对字符串匹配数组的话 [英] Match array words against string

问题描述

我已经重新格式化了这个问题，因为一个误会，这产生敌意对我和问题，并不是有意而为它我appologise。

I have reformatted this question, due to a misunderstanding, which generated ill feeling toward me and the question, not intentionally and for it I appologise.

我有我使用突出在一个字符串的单词的排列，有些话却可能显示为一个短语的一部分，所以我想这句话采取presidence的亮点：

I have an array of words which I am using to highlight words in a string, some of the words however might appear as part of a phrase, and so I would like the phrase to take presidence in the highlight:

例如：

// Array
   $seo = array("apple, apple tree, orchard");

// String 
   $description = "In my orchard I have a large Apple Tree";

// Desired effect: 
   In my <strong>orchard</strong> I have a large <strong>Apple Tree</strong>

在我自己的第一次尝试，我通过数组循环运行针对字符串preg_replace，但我正在逐渐嵌套的亮点像这样＆LT;强＆GT;＆LT;强＆GT;苹果＆LT; / STRONG＆GT ;树＆LT; / STRONG＆GT;

In my own first attempt, I looped through the array running a preg_replace against the string, but I am getting nested highlights like this <strong><strong>Apple</strong> Tree</strong>

提前感谢您的帮助。

斯图

推荐答案

首先，你不应该使用一个循环来代替每个单独的词，而是一个正则表达式的选择列表（富|酒吧|啄） 。

First, you shouldn't use a loop to replace each word individually, but a regex alternatives list (foo|bar|thingy).

  $rx_words = implode("|", array_map("preg_quote", $words));
  $text = preg_replace("/\b($rx_words)\b/i", 

（其实这preg_quote错过了第二个参数，但只要你没有期待中的关键字斜线，它会工作。）

(Actually that preg_quote misses the second param, but as long as you don't have forward slashes in the keywords, it'll work.)

那么你也可以让它具有更安全的断言：

Then you can also make it safer with an assertion:

  $text = preg_replace("/(?<!<strong>)\b($rx_words)\b/i",

所以会忽略那些已经包裹的话。这只是一个解决办法，但往往就足够了。

So it will ignore words that are already wrapped. That's just a workaround, but quite often sufficient.

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