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}

试试这个。它正在运行和测试。

for k in "${array[@]}"
do
    echo $k
done

# For accessing with the echo command: echo ${array[0]}, ${array[1]}

如果您使用的是Korn shell，则有“set-A databaseName”，否则有“declare-A databaseName”

要在所有外壳上编写脚本，

 set -A databaseName=("db1" "db2" ....) ||
        declare -a databaseName=("db1" "db2" ....)
# now loop 
for dbname in "${arr[@]}"
do
   echo "$dbname"  # or whatever

done

它应该在所有外壳上工作。

循环遍历数组的方式取决于换行符的存在。使用分隔数组元素的换行符，数组可以称为“$array”，否则应称为“$｛array[@]}”。以下脚本将明确说明：

#!/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

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}

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。