一次更换多个字符串

Question

一次更换多个字符串

在JavaScript中有一个简单的等价物吗？

$find = array("<", ">", "\n");

$replace = array("&lt;", "&gt;", "<br/>");

$textarea = str_replace($find, $replace, $textarea); 

这是使用PHP str_replace，它允许您使用单词数组来查找和替换。我可以使用JavaScript / jQuery做这样的事情吗？

...

var textarea = $(this).val();

// string replace here

$("#output").html(textarea);

...

Answer 1

您可以使用自己的函数扩展String对象，该函数可以满足您的需要（如果缺少功能，则非常有用）：String.prototype.replaceArray&nbsp;=&nbsp;function(find,&nbsp;replace)&nbsp;{ &nbsp;&nbsp;var&nbsp;replaceString&nbsp;=&nbsp;this; &nbsp;&nbsp;for&nbsp;(var&nbsp;i&nbsp;=&nbsp;0;&nbsp;i&nbsp;<&nbsp;find.length;&nbsp;i++)&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;replaceString&nbsp;=&nbsp;replaceString.replace(find[i],&nbsp;replace[i]); &nbsp;&nbsp;} &nbsp;&nbsp;return&nbsp;replaceString;};对于全局替换，您可以使用正则表达式：String.prototype.replaceArray&nbsp;=&nbsp;function(find,&nbsp;replace)&nbsp;{ &nbsp;&nbsp;var&nbsp;replaceString&nbsp;=&nbsp;this; &nbsp;&nbsp;var&nbsp;regex;&nbsp; &nbsp;&nbsp;for&nbsp;(var&nbsp;i&nbsp;=&nbsp;0;&nbsp;i&nbsp;<&nbsp;find.length;&nbsp;i++)&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;regex&nbsp;=&nbsp;new&nbsp;RegExp(find[i],&nbsp;"g"); &nbsp;&nbsp;&nbsp;&nbsp;replaceString&nbsp;=&nbsp;replaceString.replace(regex,&nbsp;replace[i]); &nbsp;&nbsp;} &nbsp;&nbsp;return&nbsp;replaceString;};要使用该函数，它将与您的PHP示例类似：var&nbsp;textarea&nbsp;=&nbsp;$(this).val();var&nbsp;find&nbsp;=&nbsp;["<",&nbsp;">",&nbsp;"\n"];var&nbsp;replace&nbsp;=&nbsp;["&lt;",&nbsp;"&gt;",&nbsp;"<br/>"];textarea&nbsp;=&nbsp;textarea.replaceArray(find,&nbsp;replace);

Answer 2

常见的错误此页面上的几乎所有答案都使用累积替换，因此在替换字符串本身可以替换时会遇到相同的缺陷。以下是这种模式失败的几个例子（h / t @KurokiKaze @derekdreery）：function&nbsp;replaceCumulative(str,&nbsp;find,&nbsp;replace)&nbsp;{ &nbsp;&nbsp;for&nbsp;(var&nbsp;i&nbsp;=&nbsp;0;&nbsp;i&nbsp;<&nbsp;find.length;&nbsp;i++) &nbsp;&nbsp;&nbsp;&nbsp;str&nbsp;=&nbsp;str.replace(new&nbsp;RegExp(find[i],"g"),&nbsp;replace[i]); &nbsp;&nbsp;return&nbsp;str;};//&nbsp;Fails&nbsp;in&nbsp;some&nbsp;cases:console.log(&nbsp;replaceCumulative(&nbsp;"tar&nbsp;pit",&nbsp;['tar','pit'],&nbsp;['capitol','house']&nbsp;)&nbsp;);console.log(&nbsp;replaceCumulative(&nbsp;"you&nbsp;&&nbsp;me",&nbsp;['you','me'],&nbsp;['me','you']&nbsp;)&nbsp;);解function&nbsp;replaceBulk(&nbsp;str,&nbsp;findArray,&nbsp;replaceArray&nbsp;){ &nbsp;&nbsp;var&nbsp;i,&nbsp;regex&nbsp;=&nbsp;[],&nbsp;map&nbsp;=&nbsp;{};&nbsp; &nbsp;&nbsp;for(&nbsp;i=0;&nbsp;i<findArray.length;&nbsp;i++&nbsp;){&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;regex.push(&nbsp;findArray[i].replace(/([-[\]{}()*+?.\\^$|#,])/g,'\\$1')&nbsp;); &nbsp;&nbsp;&nbsp;&nbsp;map[findArray[i]]&nbsp;=&nbsp;replaceArray[i];&nbsp; &nbsp;&nbsp;} &nbsp;&nbsp;regex&nbsp;=&nbsp;regex.join('|'); &nbsp;&nbsp;str&nbsp;=&nbsp;str.replace(&nbsp;new&nbsp;RegExp(&nbsp;regex,&nbsp;'g'&nbsp;),&nbsp;function(matched){ &nbsp;&nbsp;&nbsp;&nbsp;return&nbsp;map[matched]; &nbsp;&nbsp;}); &nbsp;&nbsp;return&nbsp;str;}//&nbsp;Test:console.log(&nbsp;replaceBulk(&nbsp;"tar&nbsp;pit",&nbsp;['tar','pit'],&nbsp;['capitol','house']&nbsp;)&nbsp;);console.log(&nbsp;replaceBulk(&nbsp;"you&nbsp;&&nbsp;me",&nbsp;['you','me'],&nbsp;['me','you']&nbsp;)&nbsp;);注意：这是@ elchininet的更加兼容的变化的解决方案，它使用map()和Array.indexOf()，因而不会在IE8和更早的版本。@ elchininet的实现对PHP来说更真实str_replace()，因为它还允许字符串作为查找/替换参数，并且如果有重复项将使用第一个查找数组匹配（我的版本将使用最后一个）。我在这个实现中没有接受字符串，因为这个案例已经由JS的内置处理了String.replace()。

Answer 3

text&nbsp;=&nbsp;text.replace(/</g,&nbsp;'&lt;').replace(/>/g,&nbsp;'&gt;').replace(/\n/g,&nbsp;'<br/>');