与当前行的子字符串相关联的末端
我已经搜寻了类似的东西，似乎无法提出解决方案。我有一些看起来像这样的文本：
command.Parameters.Add("@Id
command.Parameters.Add("@IsDeleted
command.Parameters.Add("@MasterRecordId
command.Parameters.Add("@Name
...
我希望文字最终会这样：
command.Parameters.Add("@Id", acct.Id);
command.Parameters.Add("@IsDeleted", acct.IsDeleted);
command.Parameters.Add("@MasterRecordId", acct.MasterRecordId);
command.Parameters.Add("@Name", acct.Name);
...
如您所见，我本质上想将界限的结尾附加到：“，acct。＆lt;第二个“＆gt;）;
我正在尝试：
找到什么：（？＆lt; =@）。+？（？= \ r） - 这有效，它找到了适当的单词。
替换：\ 1“，acct。\ 1）; - 这不是。它将行更改为（for ID）：
command.Parameters.Add("@", acct.
不确定我在做什么错。我认为> \ 1应该是“查找什么”框中的“捕获”，但我猜不是吗？
I've scoured Stack Overflow for something just like this and can't seem to come up with a solution. I've got some text that looks like this:
command.Parameters.Add("@Id
command.Parameters.Add("@IsDeleted
command.Parameters.Add("@MasterRecordId
command.Parameters.Add("@Name
...
And I would like the text to end up like this:
command.Parameters.Add("@Id", acct.Id);
command.Parameters.Add("@IsDeleted", acct.IsDeleted);
command.Parameters.Add("@MasterRecordId", acct.MasterRecordId);
command.Parameters.Add("@Name", acct.Name);
...
As you can see, I essentially want to append the end of the line with: ", acct.<word between @ and second ">);
I'm trying this:
Find What: (?<=@).+?(?=\r) - This works, it finds the appropriate word.
Replace: \1", acct.\1); - This doesn't. It changes the line to (for Id):
command.Parameters.Add("@", acct.
Not sure what I'm doing wrong. I thought that \1 is supposed to be the "capture" from the "Find what" box, but it's not I guess?
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\ 1BackReference仅在您的模式中有一个捕获组时才能使用：如果您不使用捕获组，则应使用
$＆＆＆＆＆＆＆amp;而不是<<代码> \ 1 作为整个比赛的反向注册。此外，需要逃脱替换字符串中的括号。因此，替换字符串应该是：您可能还需要使用
$而不是lookahead（？= \ r），以防最后一行不紧随其后EOL角色。说了这么多，我个人更喜欢在做正则替代时更加明确/严格，以免弄乱其他线（即误报）。因此，我会选择这样的事情：
查找：
（\ bcommand \ .parameters \ .add \（“@）（\ w+）（\ w+）$替换：
\ 1 \ 1 \ 2”，acct。 \ 2 \）;请注意，
\ w只能匹配单词字符，这可能是这里所需的行为。如果您认为标识符可能具有其他字符，请随时用角色类替换。The
\1backreference will only work if you have a capturing group in your pattern:If you're not using a capturing group, you should use
$&instead of\1as a backreference for the entire match. Additionally, parentheses in the replacement string need to be escaped. So, the replacement string should be:You might also want to use
$instead of the Lookahead(?=\r)in case the last line isn't followed by an EOL character.Having said all that, I personally prefer to be more explicit/strict when doing regex substitution to avoid messing up other lines (i.e., false positives). So I would go with something like this:
Find:
(\bcommand\.Parameters\.Add\("@)(\w+)$Replace:
\1\2", acct.\2\);Note that
\wwill only match word characters, which is likely the desired behavior here. Feel free to replace it with a character class if you think your identifiers might have other characters.您也可以省略lookbehind，并匹配 @，然后使用
\ k清除当前的匹配缓冲区。该行的其余部分
然后，您可以使用
。+匹配 该行的其余部分。在替换中，使用
$ 0使用完整匹配，请参见a REGEX DEMO 对于比赛：
找到：
替换为：
是的，您还可以将模式写为一种：
You could also omit the lookbehind, and match the @ and then use
\Kto clear the current match buffer.Then you can match the rest of the line using
.+Note that you don't have to make the quantifier non greedy
.*?as you are matching the rest of the line.In the replacement, use the full match using
$0See a regex demo for the matches:
Find what:
Replace with:
If there must be a newline to the right, you might also write the pattern as one of: