正则表达式只匹配行首字符

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Try ^CTR.\*, which literally means start of line, CTR, anything.

This will be case-sensitive, and setting non-case-sensitivity will depend on your programming language, or use ^[Cc][Tt][Rr].\* if cross-environment case-insensitivity matters.

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^CTR.*$

matches a line starting with CTR.

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^CTR

or

^CTR.*

edit:

To be more clear: ^CTR will match start of line and those chars. If all you want to do is match for a line itself (and already have the line to use), then that is all you really need. But if this is the case, you may be better off using a prefab substr() type function. I don't know, what language are you are using. But if you are trying to match and grab the line, you will need something like .* or .*$ or whatever, depending on what language/regex function you are using.

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Not sure how to apply that to your file on your server, but typically, the regex to match the beginning of a string would be :

^CTR

The ^ means beginning of string / line

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Beginning of line or beginning of string?

Start and end of string

/^CTR.*$/

/ = delimiter
^ = start of string
CTR = literal CTR
$ = end of string
.* = zero or more of any character except newline

Start and end of line

/^CTR.*$/m

/ = delimiter
^ = start of line
CTR = literal CTR
$ = end of line
.* = zero or more of any character except newline
m = enables multi-line mode, this sets regex to treat every line as a string, so ^ and $ will match start and end of line

While in multi-line mode you can still match the start and end of the string with \A\Z permanent anchors

/\ACTR.*\Z/m

\A = means start of string
CTR = literal CTR
.* = zero or more of any character except newline
\Z = end of string
m = enables multi-line mode

As such, another way to match the start of the line would be like this:

/(\A|\r|\n|\r\n)CTR.*/

or

/(^|\r|\n|\r\n)CTR.*/

\r = carriage return / old Mac OS newline
\n = line-feed / Unix/Mac OS X newline
\r\n = windows newline

Note, if you are going to use the backslash \ in some program string that supports escaping, like the php double quotation marks "" then you need to escape them first

so to run \r\nCTR.* you would use it as "\\r\\nCTR.*"

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(?i)^[ \r\n]*CTR


(?i) -- case insensitive -- Remove if case sensitive.
[ \r\n]  -- ignore space and new lines
* -- 0 or more times the same
CTR - your starts with string.

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There's are ambiguities in the question.

What is your input string? Is it the entire file? Or is it 1 line at a time? Some of the answers are assuming the latter. I want to answer the former.

What would you like to return from your regular expression? The fact that you want a true / false on whether a match was made? Or do you want to extract the entire line whose start begins with CTR? I'll answer you only want a true / false match.

To do this, we just need to determine if the CTR occurs at either the start of a file, or immediately following a new line.

/(?:^|\n)CTR/

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Regex symbol to match at beginning of a line:

Add the string you're searching for (CTR) to the regex like this:

^CTR

Example: regex

That should be enough!

However, if you need to get the text from the whole line in your language of choice, add a "match anything" pattern .*:

^CTR.*

Example: more regex

If you want to get crazy, use the end of line matcher

Add that to the growing regex pattern:

^CTR.*$

Example: lets get crazy

Note: Depending on how and where you're using regex, you might have to use a multi-line modifier to get it to match multiple lines. There could be a whole discussion on the best strategy for picking lines out of a file to process them, and some of the strategies would require this:

Multi-line flag m (this is specified in various ways in various languages/contexts)

/^CTR.*/gm

Example: we had to use m on regex101