Question

我有js和这样的不同字符串：

Tue Aug 11 2015 between  4:00 PM and  5:00 PM

单词between和and未更改但有时我可以在单词之间使用不同数量的空格来获取此字符串

Tue Aug 11 2015 between   4:00 PM and   5:00 PM  (3 spaces)

或

Tue Aug 11 2015 between 4:00 PM and 5:00 PM (1 spaces)

是否可以为此字符串创建正则表达式？

  string re1="((?:Monday|Tuesday|Wednesday|Thursday|Friday|Saturday|Sunday|Tues|Thur|Thurs|Sun|Mon|Tue|Wed|Thu|Fri|Sat))";  // Day Of Week 1
  string re2="(\\s+)";  // White Space 1
  string re3="((?:Jan(?:uary)?|Feb(?:ruary)?|Mar(?:ch)?|Apr(?:il)?|May|Jun(?:e)?|Jul(?:y)?|Aug(?:ust)?|Sep(?:tember)?|Sept|Oct(?:ober)?|Nov(?:ember)?|Dec(?:ember)?))"; // Month 1
  string re4="(\\s+)";  // White Space 2
  string re5="((?:(?:[0-2]?\\d{1})|(?:[3][01]{1})))(?![\\d])";  // Day 1
  string re6="(\\s+)";  // White Space 3
  string re7="((?:(?:[1]{1}\\d{1}\\d{1}\\d{1})|(?:[2]{1}\\d{3})))(?![\\d])";    // Year 1
  string re8="(\\s+)";  // White Space 4
  string re9="(\"between\")";   // Double Quote String 1
  string re10="(\\s+)"; // White Space 5
  string re11="((?:(?:[0-1][0-9])|(?:[2][0-3])|(?:[0-9])):(?:[0-5][0-9])(?::[0-5][0-9])?(?:\\s?(?:am|AM|pm|PM))?)"; // HourMinuteSec 1
  string re12="(\\s+)"; // White Space 6
  string re13="(\"and\")";  // Double Quote String 2
  string re14="(\\s+)"; // White Space 7
  string re15="((?:(?:[0-1][0-9])|(?:[2][0-3])|(?:[0-9])):(?:[0-5][0-9])(?::[0-5][0-9])?(?:\\s?(?:am|AM|pm|PM))?)"; // HourMinuteSec 2

如何简化此行的正则表达式？

Answer 1

试试这个：

string.replace(/\s+/g,' ').trim();

它将删除所有额外空间，每次只留一个空格。所以，如果你有3个空格，就像你说它将它转换为1个空格

Answer 2

以下是我尝试重复使用您的代码：

＆＃13;

var re1="((?:Monday|Tuesday|Wednesday|Thursday|Friday|Saturday|Sunday|Tues|Thur|Thurs|Sun|Mon|Tue|Wed|Thu|Fri|Sat))";  // Day Of Week 1
var re2="\\s+";  // White Space 1
var re3="((?:Jan(?:uary)?|Feb(?:ruary)?|Mar(?:ch)?|Apr(?:il)?|May|Jun(?:e)?|Jul(?:y)?|Aug(?:ust)?|Sep(?:tember)?|Sept|Oct(?:ober)?|Nov(?:ember)?|Dec(?:ember)?))"; // Month 1
var re5="((?:(?:[0-2]?\\d)|(?:3[01])))(?!\\d)";  // Day 1
var re7="(\\b(?:1\\d{3}|2\\d{3})\\b)";    // Year 1
var re11="((?:[0-1][0-9]|2[0-3]|[0-9]):[0-5][0-9](?::[0-5][0-9])?(?:\\s*(?:am|AM|pm|PM))?)"; // HourMinuteSec 
var reDay = "\\b((?:0?\\d|[12]\\d|3[01]))\\b";

var s = "Tue Aug 11 2015 between  4:00 PM and  5:00 PM";
var rx = RegExp(re1 + re2 + re3 + re2 + reDay + re2 + re7 + re2 + "between" + re2 + re11 + re2 + "and" + re2 + re11, 'i');
if ((m = rx.exec(s)) !== null) {
  document.write("Day of week: " + m[1] + "<br/>");
  document.write("Month: " + m[2] + "<br/>");
  document.write("Day: " + m[3] + "<br/>");
  document.write("Year: " + m[4] + "<br/>");
  document.write("From: " + m[5] + "<br/>");
  document.write("Till: " + m[6]);
  
}

＆＃13;

请注意，我没有捕获空格（已移除的括号），为reDay添加了\b\d{1,2}\b天，只用\s?捕获两个数字作为整个单词，并且我已经倾斜了一些正则表达式（删除了不必要的括号）并通过将\s*更改为?来修复时间正则表达式。看起来这是主要问题，因为*代表 0或1次出现，set.seed(1) #Create the clusters library(doParallel) cl <- makeCluster(detectCores()) registerDoParallel(cl) #Export the environment variables to each cluster clusterExport(cl,ls()) #Load the library "rgeos" to each cluster clusterEvalQ(cl, library(rgeos)) #Split the data ID.Split<-clusterSplit(cl,unique(poly1$ID)) #Define a function that calculates the distance of one ID in relation to the poly2 a<-function(x) gDistance(spgeom1 = poly1[x,], spgeom2 = poly2, byid=TRUE) #Run the function in each cluster system.time(m<-clusterApply(cl, x=ID.Split, fun=a)) #Cluster close stopCluster(cl) #Merge the results output<- do.call("cbind", m)表示 0或更多次出现。

JS：字符串的正则表达式

2 个答案: