读取 bash 数组时（未读入）
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while read bash array (not into)
提问by Cocoa Puffs
I'm trying to use an array with while read, but the entire array is output at once.
我正在尝试将数组与 一起使用while read，但会立即输出整个数组。
#!/bin/bash
declare -a arr=('1:one' '2:two' '3:three');
while read -e it ; do
echo $it
done <<< ${arr[@]}
It should output each value separately (but doesn't), so maybe while read isn't the hot ticket here?
它应该单独输出每个值（但不是），所以也许读取时不是这里的热门单？
回答by John1024
For this case, it is easier to use a forloop:
对于这种情况，使用for循环更容易：
$ declare -a arr=('1:one' '2:two' '3:three')
$ for it in "${arr[@]}"; do echo $it; done
1:one
2:two
3:three
The while readapproach is very useful (a) When you want to read data from a file, and (b) when you want to read in a nul or newline separated string. In your case, however, you already have the data in a bashvariable and the forloop is simpler.
该while read方法非常有用（a）当您想从文件中读取数据时，以及（b）当您想读取空或换行符分隔的字符串时。但是，在您的情况下，您已经将数据保存在bash变量中，并且for循环更简单。
回答by sumitya
possible by while loop
可能通过 while 循环
#!/bin/bash
declare -a arr=('1:one' '2:two' '3:three');
len=${#arr[@]}
i=0
while [ $i -lt $len ]; do
echo "${arr[$i]}"
let i++
done
