我正在寻找这个模式
<!-- Footer part at bottom of page-->
<div id="footer">
<div class="row col-md-2 col-md-offset-5">
<p class="text-muted">© 2014. Core Team</p>
</div>
<div id="downloadlinks">
<!-- downloadlinks go here-->
</div>
</div>
，并将其替换为以下模式用于许多.html文件
<!-- Footer part at bottom of page-->
<div id="footer">
<div class="row col-md-2 col-md-offset-5">
<?php
$year = date("Y");
echo "<p class='text-muted'>© $year. Core Team</p>";
?>
</div>
<div id="downloadlinks">
<!-- downloadlinks go here-->
</div>
</div>
注意区别在于这个
<p class="text-muted">© 2014. Core Team</p>
被
取代 <?php
$year = date("Y");
echo "<p class='text-muted'>© $year. Core Team</p>";
?>
我想用sed来做，但有了最初的尝试，我的困难是我可能或可能或可能不需要转义的字符。还有制表符或新行在php代码，我希望它出现在这里。
有很多文件要做，所以我想自动化它，但它可能更快，只是手工做(复制和粘贴)。但在这种情况下，sed可能是错误的方法。有人能告诉我正确的方向吗?在这个阶段，我对其他语言(例如php, python, bash)开放，以找到解决方案。
然后我计划用以下命令将每个。html文件重命名为。php:
for i in *.html; do mv "$i" "${i%.*}.php"; done;
EDIT1
根据下面的awk答案，我可以让它在这个版本下工作
$ awk -Wversion 2>/dev/null || awk --version
GNU Awk 4.1.1, API: 1.1 (GNU MPFR 3.1.2, GNU MP 6.0.0)
Copyright (C) 1989, 1991-2014 Free Software Foundation.
但是在这个版本上我得到了不同的输出。它似乎打印出3个文件，旧的，新的和文件。这个版本容易修改吗?
root@4461f768e343:/github/find_pattern# awk -Wversion 2>/dev/null || awk --version
mawk 1.3.3 Nov 1996, Copyright (C) Michael D. Brennan
root@4461f768e343:/github/find_pattern#
root@4461f768e343:/github/find_pattern#
root@4461f768e343:/github/find_pattern# awk -v RS='^$' -v ORS= 'ARGIND==1{old=$0;next} ARGIND==2{new=$0;next} s=index($0,old){ $0 = substr($0,1,s-1) new substr($0,s+length(old))} 1' old new file
<!-- Footer part at bottom of page-->
<div id="footer">
<div class="row col-md-2 col-md-offset-5">
<p class="text-muted">© 2014. Core Team</p>
</div>
<div id="downloadlinks">
<!-- downloadlinks go here-->
</div>
</div><!-- Footer part at bottom of page-->
<div id="footer">
<div class="row col-md-2 col-md-offset-5">
<?php
$year = date("Y");
echo "<p class='text-muted'>© $year. Core Team</p>";
?>
</div>
<div id="downloadlinks">
<!-- downloadlinks go here-->
</div>
</div>some pile of text
or other
<!-- Footer part at bottom of page-->
<div id="footer">
<div class="row col-md-2 col-md-offset-5">
<p class="text-muted">© 2014. Core Team</p>
</div>
<div id="downloadlinks">
<!-- downloadlinks go here-->
</div>
</div>
and more maybe.root@4461f768e343:/github/find_pattern#
您可以使用replace
html_files = ['a.html', ...]
copyright = '<p class="text-muted">© 2014. Core Team</p>'
new_copyright = """ <?php
$year = date("Y");
echo "<p class='text-muted'>© $year. Core Team</p>";
?>"""
for html_file_path in html_files:
with open(html_file_path) as html_file:
html = html_file.read()
if copyright in html:
php_file_path = html_file_path.replace('.html', '.php')
with open(php_file_path, "w") as php_file:
php = html.replace(copyright, new_copyright)
php_file.write(php)
注意这不会覆盖你的html文件，这是有用的，如果脚本有一个错误。
sed用于在单独的行上进行简单的替换，因此您的任务肯定不是sed的任务。如果文件的格式都很好，可以使用awk:
$ cat old
<!-- Footer part at bottom of page-->
<div id="footer">
<div class="row col-md-2 col-md-offset-5">
<p class="text-muted">© 2014. Core Team</p>
</div>
<div id="downloadlinks">
<!-- downloadlinks go here-->
</div>
</div>
.
$ cat new
<!-- Footer part at bottom of page-->
<div id="footer">
<div class="row col-md-2 col-md-offset-5">
<?php
$year = date("Y");
echo "<p class='text-muted'>© $year. Core Team</p>";
?>
</div>
<div id="downloadlinks">
<!-- downloadlinks go here-->
</div>
</div>
.
$ cat file
some pile of text
or other
<!-- Footer part at bottom of page-->
<div id="footer">
<div class="row col-md-2 col-md-offset-5">
<p class="text-muted">© 2014. Core Team</p>
</div>
<div id="downloadlinks">
<!-- downloadlinks go here-->
</div>
</div>
and more maybe.
.
$ awk -v RS='^$' -v ORS= 'ARGIND==1{old=$0;next} ARGIND==2{new=$0;next} s=index($0,old){ $0 = substr($0,1,s-1) new substr($0,s+length(old))} 1' old new file
some pile of text
or other
<!-- Footer part at bottom of page-->
<div id="footer">
<div class="row col-md-2 col-md-offset-5">
<?php
$year = date("Y");
echo "<p class='text-muted'>© $year. Core Team</p>";
?>
</div>
<div id="downloadlinks">
<!-- downloadlinks go here-->
</div>
</div>
and more maybe.
上面的代码使用GNU awk处理多字符RS和ARGIND。如果您想为许多文件执行此操作，您可以使用:
find . -type f -name '*.php' -exec awk -i inplace -v RS='^$' -v ORS= 'ARGIND==1{old=$0;print;next} ARGIND==2{new=$0;print;next} s=index($0,old){ $0 = substr($0,1,s-1) new substr($0,s+length(old))} 1' old new {} ';
或类似.