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Questions - qregularexpression

QRegularExpression find and capture all quoted and non-quoated parts in string

I am fairly new to using regexes. I got a string which can contain quoted and not quoted substrings. Here are examples of how they could look: "path/to/program.exe" -a -b -c "path/to/pr...
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k_k

c++

regex

qt

qregularexpression

Votes: 0

Answers: 1

Latest Answer

You assume that the groups you need to get value from will change their IDs with each new match, while, in fact, all the groups IDs are set in the pattern itself. I suggest removing all groups and jus...
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Wiktor Stribiżew

replace exact null string either startswith dot charector . or endswith . dot charector or its not appended to any other string ,it should b separate

I was trying replace null , .null or null. string from below code snippet static String str1="WTDocument: null ( 0000000043 ) A.1"; static String str2= "WTPart: null ( WHEEL_A1 )...
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Dungeon Master

java

regex

string

replaceall

qregularexpression

Votes: 0

Answers: 1

Latest Answer

Finally got it working: String str1="WTDocument: null ( 0000000043 ) A.1"; String str2= "WTPart: null ( WHEEL_A1 ) null.null"; String str3= "WTPart: null ( WHEEL_A1 )...
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Curious Mind

Regex extract parts of text after a string is matched

I have string field "event=[someevent1] ID=[000001] text=[Sample Text 123]" I need just the part inside text that is Sample Text 123 How to extract this using regex? I have tried (text=).\*$...
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Ashesh Nath Mishra

regex

regex-lookarounds

regex-group

qregularexpression

Votes: 0

Answers: 1

Latest Answer

You can use text=\[([^\]\[]*) See this regex demo. Details: text=\[ - a text=[ substring ([^\]\[]*) - Group 1: zero or more chars other than square brackets. A variation of the pattern with a lookb...
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Wiktor Stribiżew

Is 0000*1* a regular language?

If 2 L spanned from the bit alphabet. L1 = 000 L2 = 0 *1 * If we concatenate them L1L2 = 0000 *1 * I believe these 2 languages are regular because of a trivial DFA. But if you pump on L1... it would t...
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Advancedip

computation-theory

qregularexpression

Votes: 0

Answers: 1

Latest Answer

The concatenation of two regular languages is always a regular language. The simplest way to prove this is to imagine NFAs for each language, and then imagine hooking these up in sequence by having al...
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Patrick87

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