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I need a regex to extract the number 330 on ...163&angleKreludor=330&viewID=.... This number vary from day to day...it could be an Integer with up to 3 digits, but it can be a double number like 127.57 with 2 decimal places ... so I would need to capture anything between angleKreludor= and &viewID.... Here is the complete HTML:
var swf = new SWFObject('http://images.lulaser.com/shenkuu/lunar/shenkuu_calendar_v1.swf?angleNeopia=163&angleKreludor=330&viewID=2&lang=pt', 'flash_36175654223', '550', '500', '6', '#FFFFFF');
swf.addParam('quality', 'high');
swf.addParam('scale', 'exactfit');
swf.addParam('menu', 'false');
swf.addParam('allowScriptAccess', 'always');
swf.addParam('swLiveConnect', 'true');
swf.addParam('bgcolor', 'white');
swf.write();
P.S: This is needed to use in Javascript code in Selenium IDE ... I tried in the past and Selenium IDE does not accept look-aheads nor look-behinds
You can search for digits with positive look-behind for the angleKreludor= key
(?<=angleKreludor=)(\d+)
DEMO
For JavaScript use non-captuinrg group
(?:angleKreludor=)(\d+)
var s = 'http://images.lulaser.com/shenkuu/lunar/shenkuu_calendar_v1.swf?angleNeopia=163&angleKreludor=330&viewID=2&lang=pt';
var nr = s.match(/(?:angleKreludor=)(\d+)/);
console.log(nr[1]);
DEMO
Related
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I have a String pattern: "hh:mm:ss dd:MM:yyyy to hh:mm:ss dd:MM:yyyy" and I want to extract date String from it.
Example:
S = "00:00:00 19/08/2022 to 23:59:59 19/08/2022"
Split into S1 = "00:00:00 19/08/2022" and S2 = "23:59:59 19/08/2022".
I'm trying to use String.split function but can't figure out the regex yet. Can somebody help?
I'm using Java 8.
Just split on \s+to\s+:
String pattern = "00:00:00 19/08/2022 to 23:59:59 19/08/2022";
String[] parts = pattern.split("\\s+to\\s+");
System.out.println(Arrays.toString(parts));
This prints:
[00:00:00 19/08/2022, 23:59:59 19/08/2022]
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Data:
BUY 2 FOR 5(STORES)
BUY 2 FOR 10(STORES)
What I tried:
regexp_extract(DATA, '.*? (\\d+) .*$', 2)
Desired result:
5
10
Like this:
regexp_extract(DATA, '^[^0-9]+?\\d+[^0-9]+?(\\d+)', 1);
or
regexp_extract(DATA, '^\\D+?\\d+\\D+?(\\d+)', 1);
Regex means: one or more Non-digits at the beginning, one of more digits, one or more non-digits, and finally the capturing group of digits, you need to extract the group number one.
One more solution is to split string by non-didits and take 2nd element:
select split(DATA, '[^0-9]+')[2];
Or even simpler:
select split(DATA, '\\D+')[2]; --\\D+ means one or more non-digits
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I don't really know how to use regex, and I have a task to get bulk image downloader to find a set amount of pages for example pages 1-20 to link crawl.
This is the URL:
/index.php?app=core&module=search&do=viewNewContent&period=month&userMode=&search_app=forums&sid=ceb2a9ba4039e4a06d3a6775aa735f2d&search_app_filters[forums][searchInKey]=&st=400
Its page (the st param) is incremented in +25 so the following page would be:
/index.php?app=core&module=search&do=viewNewContent&period=month&userMode=&search_app=forums&sid=ceb2a9ba4039e4a06d3a6775aa735f2d&search_app_filters[forums][searchInKey]=&st=425
How can I match and replace the page number with the next consecutive page number?
You can just capture the last digits and use whatever language you're writing in to increment that by 25:
/(\/index\.php.+?)(\d+)$/
This will give you the URL in $1 and the page number in $2 or matches[2] (however your language of choice represents the first "capture"). With that, you can increment it.
This Ruby example will do that:
matches = url.match(/(\/index\.php.+?)(\d+)$/)
page = matches[2].to_i # Convert the page number to integer
page = page + 25 # Calculate the new page number
new_url = matches[1] + (page).to_s # Merge in the new page number
That should do it for this format of URL.
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Pulling some trading data and having issues using regex to separate tickers and percentage of holding
Inputs
"94324.13%"
"007007.13%"
"0354202.91%"
Desired Output
"9432|4.13%" (ticker is 4 numbers)
"00700|7.13%" (ticker is 5 numbers)
"035420|2.91%" (ticker is 6 numbers)
Main issue is that the number of digits the ticker is may vary anywhere from 4-6 digits.
With the given information it is not possible to have a 100% accurate split of the two parts. For instance:
123410.05%
... could split in either of the following two:
1234|10.05%
12341|0.05%
And if percentages might not have a zero before the decimal point, then this would also be a possible split:
123410|.05%
The following regex replace will assume the percentage has one digit before the decimal point, and possibly a minus sign:
Find:
/^(\d{4,6})(\-?\d.*)$/gm
Replace:
\1|\2
See it on regex101.com.
I'd like to try this regex
(\d{4,6})(\d+\.\d{1,2}%)
Here is full demo:
Python:
data = "007007.13%"
rx = re.compile(r"(\d{4,6})(\d+\.\d{1,2}%)")
formated_text = rx.sub(r'\1|\2', data)
print formated_text
#it will print
00700|7.13%
You can look demo in python here
Javascript:
var re = /(\d{4,6})(\d+\.\d{1,2}%)/g;
var str = '"007007.13%"';
var subst = '$1|$2';
var result = str.replace(re, subs);
Demo in Javascript
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I want to use regex so as to obtain specific information from the text and I give an example with a semi-pseudocode ~ you can also reply me with semi-pseudocode:
list=["orange","green","grey"]
text= "The Orange is orange"
for word in list:
if word == re.compile(r'word, text):
capture Orange in order to have the noun
Beware! My question focuses whether there is a possibility to use variables (as word up above) so as to make a loop and see if there are equal words in an text based on a list.
Do not focus on how to capture the Orange.
I think Biffen has the right idea, you're in a world of pain if you're using this for POS tagging. Anyway, this allows you to match words in your text variable
for word in list:
if word in text:
# Do what you want with word
If you wanted to use regex then you can build patterns from strings, use parentheses to capture. Then use group() to access captured patterns
for word in list:
pattern = re.compile(".*(" + word + ").*")
m = re.match(pattern, text)
if m:
print(m.group(1))