I am writing a parser for some Oracle commands, like
LOAD DATA
INFILE /DD/DATEN
TRUNCATE
PRESERVE BLANKS
INTO TABLE aaa.bbb
( some parameters... )
I already created a regex to match the entire command. I am now looking for a way to capture the name of the input file ("/DD/DATEN" for instance here).
My problem is that using the following regex will only return the last character of the first group ("N").
^\s*LOAD DATA\s*INFILE\s*(\w|\\|/)+\s*$
Debuggex Demo
Any ideas?
Many thanks in advance
EDIT: following #HamZa 's question, here would be the entire regex to parse Oracle LOAD DATA INFILE command (simplified though):
^\s*LOAD DATA\s*INFILE\s*((?:\w|\\|/)+)\s*((?:TRUNCATE|PRESERVE BLANKS)\s*){0,2}\s*INTO TABLE\s*((?:\w|\.)+)\s*\(\s*((\w+)\s*POSITION\s*\(\s*\d+\s*\:\s*\d+\s*\)\s*((DATE\s*\(\s*(\d+)\s*\)\s*\"YYYY-MM-DD\")|(INTEGER EXTERNAL)|(CHAR\s*\(\s*(\d+)\s*\)))\s*\,{0,1}\s*)+\)\s*$
Debuggex Demo
Let's point out the wrongdoer in your regex (\w|\\|/)+. What happens here ?
You're matching either a word character or a back/forwardslash and putting it in group 1 (\w|\\|/) after that you're telling the regex engine to do this one or more times +. What you actually want is to match those characters several times before grouping them. So you might use a non-matching group (?:) : ((?:\w|\\|/)+).
You might notice that you could just use a character class after all ([\w\\/]+). Hence, your regex could look like
^\s*LOAD DATA\s*INFILE\s*([\w\\/]+)\s*$
On a side note: that end anchor $ will cause your regex to fail if you're not using multiline mode. Or is it that you intentionally didn't post the full regex :) ?
Not tested but...
^\s*LOAD DATA\s*INFILE\s*(\S+)\s*$
Related
I'm having an issue with Regex.
I'm trying to match T0000001 (2, 3 and so on).
However, some of the lines it searches has what I can describe as positioners. These are shown as a question mark, followed by 2 digits, such as ?21.
These positioners describe a new position if the document were to be printed off the website.
Example:
T123?214567
T?211234567
I need to disregard ?21 and match T1234567.
From what I can see, this is not possible.
I have looked everywhere and tried numerous attempts.
All we have to work off is the linked image. The creators cant even confirm the flavour of Regex it is - they believe its Python but I'm unsure.
Regex Image
Update
Unfortunately none of the codes below have worked so far. I thought to test each code in live (Rather than via regex thinking may work different but unfortunately still didn't work)
There is no replace feature, and as mentioned before I'm not sure if it is Python. Appreciate your help.
Do two regex operations
First do the regex replace to replace the positioners with an empty string.
(\?[0-9]{2})
Then do the regex match
T[0-9]{7}
If there's only one occurrence of the 'positioners' in each match, something like this should work: (T.*?)\?\d{2}(.*)
This can be tested here: https://regex101.com/r/XhQXkh/2
Basically, match two capture groups before and after the '?21' sequence. You'll need to concatenate these two matches.
At first, match the ?21 and repace it with a distinctive character, #, etc
\?21
Demo
and you may try this regex to find what you want
(T(?:\d{7}|[\#\d]{8}))\s
Demo,,, in which target string is captured to group 1 (or \1).
Finally, replace # with ?21 or something you like.
Python script may be like this
ss="""T123?214567
T?211234567
T1234567
T1234434?21
T5435433"""
rexpre= re.compile(r'\?21')
regx= re.compile(r'(T(?:\d{7}|[\#\d]{8}))\s')
for m in regx.findall(rexpre.sub('#',ss)):
print(m)
print()
for m in regx.findall(rexpre.sub('#',ss)):
print(re.sub('#',r'?21', m))
Output is
T123#4567
T#1234567
T1234567
T1234434#
T123?214567
T?211234567
T1234567
T1234434?21
If using a replace functionality is an option for you then this might be an approach to match T0000001 or T123?214567:
Capture a T followed by zero or more digits before the optional part in group 1 (T\d*)
Make the question mark followed by 2 digits part optional (?:\?\d{2})?
