Regexp - recognize trailing 0 for decimal numbers? - sql

I need a regular expression which can find a trail of "0" in the decimal space.
F.e. following format should be recognized:
1.0
1.00
1.000
etc...
Is there somekind of "wildcard" for that?
Any idea?
Thanks,
KS

Please see the following regular expression that matches trailing zeros on the end of any decimal. Note that this does match trailing zeros in the case where there is a meaningful decimal value, but zeros occur after.
https://regex101.com/r/BJvLrO/1/
When working with regex, it is very valuable to always use a tool like https://regex101.com or https://regexr.com. Both of these tools will help you truly understand the Regular Expression. Try hovering your mouse over the different elements of the regex in my example and the tool will describe each part. You can also read the "Explanation" section to on the right side.

Related

regex decimal with negative

with the help of this thread I have tweak a regex for my use.
Decimal number regular expression, where digit after decimal is optional
So far I have this /^-?[1-9]$|^,\d+$|^0,\d$|^[1-9]\d*,\d*$
It works for
-12
0
1
It does not work for
-1,
-1,1
what I want is:
-1,1
may a kind soul explain what I am missing or doing wrong with this regex pls? Thank you very much!
I am tweaking it with : https://regexr.com/6sjkh but it's been 2 hours and I still don't know what I am missing. I don't know if language is important but I am in visual basic.
The thread you are quoting in your question has already some good answers. However if you do not need to also capture + and do not care about numbers without the leading zero before the comma and no thousand separator I would recommend the following:
/^-?\d+(,\d+)?/
The ^ indicates the beginning so that no other text in front is allowed
The -? is allowing a negative sign ( you need to escape this with \ as - can be a special character
\d+ expects 1 or more digits
The last part in parentheses covers the fractional part which as a whole can occure 0 or one time due to the question mark at the end
, is the decimal separator (if you want to have dots you need to put her ., as a dot is the special character for any character and also needs to be escaped if you look for a real dot)
\d+ is again one or more digits after the comma
I hope this helps a bit. Besides this I can also recommend using an online regular expression tool like https://regex101.com/

REGEX Extract Amount Without Currency

SELECT
ocr_text,
bucket,
REGEXP_EXTRACT('-?[0-9]+(\.[0-9]+)?', ocr_text)
FROM temp
I am trying to extract amounts from a string that will not have currency present. Any number that does not have decimals should not match. Commas should be allowed assuming they follow the correct rules (at hundreds marker)
56 no (missing decimals)
56.45 yes
120 no (missing decimals)
120.00 yes
1200.00 yes
1,200.00 yes
1,200 no (missing decimals)
1200 no (missing decimals)
134.5 no (decimal not followed by 2 digits)
23,00.00 no (invalid comma location)
I'm a noob to REGEX so I know my above statement already does not meet the criteria i've listed. However, i'm already stuck getting the error (INVALID_FUNCTION_ARGUMENT) premature end of char-class on my REGEX_EXTRACT line
Can someone point me in the right direction? How can I resolve my current issue? How can I modify to correctly incorporate the other criteria listed?
Here is a general regex pattern for a positive/negative number with two decimal places and optional thousands comma separators:
(?<!\S)(?:-?[0-9]{1,3}(,[0-9]{3})*(\.[0-9]{2})|-?[0-9]+(\.[0-9]{2}))(?!\S)
Demo
Your updated query:
SELECT
ocr_text,
bucket,
REGEXP_EXTRACT(ocr_text, '(?<!\S)(?:-?[0-9]{1,3}(,[0-9]{3})*(\.[0-9]{2})|-?[0-9]+(\.[0-9]{2}))(?!\S)')
FROM temp;
From the Presto docs I read, it supposedly supports Java's regex syntax. In the event that lookarounds are not working, you may try this version:
SELECT
ocr_text,
bucket,
REGEXP_EXTRACT(ocr_text, '(\s|^)(?:-?[0-9]{1,3}(,[0-9]{3})*(\.[0-9]{2})|-?[0-9]+(\.[0-9]{2}))(\s|$)')
FROM temp;
REGEXP_EXTRACT('^[-]?(\d*.\d*)', ocr_text)
Pattern: ^[-]?(\d*\.\d*)
Explanation:
^ - Start of line
[-]? - With or without negative dash (-)
\d* - 0 or more digits
\. - a decimal (escaped, because in regex decimals are considered special characters)
\d* - 0 or more digits (the decimal part);
$ - End of the line.
Bonus tip: There are helpful tools online to test your regex!
The Below code works to extract the value like all numbers but it catches all, only specific to certain alphabets its not working well. Anyone, please suggest well.
-?\d+\.?\d*
I have done work on NLP using Regex.

