regex

问题 I am trying to extract the content between two lines of string. Here is how the text looks Hhhshhajsjsjsj Hshhejjsjsmk Hahjajajajajja Message-ID: b123467 abc def Kjhshjsjs Received: Hdjjddjdjdjdjd I need to keep the text between ‘Message-ID’ and ‘Received:’ I tried ‘@“(?:\G(?!\A)[X-Message-ID:)\r?\n(.*)(?>\r?\nReceived:(?=S\r?$))?’ I have got an error ‘can’t find the text’ 回答1: Your problem would be straightforward if you only had a single occurrence of this text, and you only needed to

问题 I have a statement as follows: String n = \ul\insrsid14762702 Symptom}{\insrsid14762702\ul\insrsid14762702 Acid}{\insrsid14762702\ul\insrsid14762702 Nonacid}{\insrsid14762702\ul\insrsid14762702 All}{\insrsid14762702 Reg\tab 100%\tab 100%\tab 100% Stg pain\tab 100%\tab 83%\tab 100% F pain\tab 72%\tab 0%\tab 67% I would like to see how many tabs there are but I always get zero returned for the size of the arrayList I thought I was adding to. My code: Pattern patternTabs = Pattern.compile("\\tab

问题 I have a list like this: map_tmp <- list("ABC", c("EGF", "HIJ"), c("KML", "ABC-IOP"), "SIN", "KMLLL") > grep("ABC", map_tmp) [1] 1 3 > grep("^ABC$", map_tmp) [1] 1 # by using regex, I get the index of "ABC" in the list > grep("^KML$", map_tmp) [1] 5 # I wanted 3, but I got 5. Claiming the end of a string by "$" didn't help in this case. > grep("^HIJ$", map_tmp) integer(0) # the regex do not return to me the index of a string inside the vector How can I get the index of a string (exact match)

问题 I have a statement as follows: String n = \ul\insrsid14762702 Symptom}{\insrsid14762702\ul\insrsid14762702 Acid}{\insrsid14762702\ul\insrsid14762702 Nonacid}{\insrsid14762702\ul\insrsid14762702 All}{\insrsid14762702 Reg\tab 100%\tab 100%\tab 100% Stg pain\tab 100%\tab 83%\tab 100% F pain\tab 72%\tab 0%\tab 67% I would like to see how many tabs there are but I always get zero returned for the size of the arrayList I thought I was adding to. My code: Pattern patternTabs = Pattern.compile("\\tab

问题 What I'm trying to accomplish I'm trying to create a function to use string interpolation within VBA. The issue I'm having is that I'm not sure how to replace "\n" with a vbNewLine , as long as it does not have the escape character "\" before it? What I have found and tried VBScript does not have a negative look behind as far as I can research. Below has two examples of Patterns that I have already tried: Private Sub testingInjectFunction() Dim dict As New Scripting.Dictionary dict("test") =

问题 I am trying to replace 2.0 to stack, but the following code replace 2008 to 2.08 Following is my code: $string = 'The story is inspired by the Operation Batla House that took place in 2008 '; $tag = '2.0'; $pattern = '/(\s|^)'.($tag).'(?=[^a-z^A-Z])/i'; echo preg_replace($pattern, '2.0', $string); 回答1: Use preg_quote and make sure you pass the regex delimiter as the second argument: $string = 'The story is inspired by the Operation Batla House that took place in 2008 '; $tag = '2.0'; $pattern

问题 What I'm trying to accomplish I'm trying to create a function to use string interpolation within VBA. The issue I'm having is that I'm not sure how to replace "\n" with a vbNewLine , as long as it does not have the escape character "\" before it? What I have found and tried VBScript does not have a negative look behind as far as I can research. Below has two examples of Patterns that I have already tried: Private Sub testingInjectFunction() Dim dict As New Scripting.Dictionary dict("test") =

问题 I have a data frame, in which I need to find all the possible matches rows which match with terms . My code is texts = ['foo abc', 'foobar xyz', 'xyz baz32', 'baz 45','fooz','bazzar','foo baz'] terms = ['foo','baz','foo baz'] # create df df = pd.DataFrame({'Match_text': texts}) #cretae pattern pat = r'\b(?:{})\b'.format('|'.join(terms)) # use str.contains to find matchs df = df[df['Match_text'].str.contains(pat)] #create pattern p = re.compile(pat) #search for pattern in the column results =

问题 I am trying to replace 2.0 to stack, but the following code replace 2008 to 2.08 Following is my code: $string = 'The story is inspired by the Operation Batla House that took place in 2008 '; $tag = '2.0'; $pattern = '/(\s|^)'.($tag).'(?=[^a-z^A-Z])/i'; echo preg_replace($pattern, '2.0', $string); 回答1: Use preg_quote and make sure you pass the regex delimiter as the second argument: $string = 'The story is inspired by the Operation Batla House that took place in 2008 '; $tag = '2.0'; $pattern

问题 As an example, my string might look like this: first_name:Tom last_name:Jones email:tom@somedomain.com I would want my array to look like the following: Array ( ['first_name'] => 'Tom', ['last_name'] => 'Jones', ['email'] => 'tom@somedomain.com' ) This will then be used to search the database based on the column and value. So I will first need to retrieve the key (such as first_name) and the value (like Tom) for searching. I have the following regex: ((?:[a-z][a-z0-9_]*)) This works. It finds