames="RA
RB
R C
RD"
# To allow for other whitespace in the string:
# 1. add double quotes around the list variable, or
# 2. see the IFS note (under 'Side Notes')
for databaseName in "$listOfNames" # <-- Note: Added "" quotes.
do
echo "$databaseName" # (i.e. do action / processing of $databaseName here...)
done
# Outputs
# RA
# RB
# R C
# RD
B.名称中无空格:
listOfNames="RA
RB
R C
RD"
for databaseName in $listOfNames # Note: No quotes
do
echo "$databaseName" # (i.e. do action / processing of $databaseName here...)
done
# Outputs
# RA
# RB
# R
# C
# RD
笔记
在第二个示例中,使用listOfNames=“RA RB R C RD”具有相同的输出。
其他引入数据的方法包括:
stdin(如下所列),变量,数组(接受的答案),文件。。。
从stdin读取
# line delimited (each databaseName is stored on a line)
while read databaseName
do
echo "$databaseName" # i.e. do action / processing of $databaseName here...
done # <<< or_another_input_method_here
可以在脚本中指定bash IFS“字段分隔符到行”[1]分隔符,以允许其他空格(即IFS='\n',或MacOS IFS='\r')我也喜欢接受的答案:)--我将这些片段作为其他有用的方式来回答这个问题。包括#/脚本文件顶部的bin/bash指示执行环境。我花了几个月的时间才弄清楚如何简单地编写代码:)
其他来源(读取循环时)
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