如何在bash中使用全局数组？

Question

我正在搜索SO和谷歌搜索我的问题2天,但我没有运气.我的问题是我有一个数组,我用0来启动它的元素,并在\ while循环范围内修改其元素的值.但是当我在循环外部打印数组的值时,我发现它的初始值都是0!

这是我的代码:

#!/bin/bash
#
# This is a script to reformat CSV files into pretty-formatted txt files.

clear

echo 'This is a script to reformat CSV files into pretty-formatted txt files.'
echo

# Get number of elements.
elements=$(cat CSV_File | sed -n 1'p' | tr ',' '\n' | wc -l)
# "cat CSV_File" for getting text of the file.
# "sed -n 1'p'" for getting only first line.
# "tr ',' '\n'" for replacing ',' with '\n'.
# "wc -l" for getting number of lines (i.e number of elements).

# Get number of lines.
lines=$(cat CSV_File | wc -l)
# "cat CSV_File" for getting text of the file.
# "wc -l" for getting number of lines.

# Initiate an array for elements' lengths.
for ((i=1; $i < (( $elements + 1 )) ;i++))
do lengths[$i]="0"
done

# Find tallest element of each column and put the values into 'lengths' array.
for ((i=1; $i <= $lines ;i++))
do
   j="1"
   cat CSV_File | sed -n ${i}'p' | tr ',' '\n' | while read element
   do
      if [ ${lengths[$j]} -le ${#element} ]
      then lengths[$j]=${#element}
      fi
      (( j++ ))
   done
done

echo ${lengths[1]} # It should print 5 not 0
echo ${lengths[2]} # It should print 4 not 0
echo ${lengths[3]} # It should print 19 not 0
echo ${lengths[4]} # It should print 11 not 0
echo ${lengths[5]} # It should print 2 not 0
echo ${lengths[6]} # It should print 5 not 0
echo ${lengths[7]} # It should print 14 not 0
echo

...

这是CSV_File内容:

Jones,Bill,235 S. Williams St.,Denver,CO,80221,(303) 244-7989
Smith,Tom,404 Polk Ave.,Los Angeles,CA,90003,(213) 879-5612

请注意,当我放入echoif语句时,它会输出正确的值:

if [ ${lengths[$j]} -le ${#element} ]
then lengths[$j]=${#element}
     echo ${lengths[$j]}
fi

我在做什么错了？我错过了什么吗？

Answer 1

它| while read element; do ... done在子shell中运行,因此当子shell退出时,它对全局的更新将丢失.

一种解决方案是使用bash的"进程替换"来获取while循环的输入以在子shell中运行.请参阅: 在bash中读取多行而不生成新的子shell？