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如何将旋转的日志重新连接在一起

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如何将旋转的日志重新连接在一起

如何将旋转的日志重新连接在一起以制作原始文件？

huali-access.log     huali-access.log.15  huali-access.log.21  huali-access.log.28  huali-access.log.34  huali-access.log.40  huali-access.log.47  huali-access.log.6
huali-access.log.1   huali-access.log.16  huali-access.log.22  huali-access.log.29  huali-access.log.35  huali-access.log.41  huali-access.log.48  huali-access.log.7
huali-access.log.10  huali-access.log.17  huali-access.log.23  huali-access.log.3   huali-access.log.36  huali-access.log.42  huali-access.log.49  huali-access.log.8
huali-access.log.11  huali-access.log.18  huali-access.log.24  huali-access.log.30  huali-access.log.37  huali-access.log.43  huali-access.log.5   huali-access.log.9
huali-access.log.12  huali-access.log.19  huali-access.log.25  huali-access.log.31  huali-access.log.38  huali-access.log.44  huali-access.log.50
huali-access.log.13  huali-access.log.2   huali-access.log.26  huali-access.log.32  huali-access.log.39  huali-access.log.45  huali-access.log.51
huali-access.log.14  huali-access.log.20  huali-access.log.27  huali-access.log.33  huali-access.log.4   huali-access.log.46  huali-access.log.52

答案1

像这样：

cat huali-access.log* > merged-huali-access.log

或者确保其按时间顺序排列：

echo -n "" > merged-huali-access.log # creating new file and making sure its empty
for i in {1..52}
do
    cat huali-access.log.${i} >> merged-huali-access.log
done
cat huali-access.log >> merged-huali-access.log

答案2

如果文件设置了正确的修改时间（例如，你没有在没有注意保留修改时间的情况下复制它们），则可以使用

 cat $(ls -t huali-access.log*) > output.log

ls 中的选项-t将按修改时间对其进行排序。

答案3

来自@mauro.stettler，修复了文件顺序问题，并使其变得通用：

for LOG in *.log; do 
    ( for i in {100..1}; do 
       F=${LOG}.${i}; 
       [ -e $F ] && cat $F; 
    done ; cat ${LOG} ) > aggregated_${LOG};
done

或者您可以在访问日志上使用，它不像其他的那么通用，无法应用于任何日志，并且不能在不同的月份中使用：

cat accesslog.log* | sort -nk 4 > aggregated_accesslog.log

答案4

ls -1t *.access.log* | xargs zcat >  access.all.log

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