bash shell嵌套for循环

Question

我想编写一个嵌套的for循环,它必须在bash shell提示符下工作.在单行命令中嵌套for循环.

例如,

for i in a b; do echo $i; done
a
b

在上面的示例中,for循环在单行命令中执行.像这样我在shell提示符中尝试了嵌套的for循环.它不起作用.这该怎么做.请告诉我这个.

Answer 1

这不是嵌套循环,只是一个循环.嵌套版本也可以使用:

# for i in a b; do for j in a b; do echo $j; done; done
a
b
a
b

Answer 2

一行(需要半冒号):

for i in 0 1 2 3 4 5 6 7 8 9; do for j in 0 1 2 3 4 5 6 7 8 9; do echo "$i$j"; done; done

格式化易读性(不需要分号):

for i in 0 1 2 3 4 5 6 7 8 9
do
    for j in 0 1 2 3 4 5 6 7 8 9
    do 
        echo "$i$j"
    done
done

关于如何在多行上布置shell代码有不同的看法; 这是关于我通常使用的内容,除非我将下一个操作放在同一行do(这里保存两行).

Answer 3

#!/bin/bash
# loop*figures.bash

for i in 1 2 3 4 5  # First loop.
do
    for j in $(seq 1 $i)
    do
        echo  -n "*" 
    done
    echo 
done
echo
# outputs
# *
# **
# ***
# ****
# *****

for i in 5 4 3 2 1 # First loop.
do
    for j in $(seq -$i -1)
    do
        echo  -n "*" 
    done
    echo 
done

# outputs
# *****
# ****
# ***
# **
# *

for i in 1 2 3 4 5  # First loop.
do
    for k in $(seq -5 -$i)
    do
        echo -n ' '
    done
    for j in $(seq 1 $i)
    do
        echo  -n "* " 
    done
    echo 
done
echo

# outputs
#     * 
#    * * 
#   * * * 
#  * * * * 
# * * * * * 

for i in 1 2 3 4 5  # First loop.
do
    for j in $(seq -5 -$i)
    do
        echo  -n "* " 
    done
    echo 
    for k in $(seq 1 $i)
    do
        echo -n ' '
    done
done
echo

# outputs
# * * * * * 
#  * * * * 
#   * * * 
#    * * 
#     *


exit 0