Capture one or more digits after in group 2 (\d+).
Then in the replacement you could use group1group2 \1\2.
Using word boundaries \b (Or use assertions for the start and the end of the line ^ $) this could look like:
\b(T\d*)(?:\?\d{2})?(\d+)\b
Example Python
Is the below what you want?
Use RegExReplace with multiline tag (m) and enable replace all occurrences!
Pattern = (T\d*)\?\d{2}(\d*)
replace = $1$2
Usage Example:
I am hoping someone can assist with the REGEX I am trying to do. I just want to be able to capture the first group of characters immediately after either "Job" or "Job -".
EXAMPLE:
Job PXDFUH34 RE443 JRA99
Job - W0WEIN12SD UIS90 TYPSOS48
I want to only capture PXDFUH34 and W0WEIN12SD in this example.
UPDATE
I was able to use this to capture what I needed.
\s(\w+)\s
However, I ran into a special character (#) that this regex doesn't like. How do I account for # now?
EXAMPLE:
Job R#DFUH34 RE143 JRU89
Job - W0WEIN12SD# UIS10 TTPSOS45
Try this regex:
Job\b[\s-]*(\S+)
It means:
Look for Job and a limit \b - to avoid text like Jobless
and [\s-] spaces and hyphens * as many as possible you can find,
and then group ()
the first word \S+.
Regex live here.
Hope it helps.
Try this regex
^Job\s\-?\s?\K[^\s]*\b
On the basis of #alanmoore comments this is the alternative
^Job\s\-?\s?([^\s]*)\b
Working Regex
I have to parse a lot of content with a regular expression.
The content might, for example, be:
14-08-2015 14:18 : Example : Hello =) How are you?
What are you doing?
14-08-2015 14:19: Example2 : I'm fine thanks!
I have this regular expression that will of course return 2 matches, and the groups that I need - data, hour, name, multi line message:
(\d{2}-\d{2}-\d{4})\s?(\d{2}:\d{2})\s?:([^:]+):([^\d]+)
The problem is that if a number is written inside the message this will not be OK, because the regex will stop getting more characters.
For example in this case this will not work:
14-08-2015 14:18 : Example : Hello =) How are you?
What are you 2 doing?
14-08-2015 14:19: Example2 : I'm fine thanks!
How do I get all the characters until a new date/hour is found?
The problem is with your final capturing group ([^\d]+).
Instead you can use ((?:(?!\d{2}-\d{2}-\d{4})[\s\S])+)
The outer parenthesis: ((?:(?!\d{2}-\d{2}-\d{4})[\s\S])+) indicate a capturing group
The next set of parenthesis: ((?:(?!\d{2}-\d{2}-\d{4})[\s\S])+) indicate a non-capturing group that we want to match 1 to infinite amount of times.
Inside we have a negative look ahead: ((?:(?!\d{2}-\d{2}-\d{4})[\s\S])+). This says that whatever we are matching cannot include a date.
What we actually capture: ((?:(?!\d{2}-\d{2}-\d{4})[\s\S])+) means we capture every character including a new line.
The entire regex that works looks like this:
(\d{2}-\d{2}-\d{4})\s?(\d{2}:\d{2})\s?:([^:]+):((?:(?!\d{2}-\d{2}-\d{4})[\s\S])+)
https://regex101.com/r/wH5xR2/2
Use a lookahead for dates and get everything up to that.
/^(\d{2}-\d{2}-\d{4})\s?(\d{2}:\d{2})\s?:([^:]+):\s?((?:(?!^\d{2}-\d{2}-\d{4}\s?\d{2}:\d{2}).)*)/sm
I've edited you regex in two ways:
Added ^to the front, ensuring you only start from timestamps on their own line, which should filter out most issues with people posting timestamps
Replaced the last capturing group with ((?:(?!^\d{2}-\d{2}-\d{4}\s?\d{2}:\d{2}).)*)
(?!^\d{2}-\d{2}-\d{4}\s?\d{2}:\d{2}) is a negative lookahead, with date
(?:(lookahead).)* Looks for any amount of characters that aren't followed by a date anchored to the start of a line.
((?:(lookahead).)*) Just captures the group for you.
