Arrays 在Bash中循环字符串数组?
我想写一个循环15个字符串的脚本(可能是数组?) 比如:Arrays 在Bash中循环字符串数组?,arrays,bash,shell,Arrays,Bash,Shell,我想写一个循环15个字符串的脚本(可能是数组?) 比如: for databaseName in listOfNames then # Do something end 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。您可以这样使用它: ## declare an array variable declare -a arr=("e
for databaseName in listOfNames
then
# Do something
end
当然,这是可能的
for databaseName in a b c d e f; do
# do something like: echo $databaseName
done
有关详细信息,请参阅。您可以这样使用它:
## declare an array variable
declare -a arr=("element1" "element2" "element3")
## now loop through the above array
for i in "${arr[@]}"
do
echo "$i"
# or do whatever with individual element of the array
done
# You can access them using echo "${arr[0]}", "${arr[1]}" also
也适用于多行数组声明
declare -a arr=("element1"
"element2" "element3"
"element4"
)
声明数组不适用于Korn shell。对Korn shell使用以下示例:
promote_sla_chk_lst="cdi xlob"
set -A promote_arry $promote_sla_chk_lst
for i in ${promote_arry[*]};
do
echo $i
done
试试这个。它正在工作并经过测试
for k in "${array[@]}"
do
echo $k
done
# For accessing with the echo command: echo ${array[0]}, ${array[1]}
本着与4ndrew的回答相同的精神:
listOfNames="RA
RB
R C
RD"
# To allow for other whitespace in the string:
# 1. add double quotes around the list variable, or
# 2. see the IFS note (under 'Side Notes')
for databaseName in "$listOfNames" # <-- Note: Added "" quotes.
do
echo "$databaseName" # (i.e. do action / processing of $databaseName here...)
done
# Outputs
# RA
# RB
# R C
# RD
注释
for Item in Item1 Item2 Item3 Item4 ;
do
echo $Item
done
for Item in Item1 \
Item2 \
Item3 \
Item4
do
echo $Item
done
listOfNames=“RA RB R C RD”
具有相同的输出- 标准DIN(如下所列)
- ,
- (被接受的答案)
#行分隔(每个数据库名存储在一行中)
读取数据库名时
做
echo“$databaseName”#即在此处执行$databaseName的操作/处理。。。
完成#您可以使用${arrayName[@]}
#!/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
这些答案中没有一个包含计数器
#!/bin/bash
## declare an array variable
declare -a array=("one" "two" "three")
# get length of an array
arraylength=${#array[@]}
# use for loop to read all values and indexes
for (( i=0; i<${arraylength}; i++ ));
do
echo "index: $i, value: ${array[$i]}"
done
如果您使用的是Korn shell,则有“设置-A databaseName”,否则有“声明-A databaseName”
要在所有shell上编写脚本
set -A databaseName=("db1" "db2" ....) ||
declare -a databaseName=("db1" "db2" ....)
# now loop
for dbname in "${arr[@]}"
do
echo "$dbname" # or whatever
done
它应该适用于所有外壳。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
每个Bash脚本/会话的可能第一行:
say() { for line in "${@}" ; do printf "%s\n" "${line}" ; done ; }
使用例如:
$ aa=( 7 -4 -e ) ; say "${aa[@]}"
7
-4
-e
可以考虑:echo
将-e
解释为此处的选项
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 $databaseName
done
这与user2533809的回答类似,但每个文件都将作为单独的命令执行
#!/bin/bash
names="RA
RB
R C
RD"
while read -r line; do
echo line: "$line"
done <<< "$names"
#/bin/bash
names=“RA
铷
R C
RD“
而read-r行;做
回音行:“$line”
完成我循环浏览我的项目数组,以进行git pull
更新:
#!/bin/sh
projects="
web
ios
android
"
for project in $projects do
cd $HOME/develop/$project && git pull
end
单线循环
declare -a listOfNames=('db_a' 'db_b' 'db_c')
for databaseName in ${listOfNames[@]}; do echo $databaseName; done;
您将得到这样的输出
db_a
db_b
db_c
是
for Item in Item1 Item2 Item3 Item4 ;
do
echo $Item
done
for Item in Item1 \
Item2 \
Item3 \
Item4
do
echo $Item
done
输出:
index: 0, value: one
index: 1, value: two
index: 2, value: three
Item1