How to include apostrophe in character set for REGEXP_SUBSTR()

The IBM i implementation of regex uses apostrophes (instead of e.g. slashes) to delimit a regex string, i.e.:
... where REGEXP_SUBSTR(MYFIELD,'myregex_expression')
If I try to use an apostrophe inside a [group] within the expression, it always errors - presumably thinking I am giving a closing quote. I have tried:
- escaping it: \'
- doubling it: '' (and tripling)
No joy. I cannot find anything relevant in the IBM SQL manual or by google search.
I really need this to, for instance, allow names like O'Leary.
Thanks to Wiktor Stribizew for the answer in his comment.
There are a couple of "gotchas" for anyone who might land on this question with the same problem. The first is that you have to give the (presumably Unicode) hex value rather than the EBCDIC value that you would use, e.g. in ordinary interactive SQL on the IBM i. So in this case it really is \x27 and not \x7D for an apostrophe. Presumably this is because the REGEXP_ ... functions are working through Unicode even for EBCDIC data.
The second thing is that it would seem that the hex value cannot be the last one in the set. So this works:
^[A-Z0-9_\+\x27-]+ ... etc.
But this doesn't
^[A-Z0-9_\+-\x27]+ ... etc.
I don't know how to highlight text within a code sample, so I draw your attention to the fact that the hyphen is last in the first sample and second-to-last in the second sample.
If anyone knows why it has to not be last, I'd be interested to know. [edit: see Wiktor's answer for the reason]
btw, using double quotes as the string delimiter with an apostrophe in the set didn't work in this context.
A single quote can be defined with the \x27 notation:
^[A-Z0-9_+\x27-]+
^^^^
Note that when you use a hyphen in the character class/bracket expression, when used in between some chars it forms a range between those symbols. When you used ^[A-Z0-9_\+-\x27]+ you defined a range between + and ', which is an invalid range as the + comes after ' in the Unicode table.

Objective C parse string for middle chars

This is a bit of a puzzler for me. I have a string that looks like:
fanspd<fanspd>3</fanspd>
doorinprocess<doorinprocess>0</doorinprocess>
timeremaining<timeremaining>0</timeremaining>
macaddr<macaddr>60:CB:FB:99:99:C1</macaddr>
ipaddr<ipaddr>10.0.0.6</ipaddr>
model<model>4.4eWHF</model>
softver: <softver>2.14.2</softver>
interlock1: <interlock1>0</interlock1>
interlock2: <interlock2>0</interlock2>
cfm: <cfm>2200</cfm>
power: <power>120</power>
inside: <house_temp>-99</house_temp>
<DNS1>10.0.0.1</DNS1>
attic: <attic_temp>76</attic_temp>
OA: <oa_temp>-99</oa_temp>
server response: <server_response>Ó£àêEE²ç©þ]kõ «jsÐ</server_response>
DIP Switches: <DIPS>11100</DIPS>
Remote Switch: <switch2>1111</switch2>
Setpoint:<Setpoint>0</Setpoint>
The string includes the "/n" so I have split it into corrisponding lines that look like
fanspd<fanspd>0</fanspd>
All I really want is the char(s) in the middle of the line. In the above example it would be 0.
I can match everything with regular expressions but by doing the following:
(.*)(<[a-z]+>)(.*)(</[a-z]+>)
But what I'd like is something more that would exclude or strip away or remove all the junk and grab the middle chars.
(!(.*)(!<[a-z]+>))(.*)(!(</[a-z]+>))
I've tried this and it does not work. I've also thought of doing another [NSstring componentsSeparatedByString:#"(with either < or or >"] but that would leave be with more parsing yet to do and I think there should be a way to get just the chars inbetween the tags with either regular expressions or string compare or some such way to parse out the
Any suggestions or help would be greatly appreciated.
Thanks
Two things.
Your regular expression does not escape the forward slash.
Your regular expression seems overly complicated for what you are trying to do.
If all you want is that lone middle character with regular expressions,
Try this:
<[a-z]+>(.*)<\/[a-z]+>
Here's a great tool to play around with:
http://rubular.com
Heck you could probably even get away with:
<[a-z]+>(.*)<\/
EDIT:
I figured out your problem partially, some of the tags part way down contain characters other than a through z. So here you go:
<.+>(.*)<\/.+>

Regular expression for date in selenium ide

I tried to create a regular expression for the below date format, where as I am unable to justify that, please help me out.
04-Apr-2013 [10:58:13 GMT+05:30]
This is what I came up with:
\\d{2}-\\w{3}-d{4} [\\d{2}:d{2}:d{2} \\w{3}+\\d{2}:d{2}]
Correct me where i have gone wrong.
Thanks
You have to escape the square brackets as they have a special meaning in regular expressions and also some of your digit indicators doesn't have backslash prefix. I've tested the following regex and it worked for me:
regexp:\d{2}-\w{3}-\d{4} \[\d{2}:\d{2}:\d{2} \w{3}\+\d{2}:\d{2}\]