用sed修改配置文件id = 1timeA = "2012-03-04","2012-03-05","2013-03-0
来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/11/07 16:38:17
用sed修改配置文件
id = 1
timeA = "2012-03-04","2012-03-05","2013-03-06"
timeB = "2012-03-04","2012-03-05","2013-03-06"
我想修改timeB的时间,且保留空格,我该怎么写?
我默认写的sed -i "s/timeB.*=/timeB =\"2012-04-04\",\"2012-04-05\",\"2013-04-06\"/g" file
但是我的空格丢了?有什么高招只修改 = 号后面的值?
id = 1
timeA = "2012-03-04","2012-03-05","2013-03-06"
timeB = "2012-03-04","2012-03-05","2013-03-06"
我想修改timeB的时间,且保留空格,我该怎么写?
我默认写的sed -i "s/timeB.*=/timeB =\"2012-04-04\",\"2012-04-05\",\"2013-04-06\"/g" file
但是我的空格丢了?有什么高招只修改 = 号后面的值?
sed -i '/timeB/s/\=.*/\= \"2012-04-04\",\"2012-04-05\",\"2013-04-06\"/g' file
sed "s/.* ([0-9][0-9]*)/\1/
where id='"&session("id")&
搞不明白这句sed -e 'N;s/.*/[&]/'和awk '{ $0=$1 "\n" $2; if (/line.1
Shell脚本,sed 's/^.*credit=//g'|sed 's/\s.*$//g'
英语翻译update table set ID=left(ID,charindex('-',ID)-1)+'-'+rig
帮忙详细解释一下这句shell语句(path="`echo $0 | sed 's,//*,/,g'`")的意思,
$id = isset($_GET['id']) ? $_GET['id'] : ''; 这句是什么意思?
android:id="@+id/btn_save"是什么意思?
怎么修改apple id
0 qbox" id="question-box">
shell sed用法 sed -e 's/\(.*\)/rename \1 ..\/\1/'
select Field_Content,ID*1 as ID,[index]=identity(int,1,1) in