Item2
Item3
Item4
Item 1
Item 2
Item 3
Item 4
Item1
Item2
Item3
Item4
Item1 Item2 Item3
Item1
Item2
Item3
element 1
element 2
element 3
Argentina is in America
Vietnam is in Asia
France is in Europe
Item 1
Item 2
Item 3
保留空间;单引号或双引号列表条目和双引号列表扩展。
for Item in 'Item 1' 'Item 2' 'Item 3' 'Item 4' ;
do
echo "$Item"
done
输出:
index: 0, value: one
index: 1, value: two
index: 2, value: three
Item1
Item2
Item3
Item4
Item 1
Item 2
Item 3
Item 4
Item1
Item2
Item3
Item4
Item1 Item2 Item3
Item1
Item2
Item3
element 1
element 2
element 3
Argentina is in America
Vietnam is in Asia
France is in Europe
Item 1
Item 2
Item 3
多行列表
for Item in Item1 Item2 Item3 Item4 ;
do
echo $Item
done
for Item in Item1 \
Item2 \
Item3 \
Item4
do
echo $Item
done
输出:
index: 0, value: one
index: 1, value: two
index: 2, value: three
Item1
Item2
Item3
Item4
Item 1
Item 2
Item 3
Item 4
Item1
Item2
Item3
Item4
Item1 Item2 Item3
Item1
Item2
Item3
element 1
element 2
element 3
Argentina is in America
Vietnam is in Asia
France is in Europe
Item 1
Item 2
Item 3
简单列表变量
List=( Item1 Item2 Item3 )
或
显示列表变量:
echo ${List[*]}
for Item in ${List[*]}
do
echo $Item
done
Loop(){
for item in ${*} ;
do
echo ${item}
done
}
Loop ${List[*]}
declare -a List=(
"element 1"
"element 2"
"element 3"
)
for entry in "${List[@]}"
do
echo "$entry"
done
declare -A continent
continent[Vietnam]=Asia
continent[France]=Europe
continent[Argentina]=America
for item in "${!continent[@]}";
do
printf "$item is in ${continent[$item]} \n"
done
`
输出:
index: 0, value: one
index: 1, value: two
index: 2, value: three
Item1
Item2
Item3
Item4
Item 1
Item 2
Item 3
Item 4
Item1
Item2
Item3
Item4
Item1 Item2 Item3
Item1
Item2
Item3
element 1
element 2
element 3
Argentina is in America
Vietnam is in Asia
France is in Europe
Item 1
Item 2
Item 3
循环浏览列表:
echo ${List[*]}
for Item in ${List[*]}
do
echo $Item
done
Loop(){
for item in ${*} ;
do
echo ${item}
done
}
Loop ${List[*]}
declare -a List=(
"element 1"
"element 2"
"element 3"
)
for entry in "${List[@]}"
do
echo "$entry"
done
declare -A continent
continent[Vietnam]=Asia
continent[France]=Europe
continent[Argentina]=America
for item in "${!continent[@]}";
do
printf "$item is in ${continent[$item]} \n"
done
`
输出:
index: 0, value: one
index: 1, value: two
index: 2, value: three
Item1
Item2
Item3
Item4
Item 1
Item 2
Item 3
Item 4
Item1
Item2
Item3
Item4
Item1 Item2 Item3
Item1
Item2
Item3
element 1
element 2
element 3
Argentina is in America
Vietnam is in Asia
France is in Europe
Item 1
Item 2
Item 3
创建一个函数来浏览列表:
echo ${List[*]}
for Item in ${List[*]}
do
echo $Item
done
Loop(){
for item in ${*} ;
do
echo ${item}
done
}
Loop ${List[*]}
declare -a List=(
"element 1"
"element 2"
"element 3"
)
for entry in "${List[@]}"
do
echo "$entry"
done
declare -A continent
continent[Vietnam]=Asia
continent[France]=Europe
continent[Argentina]=America
for item in "${!continent[@]}";
do
printf "$item is in ${continent[$item]} \n"
done
`
使用declare关键字(命令)创建列表,技术上称为数组:
echo ${List[*]}
for Item in ${List[*]}
do
echo $Item
done
Loop(){
for item in ${*} ;
do
echo ${item}
done
}
Loop ${List[*]}
declare -a List=(
"element 1"
"element 2"
"element 3"
)
for entry in "${List[@]}"
do
echo "$entry"
done
declare -A continent
continent[Vietnam]=Asia
continent[France]=Europe
continent[Argentina]=America
