在Bash中循环遍历一串字符串？

Question

我想编写一个循环遍历15个字符串的脚本(可能是数组？)这可能吗？

就像是:

for databaseName in listOfNames
then
  # Do something
end

Answer 1

你可以像这样使用它:

## 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"
                )

Answer 2

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Bash循环.

Answer 3

这些答案都不包括反...

#!/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=1; i<${arraylength}+1; i++ ));
do
  echo $i " / " ${arraylength} " : " ${array[$i-1]}
done

输出:

1  /  3  :  one
2  /  3  :  two
3  /  3  :  three

Answer 4

是

for Item in Item1 Item2 Item3 Item4 ;
  do
    echo $Item
  done

输出:

Item1
Item2
Item3
Item4

多行

for Item in Item1 \
            Item2 \
            Item3 \
            Item4
  do
    echo $Item
  done

输出:

Item1
Item2
Item3
Item4

简单列表变量

List=( Item1 Item2 Item3 )

要么

List=(
      Item1 
      Item2 
      Item3
     )

显示列表变量:

echo ${List[*]}

输出:

Item1 Item2 Item3

循环列表:

for Item in ${List[*]} 
  do
    echo $Item 
  done

输出:

Item1
Item2
Item3

创建一个函数来浏览列表:

Loop(){
  for item in ${*} ; 
    do 
      echo ${item} 
    done
}
Loop ${List[*]}

保留空间; 单引号或双引号列表条目和双引号列表扩展:

List=(' Item 1 '
      ' Item 2' 
      ' Item 3'
     )
for item in "${List[@]}"; 
  do 
    echo "$item"
  done 

输出:

 Item 1
 Item 2
 Item 3

使用declare关键字(命令)创建列表,技术上称为数组:

declare -a List=(
                 "element 1" 
                 "element 2" 
                 "element 3"
                )
for entry in "${List[@]}"
   do
     echo "$entry"
   done

输出:

element 1
element 2
element 3

创建关联数组.一本字典:

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:

 Argentina is in America
 Vietnam is in Asia
 France is in Europe

CVS变量或文件到列表中.
将内部字段分隔符从空格更改为您想要的任何内容.
在下面的示例中,它将更改为逗号

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:

Item 1
Item 2
Item 3

如果需要编号:

` 

这被称为后退.将命令放在后面.

`commend` 

它位于键盘上的第一位和标签键上方.在标准的美国英语键盘上.

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

输出是:

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

更熟悉bashes行为:

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

将列表文件读入列表并显示

List=$(cat List_entries.txt)
echo $List
echo '$List'
echo "$List"
echo ${List[*]}
echo '${List[*]}'
echo "${List[*]}"
echo ${List[@]}
echo '${List[@]}'
echo "${List[@]}"

BASH命令行参考手册:shell中某些字符或单词的特殊含义.

Answer 5

本着与4ndrew的回答一样的精神:

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.名字中没有空格:

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

1. 在第二个示例中,using使用listOfNames="RA RB R C RD"相同的输出.

其他引入数据的方法包括:

• stdin(下面列出),
• 变量,
• 一个数组(接受的答案),
• 一个档案 ......

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

2. 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',对于MacOS IFS='\r')
3. 我也喜欢接受的答案:) - 我已经将这些片段作为其他有用的方式包含在内,也可以回答这个问题.
4. 包括#!/bin/bash在脚本文件的顶部表示执行环境.
5. 我花了几个月的时间才弄清楚如何编写这个简单:)

其他来源(读取循环时)

Answer 6

你可以使用的语法 ${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

Answer 7

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
   

您可以使用具有三个表达式（C 风格）的计数器来迭代 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

Answer 8

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

Answer 9

很惊讶没有人发布这个 - 如果你在循环数组时需要元素的索引,你可以这样做:

arr=(foo bar baz)

for i in ${!arr[@]}
do
    echo $i "${arr[i]}"
done

输出:

0 foo
1 bar
2 baz

我觉得这比"传统的"循环风格(for (( i=0; i<${#arr[@]}; i++ )))要优雅得多.

(${!arr[@]}并且$i不需要引用,因为它们只是数字;有些人会建议引用它们,但这只是个人偏好.)

Answer 10

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

有一个特殊的情况下的bash:

当运行一个脚本或函数,参数在命令行通过将被分配给$@数组变量,你可以访问$1,$2,$3,等等.

这可以填充(用于测试)

set -- arg1 arg2 arg3 ...

一个循环在这个阵列可以简单地写成:

for item ;do
    echo "This is item: $item."
  done

请注意,保留的工作in不存在,也没有数组名称!

样品:

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

然后进入一个脚本:

#!/bin/bash

for item ;do
    printf "Doing something with '%s'.\n" "$item"
  done

在脚本保存此myscript.sh,chmod +x myscript.sh话

./myscript.sh arg1 arg2 arg3 ...
Doing something with 'arg1'.
Doing something with 'arg2'.
Doing something with 'arg3'.
Doing something with '...'.

功能相同:

myfunc() { for item;do cat <<<"Working about '$item'."; done ; }

然后

myfunc item1 tiem2 time3
Working about 'item1'.
Working about 'tiem2'.
Working about 'time3'.

Answer 11

listOfNames="db_one db_two db_three"
for databaseName in $listOfNames
do
  echo $databaseName
done

要不就

for databaseName in db_one db_two db_three
do
  echo $databaseName
done

Answer 12

这与 user2533809 的答案类似，但每个文件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"

Answer 13

简单方法：

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}

Answer 14

该声明数组不适用于Korn shell。将以下示例用于Korn shell：

promote_sla_chk_lst="cdi xlob"

set -A promote_arry $promote_sla_chk_lst

for i in ${promote_arry[*]};
    do
            echo $i
    done

Answer 15

尝试这个。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]}