It's not that efficient, but it works. Note the s flag for dotall (dot matches newlines) and m flag that lets ^ match at the start of line. ^ is necessary in the lookahead so that you don't stop the match in case someone posts a timestamp, and in the start to make sure you only match dates from the start of a line.
DEMO: https://regex101.com/r/rX8eH0/3
DEMO with flags in regex: https://regex101.com/r/rX8eH0/4
I have this dataset: (about 10k times)
<Id>HOW2SING</Id>
<PopularityRank>1</PopularityRank>
<Title><![CDATA[Superior Singing Method - Online Singing Course]]></Title>
<Description><![CDATA[High Quality Vocal Improvement Product With High Conversions. Online Singing Lessons Course Converts Like Crazy Using Content Packed Sales Video. You Make 75% On Every Sale Including Front End, Recurring, And 1-click Upsells!]]></Description>
<HasRecurringProducts>true</HasRecurringProducts>
<Gravity>45.9395</Gravity>
<PercentPerSale>74.0</PercentPerSale>
<PercentPerRebill>20.0</PercentPerRebill>
<AverageEarningsPerSale>74.9006</AverageEarningsPerSale>
<InitialEarningsPerSale>70.1943</InitialEarningsPerSale>
<TotalRebillAmt>16.1971</TotalRebillAmt>
<Referred>75.0</Referred>
<Commission>75</Commission>
<ActivateDate>2011-06-23</ActivateDate>
</Site>
I am trying to do the following:
Get the data from within the tags, and use it to create a URL, so in this example it should make
http://www.reviews.how2sing.domain.com
also, all other data has to go, i want to perform a REGEX function that will just give me a list of URLS.
I prefer to do it using notepad++ but i suck at regex, any help would be welome
To keep the regex relatively simple you can just use:
.*?<id>(.+?)</id>
Replace with:
http://www.reviews.\1.domain.com\n
That will search and replace all instances of Id tag and preceding text. You can then just remove the last manually.
Make sure matches newline is selected.
Regex is straightforward, only slightly tricky part is that it uses +? and *? which are non-greedy. This prevents the whole file from being matched. The () indicate a capture group that is used in the replacement, i.e. \1.
If you want to a regex that will include replacing the last part then use:
.*?(?:(<id>)?(.+?)</id>).+?(?:<id>|\Z)
This is a bit more tricky, it uses:
?:. A non-capturing group.
| OR
\Z end of file
Basically, the first time it will match everything up to the end of the first </id> and replace up to and including the next <id>. After that it will have replaced the starting <id> so everything before </id> goes in the group. On the last match it will match the end of file \Z.
If you only want the Id values, you can do:
'<Id>([^<]*)<\/Id>'
Then you can get the first captured group \1 which is the Id text value and then create a link from it.
Here is a demo:
http://regex101.com/r/jE9qN8
[UPDATE]
To get rid of all other lines, match this regex: '.*<Id>([^<]*)<\/Id>.*' and replace by first captured group \1. Note for the regex match, since there are multiple lines, you will need to have the DOTALL or /s flag activated to also match newlines.
Hope that helps.
From the following text...
Acme Inc.<SPACE>12345<SPACE or TAB>bla bla<CRLF>
... I need to extract company name + zip code + rest of the line.
Since either a TAB or a SPACE character can separate the second from the third tokens, I tried using the following regex:
FIND:^(.+) (\d{5})(\t| )(.+)$
REPLACE:\1\t\2\t\3
However, the contents of the alternative part is put in the \3 part, so the result is this:
Acme Inc.<TAB>12345<TAB><TAB or SPACE here>$
How can I tell the (Perl) regex engine that (\t| ) is an alternative instead of a token to be saved in RAM?
Thank you.
You want:
^(.+?) (\d{5})[\t ](.+)$
Since you are matching one character or the other, you can use a character class instead. Also, I made your first quantifier non-greedy (+? instead of +) to reduce the amount of backtracking the engine has to do to find the match.
In general, if you want to make capture groups not capture anything, you can add ?: to it, like:
^(.+?) (\d{5})(?:\t| )(.+)$
Use non-capturing parentheses:
^(.+) (\d{5})(?:\t| )(.+)$
One way is to use \s instead of ( |\t) which will match any whitespace char.
See Backslash-sequences for how Perl defines "whitespace".