如何将旋转的日志连接在一起以生成原始文件？
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 喜欢这个：
cat huali-access.log* > merged-huali-access.log
或按照时间顺序确定：
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
如果文件具有正确的修改时间（例如，您没有在保留修改时间的情况下复制它们），则可以使用
cat $(ls -t huali-access.log*) > output.log
ls中的-t选项将按修改时间sorting。
从@ mauro.stettler，修正了文件顺序的问题，也使它通用：
for LOG in *.log; do ( for i in {100..1}; do F=${LOG}.${i}; [ -e $F ] && cat $F; done ; cat ${LOG} ) > aggregated_${LOG}; done
或者你可以在accesslogs上使用，而不是像你可以应用到任何日志那样通用，并且不能在不同的月份中工作：
cat accesslog.log* | sort -nk 4 > aggregated_accesslog.log
这将连接所有日志文件（包括gzipped）到log.all只需要replace“日志”来使用它
l='log'; test -f ${l}.all && rm ${l}.all; ls -1tr ${l}* | xargs zcat -f >> $l.all
ls -1t *.access.log* | xargs zcat > access.all.log