Multiple words in any order using regex [duplicate]

Answer 1

You can use

(?=.*test)(?=.*long)

Source: MySQL SELECT LIKE or REGEXP to match multiple words in one record

Answer 2

I was using libpcre with C, where I could define callouts. They helped me to easily match not just words, but any subexpressions in any order. The regexp looks like:

(?C0)(expr1(?C1)|expr2(?C2)|...|exprn(?Cn)){n}

and the callout function guards that every subexpression is matched exactly once,like:

int mycallout(pcre_callout_block *b){
static int subexpr[255];
if(b->callout_number == 0){
    //callout (?C0) - clear all counts to 0
    memset(&subexpr,'\0',sizeof(subexpr));
    return 0;
}else{
    //if returns >0, match fails
    return subexpr[b->callout_number-1]++;
}
}

Something like that should be possible in perl as well.

Answer 3

Use a capturing group if you want to extract the matches: (test)|(long) Then depending on the language in use you can refer to the matched group using $1 and $2, for example.

Answer 4

Vim has a branch operator \& that allows an even terser regex when searching for a line containing any number of words, in any order.

For example,

/.*test\&.*long

will match a line containing test and long, in any order.

See this answer for more information on usage. I am not aware of any other regex flavor that implements branching; the operator is not even documented on the Regular Expression wikipedia entry.

Answer 5

Try this:

/(?i)(?:test.*long|long.*test)/

That will match either test and then long, or long and then test. It will ignore case differences.

Answer 6

without knowing what language

 /test.*long/ 

or

/long.*test/

or

/test/ && /long/