<p>　　severaltipsaboutRegularExpressions</p><p>　　1.processfor"greedy"</p><p>　　Bydefault,thequantifiersare"greedy",thatis,they</p><p>　　matchasmuchaspossible(uptothemaximumnumberofper-</p><p>　　mittedtimes),withoutcausingtherestofthepatternto</p><p>　　fail.Theclassicexampleofwherethisgivesproblemsisin</p><p>　　tryingtomatchcommentsinCprograms.Theseappearbetween</p><p>　　thesequences/*and*/andwithinthesequence,individual</p><p>　　*and/charactersmayappear.AnattempttomatchCcom-</p><p>　　mentsbyapplyingthepattern</p><p>　　/\*.*\*/</p><p>　　tothestring</p><p>　　/*firstcommand*/notcomment/*secondcomment*/</p><p>　　fails,becauseitmatchestheentirestringduetothe</p><p>　　greedinessofthe.*item.</p><p>　　However,ifaquantifierisfollowedbyaquestionmark,</p><p>　　thenitceasestobegreedy,andinsteadmatchestheminimum</p><p>　　numberoftimespossible,sothepattern</p><p>　　/\*.*?\*/</p><p>　　小结：</p><p>　　?与/U有类似功能，但同时出现彼此抵消</p><p>　　如下：</p><p>　　$a="asdf/*asdfaldsfasdf*/asfdasldf;kfldsj*/asfddsaf";</p><p>　　$pattern="/\/\*.*?\*\//";</p><p>　　//$pattern="/\/\*.*\*\//U";</p><p>　　//$pattern="/\/\*.*?\*\//U";</p><p>　　preg_match($pattern,$a,$match);</p><p>　　print_r($match);</p><p>　　?></p><p>　　2.Assertions</p><p>　　\w+(?=;)</p><p>　　matchesawordfollowedbyasemicolon,butdoesnotinclude</p><p>　　thesemicoloninthematch,and</p><p>　　foo(?!bar)</p><p>　　matchesanyoccurrenceof"foo"thatisnotfollowedby</p><p>　　"bar".Notethattheapparentlysimilarpattern</p><p>　　小结：</p><p>　　(?!)只前向判断匹配，如bar(?!foo)，而(?!foo)bar没有意义</p><p>　　(?</p>