Find Regex Give more than one result

Question

I have a code that findme info for some URLs How can i set this code for that only give me one string result?. Actually this code bring me all results, I just need one. In the image below the green rectangle is the correct result, but if the url contains the string more than one time this show me boths, red rectangle.

for idx,row in df.iterrows():
    url = row['e.URL'].replace('/v01/', '/depot/')
    x = urlopen(url)
    new = x.read()
    soup = BeautifulSoup(new, "lxml-xml")
    match = ''.join(re.findall(r"(?i)cl[a-zA-Z]{3}\d{5}", str(soup)))
    df.at[idx,'NEW_APP'] = match

The below code brings me all results:

match = ''.join(re.findall(r"(?i)cl[a-zA-Z]{3}\d{5}", str(soup)))

See image below for reference:

Change [`findall`](https://docs.python.org/3/library/re.html#re.findall) to [`search`](https://docs.python.org/3/library/re.html#re.search) — ctwheels, Oct 24 '19 at 14:38
----> 6 match = ''.join(re.search(r"(?i)cl[a-zA-Z]{3}\d{5}", str(soup))) TypeError: can only join an iterable — salem1992, Oct 24 '19 at 14:40
You can't join a single string on nothing. You would just return the match? `match = re.search(...)` — ctwheels, Oct 24 '19 at 14:40
now i set the code like this: match = (re.search(r"(?i)cl[a-zA-Z]{3}\d{5}", str(soup))) but the result is something like this, i only need the code: — salem1992, Oct 24 '19 at 14:43
Add `if match:` before your `df.at[idx, 'NEW_APP'] = match` line and get the `0`th group — ctwheels, Oct 24 '19 at 14:45

score 0 · Answer 1 · answered Oct 24 '19 at 14:44

If you do not want to have multiple matches then you can use re.search.

found = re.search(r"(?i)cl[a-zA-Z]{3}\d{5}", str(soup))
match = found.group(0) if found else ''

Or you can use the findall like you do now but only use the first match

matches = re.findall(r"(?i)cl[a-zA-Z]{3}\d{5}", str(soup))
match = matches[0] if matches else ''

Find Regex Give more than one result

1 Answers1