如何在 bash 中使用全局数组？

Question

我在 SO 和 google 中搜索了 2 天来解决我的问题，但我没有运气。我的问题是我有一个数组，我用 0 初始化它的元素，然后在 for\while 循环范围内修改它的元素值。但是当我在循环外打印数组的值时，我发现它都是 0 作为它们的初始值！！

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

...

这里是 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，当我将 echo 放在 if 语句中时，它会打印正确的值：

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

我做错了什么？我错过了什么吗？

Answer 1

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

一种解决方案是使用 bash 的“进程替换”来获取 while 循环的输入，以便在子 shell 中运行。看： Reading multiple lines in bash without spawning a new subshell?