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关于php正则表达式的两点备注

作者:爱好者    时间:2008-01-22    来源:php之家    点击:33189    本文共1篇文章 字体:[ ]

关于php正则表达式的两点备注

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