Unix Shell脚本中的变量问题

Question

我无法从循环内部传递变量.

码:

# find all output.txt that has been modified last 24h ...
PROCESSED=1
find ${PROCESSED_DIR} -mtime -1 -name "output.txt" | while read i
do
    # .. and compare those with TMP_TXT
    if diff $i ${TMP_TXT} > /dev/null   
    then
        # If both are same EXIT search loop 
        PROCESSED=0
        exit
    fi
done
echo "PROCESSED=${PROCESSED}"

这将始终输出1.任何想法如何使PROCESSED = 0？

这是在Solaris 9计算机上完成的.

Answer 1

原因是while-command在管道内运行,这意味着它在子shell中运行,因此变量赋值不会传播到调用shell.

一种解决方案是重定向输入(您可能需要先将其写入临时文件):

while read f; do
    ...
done < foo.tmp

另一种解决方案是使用while命令中的stdout传播以下值PROCESSED:

PROCESSED=`find | while read f; do
    echo "Echo other stuff on stderr" >&2 
    echo 1
done`