I have a section of code like so:
我有一段代码如下:
$this->getUrl('example/page.html')
and I need to replace it to look like the following:
我需要将其替换为如下所示:
$this->getUrl('', array('_direct' => 'example/page.html'))
My code editor (sublime text) has regular expression find and replace, but how do I do this whilst keeping the 'example/page.html' string across all occurrences?
我的代码编辑器(sublime text)有正则表达式查找和替换,但是如何在所有出现时保持'example / page.html'字符串的同时执行此操作?
So substitution are: '', array('_direct' => '
and )
at the end
所以替换是:'',array('_ direct'=>'和)最后
Thanks
2 个解决方案
#1
2
This should do the job:
这应该做的工作:
- find what:
(\$this->getUrl\()([^)]+)\)
- replace:
$1'', array('_direct' => $2))
找到什么:(\ $ this-> getUrl \()([^]] +)\)
替换:$ 1'',数组('_ direct'=> $ 2))
#2
0
Press Ctrl-H
for invoking replace bar
按Ctrl-H调用替换栏
then
find what: (?<=\$this->getUrl\()([^)]+)
Replace with: '', array('_direct' => $1)
找到什么:(?<= \ $ this-> getUrl \()([^]] +)替换为:'',array('_ direct'=> $ 1)
first find a zero-match-length with look-behind and the capture between parentheses ()
to special variable $1
首先找到带有后视的零匹配长度和括号()到特殊变量$ 1之间的捕获
#1
2
This should do the job:
这应该做的工作:
- find what:
(\$this->getUrl\()([^)]+)\)
- replace:
$1'', array('_direct' => $2))
找到什么:(\ $ this-> getUrl \()([^]] +)\)
替换:$ 1'',数组('_ direct'=> $ 2))
#2
0
Press Ctrl-H
for invoking replace bar
按Ctrl-H调用替换栏
then
find what: (?<=\$this->getUrl\()([^)]+)
Replace with: '', array('_direct' => $1)
找到什么:(?<= \ $ this-> getUrl \()([^]] +)替换为:'',array('_ direct'=> $ 1)
first find a zero-match-length with look-behind and the capture between parentheses ()
to special variable $1
首先找到带有后视的零匹配长度和括号()到特殊变量$ 1之间的捕获