for item in "${!continent[@]}";
do
printf "$item is in ${continent[$item]} \n"
done
`
输出:
index: 0, value: one
index: 1, value: two
index: 2, value: three
Item1
Item2
Item3
Item4
Item 1
Item 2
Item 3
Item 4
Item1
Item2
Item3
Item4
Item1 Item2 Item3
Item1
Item2
Item3
element 1
element 2
element 3
Argentina is in America
Vietnam is in Asia
France is in Europe
Item 1
Item 2
Item 3
创建关联数组。字典:
echo ${List[*]}
for Item in ${List[*]}
do
echo $Item
done
Loop(){
for item in ${*} ;
do
echo ${item}
done
}
Loop ${List[*]}
declare -a List=(
"element 1"
"element 2"
"element 3"
)
for entry in "${List[@]}"
do
echo "$entry"
done
declare -A continent
continent[Vietnam]=Asia
continent[France]=Europe
continent[Argentina]=America
for item in "${!continent[@]}";
do
printf "$item is in ${continent[$item]} \n"
done
`
输出:
index: 0, value: one
index: 1, value: two
index: 2, value: three
Item1
Item2
Item3
Item4
Item 1
Item 2
Item 3
Item 4
Item1
Item2
Item3
Item4
Item1 Item2 Item3
Item1
Item2
Item3
element 1
element 2
element 3
Argentina is in America
Vietnam is in Asia
France is in Europe
Item 1
Item 2
Item 3
列表中的CSV变量或文件
将内部字段分隔符从空格更改为所需的内容
在下面的示例中,它被更改为逗号
List="Item 1,Item 2,Item 3"
Backup_of_internal_field_separator=$IFS
IFS=,
for item in $List;
do
echo $item
done
IFS=$Backup_of_internal_field_separator
输出:
index: 0, value: one
index: 1, value: two
index: 2, value: three
Item1
Item2
Item3
Item4
Item 1
Item 2
Item 3
Item 4
Item1
Item2
Item3
Item4
Item1 Item2 Item3
Item1
Item2
Item3
element 1
element 2
element 3
Argentina is in America
Vietnam is in Asia
France is in Europe
Item 1
Item 2
Item 3
如果需要编号:
echo ${List[*]}
for Item in ${List[*]}
do
echo $Item
done
Loop(){
for item in ${*} ;
do
echo ${item}
done
}
Loop ${List[*]}
declare -a List=(
"element 1"
"element 2"
"element 3"
)
for entry in "${List[@]}"
do
echo "$entry"
done
declare -A continent
continent[Vietnam]=Asia
continent[France]=Europe
continent[Argentina]=America
for item in "${!continent[@]}";
do
printf "$item is in ${continent[$item]} \n"
done
`
这被称为回勾。将命令放在back ticks中
`command`
它位于键盘上的数字1旁边,或位于标准美式英语键盘上的tab键上方
List=()
Start_count=0
Step_count=0.1
Stop_count=1
for Item in `seq $Start_count $Step_count $Stop_count`
do
List+=(Item_$Item)
done
for Item in ${List[*]}
do
echo $Item
done
输出为:
Item_0.0
Item_0.1
Item_0.2
Item_0.3
Item_0.4
Item_0.5
Item_0.6
Item_0.7
Item_0.8
Item_0.9
Item_1.0
更加熟悉bash行为:
echo ${List[*]}
for Item in ${List[*]}
do
echo $Item
done
Loop(){
for item in ${*} ;
do
echo ${item}
done
}
Loop ${List[*]}
declare -a List=(
"element 1"
"element 2"
"element 3"
)
for entry in "${List[@]}"
do
echo "$entry"
done
declare -A continent
continent[Vietnam]=Asia
continent[France]=Europe
continent[Argentina]=America
for item in "${!continent[@]}";
do
printf "$item is in ${continent[$item]} \n"
done
`
在文件中创建列表
cat <<EOF> List_entries.txt
Item1
Item 2
'Item 3'
"Item 4"
Item 7 : *
"Item 6 : * "
"Item 6 : *"
Item 8 : $PWD
'Item 8 : $PWD'
"Item 9 : $PWD"
EOF
脚本或函数的隐式数组:
除了的正确答案:如果循环的基本语法为:
for var in "${arr[@]}" ;do ...$var... ;done
有一种特殊情况:
运行脚本或函数时,在命令行传递的参数将被分配给数组变量$@
,您可以通过$1
、$2
、$3
等方式访问
可通过以下方式填充(用于测试)
上的循环此数组可以简单地写入:
for item ;do
echo "This is item: $item."
done
注意中的保留工作不存在,也没有数组名称
样本:
set -- arg1 arg2 arg3 ...
for item ;do
echo "This is item: $item."
done
This is item: arg1.
