如何在 linux bash 脚本中"exec &"生成的输出前面加上时间/日期？

Question

我有以下脚本文件，从本地文件系统向s3写入文件:

#!/bin/bash
CURR_DIR=`dirname $0`
SCRIPT_NAME="$(basename $0)"
LOG_FILE=$(echo $SCRIPT_NAME | cut -f 1 -d '.')
TODAY=$(date '+%Y-%m-%d')
NOW=$(date -d "$(date +%Y-%m-%d)" +%Y"-"%m"-"%d)
LOG_PATH="$CURR_DIR"/logs/"$LOG_FILE"-$TODAY.log
LOG="[$(date '+%Y-%m-%d %H:%M:%S,%3N')]  INFO {$LOG_FILE} -"
ERROR_LOG="[$(date '+%Y-%m-%d %H:%M:%S,%3N')]  ERROR {$LOG_FILE} -"
BUCKET="s3.bucket.example"
OUT_FOLDER="path/to/folderA"
S3_PUSH="s3://$BUCKET/$OUT_FOLDER"
exec &>> $LOG_PATH
echo "$LOG Copying files to local out folder..." >> $LOG_PATH
cp /path/to/folderA/*.* /path/to/folderB
echo "$LOG Command returned code:" $?
if [ "$(ls -A path/to/folderA/)" ]; then
FILES="$(ls path/to/folderA/*)"
for file in $FILES ; do
echo "$LOG File $file found for sync" >> $LOG_PATH
echo "$LOG Pushing $file to S3 /Folder..." >> $LOG_PATH
echo -n "$LOG " ; s3cmd put -c /home/config/.s3cfg "$file" "$S3_PUSH"/
echo "$LOG Command returned code:" $?
echo "$LOG Copying $file to local backup..." >> $LOG_PATH
mv "$file" /path/to/folderA/backup/
echo "$LOG Command returned code:" $? >> $LOG_PATH
RCC=$?
if [ $? -eq 0 ]
then
echo "$LOG Command returned code:" $?
else
echo "$ERROR_LOG Command returned code:" $?
fi
done
else
echo "$LOG No files found for sync." >> $LOG_PATH
fi

输出以特定的grok模式出现，我需要将此输出解析为日志到Elastic Search，然而第27行输出如下:

[2021-09-02 08:15:25,629]  INFO {TestGrokScriptPattern} - upload: '/path/to/folderA/File.txt' -> 's3://s3.bucket.example/Path/To/Bucket/File.txt'  [1 of 1]
0 of 0     0% in    0s     0.00 B/s  done

那上传和0的0 0%…Line是由执行&第16行执行的命令

我怎么能得到输出不去下一行没有日期，时间和脚本名称之前，为了不打破我试图创建的日志模式?

Answer 1

您可以将脚本主体包装在单个块中，然后在一个地方处理整个块的输出，而不是在每行上重定向输出。然后，您可以使用流编辑器sed处理输出。例如:

if true; then # Always true. Just simplifies redirection.
echo "Doing something..."
command_with_output
command_with_more_output
echo "Done."
fi | sed "s/^/${LOG}/" > ${LOG_PATH} 2>&1

sed表达式的意思是:用LOG变量的内容替换每行开头的^。

在末尾使用2>&1也消除了对exec &>> $LOG_PATH命令的需要。