Regular expression, omit bracket

Question

Need a help in figuring out the regular expression where I need to remove all the data between {{ and }}?

Below is the coupus:

{{for|the American actor|Russ Conway (actor)}}
{{Use dmy dates|date=November 2012}}
{{Infobox musical artist &lt;!-- See Wikipedia:WikiProject_Musicians --&gt;
| birth_name          = Trevor Herbert Stanford
| birth_date          = {{birth date|1925|09|2|df=y}}
| birth_place         = [[Bristol]], [[England]], UK
| death_date          = {{death date and age|2000|11|16|1925|09|02|df=y}}
| death_place         = [[Eastbourne]], [[Sussex]], England, UK
| origin              = 
}}

record|hits]].&lt;ref name=&quot;British Hit Singles &amp; Albums&quot;/&gt;

{{reflist}}

==External links==
*[http://www.russconway.co.uk/ Russ Conway]
*{{YouTube|TnIpQhDn4Zg|Russ Conway playing Side Saddle}}

{{Authority control|VIAF=41343596}}

&lt;!-- Metadata: see [[Wikipedia:Persondata]] --&gt;
{{Persondata
| NAME              =Conway, Russ
}}
{{DEFAULTSORT:Conway, Russ}}
[[Category:1925 births]]

Below is the output with all the curly braces are removed along with the text within it:

record|hits]].&lt;ref name=&quot;British Hit Singles &amp; Albums&quot;/&gt;
==External links==
*[http://www.russconway.co.uk/ Russ Conway]
*
&lt;!-- Metadata: see [[Wikipedia:Persondata]] --&gt;
[[Category:1925 births]]

P.S - I have omitted the space in the output, I will take care of that.

score 0 · Answer 1 · answered Sep 20 '13 at 07:56

0

string.replaceAll("\\{\\{[\\s\\S]*?\\}\\}","");

will produce:





record|hits]].<ref name="British Hit Singles & Albums"/>



==External links==
*[http://www.russconway.co.uk/ Russ Conway]
*



<!-- Metadata: see [[Wikipedia:Persondata]] -->


[[Category:1925 births]]

answered Sep 20 '13 at 07:56

justhalf

8,960
3
47
74

Hi, the above expression won't take care of the nested braces...for eg in this block: {{Infobox musical artist  | birth_name = Trevor Herbert Stanford | birth_date = {{birth date|1925|09|2|df=y}} | birth_place = [[Bristol]], [[England]], UK | death_date = {{death date and age|2000|11|16|1925|09|02|df=y}} | death_place = [[Eastbourne]], [[Sussex]], England, UK | origin = }} it will miss out data after | birth_place ... – Ankit Sep 20 '13 at 08:02
You can't parse nested expression using Regex, because it's not regular. It's similar to parsing HTML using Regex, which is discussed in this post: http://stackoverflow.com/questions/1732348/regex-match-open-tags-except-xhtml-self-contained-tags – justhalf Sep 20 '13 at 08:07

Anirudha · Accepted Answer · 2013-10-01T07:42:05.893

0

This would take care of nested {{ }}

Matcher m=Pattern.compile("\\{[^{}]*\\}").matcher(input);
while(m.find())
{
    input=m.replaceAll("");
    m.reset(input);
}

edited Oct 01 '13 at 07:42

answered Sep 20 '13 at 07:56

Anirudha

32,393
7
68
89

Though it worked, I am not able to understand how are you comparing the index? – Ankit Sep 20 '13 at 08:18
`[^{}]*` would match 0 to many characters which are not `{`,`}`..So if you have nested `{{ }}`,that would only remove non nested `{{}}`...So we need to keep on replacing till we don't have `{{` in the string – Anirudha Sep 20 '13 at 08:20
Because of Texts like these {{cite news |title=Украинцы планируют убийства в Грузии? |url=km.ru/magazin/view.asp?id={5CE15A8F-9F1E-4C36-A007-0C818963B6CD} |work= |publisher=KMnews.RU |date=13 August 2008 |language=Russian |accessdate=4 December 2008 }} are going into infinite loop. It suppose the pattern {{{}}} is not matched with this @Anirudh – Ankit Oct 01 '13 at 04:13
@Kailash ur input doesn't have }} and hence it would go in loop..if }} is not present what would you like to do with the input – Anirudha Oct 01 '13 at 05:11
I need to remove nested {{ }} or { } if present..else i need to move on! @Anirudh – Ankit Oct 01 '13 at 07:01
can you help me with a similar question? http://stackoverflow.com/questions/20041891/regular-expression-to-extract-a-section-from-the-wikipedia-page @Anirudh – Ankit Nov 18 '13 at 07:46

Regular expression, omit bracket

2 Answers2