Joh*_*Doe 4 shell filenames rename sed
我想从文件中更改文件名后缀(使用bash脚本),但有时会有一个句点和一些带有两个句点的文件.
现在我用这个:
new_file=`echo ${file} | sed 's/\(.*\.log.*\)'${suf}'/\1.'${num}'/'`
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其中'new_file'是新文件名,'file'是原始文件名,'$ {suf}'是文件的后缀,$ {num}是新号码.
因此some.log必须成为some.log.1,some.log.1必须成为some.log.2.使用我的代码some.log成为some.log.1,但some.log.1仍然是some.log.1.
我希望我足够清楚.我很感激任何建议(甚至不使用sed).
更新:
@paxdiablo.我认为测试出了问题.
现在我用这段代码作为测试;
#!/usr/bin/bash
shft() {
for suff in {6..1} ; do
if [[ -f "$1.${suff}" ]] ; then
((nxt = suff + 1))
echo Moving "$1.${suff}" to "$1.${nxt}"
mv -f "$1.${suff}" "$1.${nxt}"
fi
done
echo Moving "$1" to "$1.1"
mv -f "$1" "$1.1"
}
clear
folder=~/logs/*.log
for i in {1..20}; do
echo ${i}> ~/logs/some.log
for fspec in ${folder} ; do
shft "${fspec}"
done
done
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现在每件事情都很好.对困惑感到抱歉.
如果您要翻转日志文件,并且根据您需要获得的复杂程度,我之前使用过以下段:
#!/usr/bin/bash
# rollover.sh
# Rolls over log files in the current directory.
# *.log.8 -> *.log.9
# *.log.7 -> *.log.8
# : : :
# *.log.1 -> *.log.2
# *.log -> *.log.1
shft() {
# Change this '8' to one less than your desired maximum rollover file.
# Must be in reverse order for renames to work (n..1, not 1..n).
for suff in {8..1} ; do
if [[ -f "$1.${suff}" ]] ; then
((nxt = suff + 1))
echo Moving "$1.${suff}" to "$1.${nxt}"
mv -f "$1.${suff}" "$1.${nxt}"
fi
done
echo Moving "$1" to "$1.1"
mv -f "$1" "$1.1"
}
for fspec in *.log ; do
shft "${fspec}"
#date >"${fspec}" #DEBUG code
done
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这将自动将日志文件翻转到版本9,尽管您可以更改suff for循环以允许更多.
与DEBUG加太新的文件进行检测自动创建,下面的记录表明它在行动:
pax> touch qq.log ; ./rollover.sh
Moving "qq.log" to "qq.log.1"
pax> touch "has spaces.log" ; ./rollover.sh
Moving "has spaces.log" to "has spaces.log.1"
Moving "qq.log.1" to "qq.log.2"
Moving "qq.log" to "qq.log.1"
pax> ll *log*
-rw-r--r-- 1 pax None 30 2010-09-11 20:39 has spaces.log
-rw-r--r-- 1 pax None 0 2010-09-11 20:39 has spaces.log.1
-rw-r--r-- 1 pax None 30 2010-09-11 20:39 qq.log
-rw-r--r-- 1 pax None 30 2010-09-11 20:38 qq.log.1
-rw-r--r-- 1 pax None 0 2010-09-11 20:38 qq.log.2
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这个脚本的好处在于,它可以轻松配置以处理大量历史记录(通过更改{8..1}位),使用空格处理名称,并在日志文件丢失时相对稳健地处理间隙.
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