linux-如何将值添加到bash数组？

这个问题已经在这里有了答案：            >            How to pipe input to a Bash while loop and preserve variables after loop ends                                    2个
抱歉subject clone,但是鉴于我从该答案中成功使用的技术无法在新脚本上正常工作,因此我认为现在是提出新问题的时候了.

无论如何,以下脚本非常简单：

missed_symbols=()

grep missed ${dest}/scribe.log | while read -r line; do
  symbol=$(echo "${line}" | cut -d' ' -f1)
  missed_symbols+=("$symbol")
done

grep missed ${dest}/scribe_close.log | while read -r line; do
  symbol=$(echo "${line}" | cut -d' ' -f1)
  missed_symbols+=("$symbol")
done

for symbol in "${missed_symbols[@]}"; do
  echo "Scribe missed a turn in ${symbol}"
done
echo "Scribe missed ${#missed_symbols[@]} turns today"
exit 0

在-！/ bin / bash行中添加-x告诉我,实际上,它是将每个$line的第一个单词从$missed_symbols [@]分配给$symbol.但是,当我检查${#missed_symbols [@]}时,我得到零.而且,当然,循环遍历不会产生任何输出.

那我的错字在哪里？

解决方法:

管道创建子壳.子外壳程序退出时,对子外壳程序中的外壳程序变量所做的更改会消失.

因此,与其传递到while读取循环中,不如从进程替换中重定向：

while read -r line; do
  symbol=${line%%" "*}
  missed_symbols+=("$symbol")
done < <(grep missed ${dest}/scribe.log)