This is item: arg2.
This is item: arg3.
This is item: ....
请注意,这与
for item in "$@";do
echo "This is item: $item."
done
然后转换成脚本:
将其保存在脚本myscript.sh
,chmod+x myscript.sh
,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:
令人惊讶的是,现在还没有人发布这篇文章——如果在数组中循环时需要元素的索引,可以这样做:
arr=(foo bar baz)
for i in ${!arr[@]}
do
echo $i "${arr[i]}"
done
输出:
index: 0, value: one
index: 1, value: two
index: 2, value: three
Item1
Item2
Item3
Item4
Item 1
Item 2
Item 3
Item 4
Item1
Item2
Item3
Item4
Item1 Item2 Item3
Item1
Item2
Item3
element 1
element 2
element 3
Argentina is in America
Vietnam is in Asia
France is in Europe
Item 1
Item 2
Item 3
0foo
1巴
2巴兹
我发现这比“传统”for循环样式(for((I=0;I简单方式)优雅得多:
arr=("sharlock" "bomkesh" "feluda" ) ##declare array
len=${#arr[*]} # it returns the array length
#iterate with while loop
i=0
while [ $i -lt $len ]
do
echo ${arr[$i]}
i=$((i+1))
done
#iterate with for loop
for i in $arr
do
echo $i
done
#iterate with splice
echo ${arr[@]:0:3}
我真正需要的是这样的东西:
## declare an array variable
declare -a arr=("element1" "element2" "element3")
## now loop through the above array
for i in "${arr[@]}"
do
echo "$i"
# or do whatever with individual element of the array
done
# You can access them using echo "${arr[0]}", "${arr[1]}" also
表示$(_数组)中的i;做某事;完成
例如:
对于$中的i(ps-aux | grep vlc | awk'{print$2}');do kill-9$i;完成
(将终止名称中包含vlc的所有进程)如何循环数组取决于新行字符的存在。新行字符分隔数组元素时,数组可以称为“$array”
,否则应称为“${array[@]}”
。下面的脚本将说明这一点:
#!/bin/bash
mkdir temp
mkdir temp/aaa
mkdir temp/bbb
mkdir temp/ccc
array=$(ls temp)
array1=(aaa bbb ccc)
array2=$(echo -e "aaa\nbbb\nccc")
echo '$array'
echo "$array"
echo
for dirname in "$array"; do
echo "$dirname"
done
echo
for dirname in "${array[@]}"; do
echo "$dirname"
done
echo
echo '$array1'
echo "$array1"
echo
for dirname in "$array1"; do
echo "$dirname"
done
echo
for dirname in "${array1[@]}"; do
echo "$dirname"
done
echo
echo '$array2'
echo "$array2"
echo
for dirname in "$array2"; do
echo "$dirname"
done
echo
for dirname in "${array2[@]}"; do
echo "$dirname"
done
rmdir temp/aaa
rmdir temp/bbb
rmdir temp/ccc
rmdir temp
这种方法的问题是什么?在简单的情况下,它似乎是有效的,而且比@anubhava的答案更直观。这在命令替换方面尤其有效,例如,$年的(seq 2000 2013)
。问题是,他询问了对数组的迭代。如果必须在多个位置对同一数组进行迭代,“declare”方法效果最好。这种方法更简洁,但灵活性较差。为什么这不是#1?它更简洁,只需设置一个字符串即可轻松重用数组,即DATABASES=“a b c d e f”
。尝试编辑器中的代码高亮显示,使您的代码看起来不错。很高兴知道,但这个问题是关于bash.Lotsa错误的。不能有带空格的列表项,不能有带全局字符的列表项。
对于${foo[*]}中的i
基本上总是错误的--
对于“${foo[@]}”中的i
是一种保留原始列表边界并防止全局扩展的表单。而echo需要是echo“$i”
这会给人留下这样的印象,即eol被用作字符串分隔符,因此字符串中允许有空格。但是,字符串