Question

我有一个函数可以纠正大写异常大写单词的大写：

var line = "some long string of text";
["AppleScript", "Bluetooth", "DivX", "FireWire", "GarageBand", 
 "iPhone", "iTunes", "iWeb", "iWork", "JavaScript", "jQuery", "MacBook", 
 "MySQL", "PowerBook", "PowerPoint", "QuickTime", "TextEdit", "TextMate",
 // ... 
 "Wi-Fi", "Xcode", "Xserve", "XMLHttpRequest"].forEach(function(name) {
      line = line.replace(RegExp(name, "gi"), name);
});

现在我遇到的问题是大多数输入字符串平均包含0到3个字。显然现在我正在做几十个（可能有几百个;那个数组有一种不可思议的趋势，随着时间的推移而增长）函数调用基本上什么都不做。

如何更快地编写代码并摆脱不必要的函数调用？

示例输入：

我的iphone应用程序在UIViewController下有一个用户表单。当我再次启动应用程序时，我的一些UIView会更改其位置和大小。（这些UIViews依赖于键盘位置）某处绝对是我的错。我试图弄清楚当应用程序从后台再次启动时会发生什么，并且可以完成UIView更改。

Answer 1

您可以构建包含所有单词的正则表达式，通过将每个单词括在括号中来捕获每个单词。在替换中使用它将提供足够的信息来恢复替换函数中的原始单词。

  function correct (text, words) {
    return text.replace (RegExp ('\\b(?:(' + words.join (')|(') + '))\\b', 'ig'), function (m) {
      for (var a = arguments.length - 2; a--;)
        if (arguments[a])
      return words[a-1] || m;
    });
  } 

  console.log (correct ("My iphone itunes divx firewire application has a user form under uiviewcontroller. When I start application again some of my uiview changes its positions and sizes. (These uiviews depend on keyboard position) Somewhere is definitely my fault. I try to figure what is going on when application starts again from background and where the uiview changes can be done.",
    ["AppleScript", "Bluetooth", "DivX", "FireWire", "GarageBand", 
 "iPhone", "iTunes", "iWeb", "iWork", "JavaScript", "jQuery", "MacBook", 
 "MySQL", "PowerBook", "PowerPoint", "QuickTime", "TextEdit", "TextMate",
 // ... 
 "UIViewController","UIView",
 "Wi-Fi", "Xcode", "Xserve", "XMLHttpRequest"]));
My iPhone iTunes DivX FireWire application has a user form under UIViewController. When I start application again some of my UIView changes its positions and sizes. (These UIViews depend on keyboard position) Somewhere is definitely my fault. I try to figure what is going on when application starts again from background and where the UIView changes can be done.

This turns out to be faster then the original code

JavaScript正则表达式。

1 个答案: