在没有超链接的情况下替换文本为超链接

Question

在没有超链接的情况下替换文本为超链接

6

一个更好的例子是这样的：

$string = "That is a very nice ford mustang, if only every other ford was quite as nice as this honda";

如果汽车名称匹配，我希望您能用制造商和型号的链接替换它们，或者只使用制造商，但如果有品牌和型号，则会在链接内放置链接（例如，如果使用str replace）...

$remove = array("ford mustang","ford","honda");
$replaceWith = array("<a href='fordID'>ford</a>","<a href='fordmustangID'>ford mustang</a>","<a href='hondaID'>honda</a>");

这将给出以下结果：

"That is a very nice <a href='<a href='fordmustangID'>ford mustang</a>ID'><a href='fordmustangID'>ford mustang</a></a>, if only every other <a href='fordmustangID'>ford mustang</a> was quite as nice as this <a href='hondaID'>honda</a>"

如果还没有超链接，我只希望它创建一个超链接，就像这样：

  "That is a very nice <a href='fordmustangID'>ford mustang</a>, if only every other <a href='fordID'>ford</a> was quite as nice as this <a href='hondaID'>honda</a>"

- Chris

你希望如何替换你找到的单词？用“鱿鱼”代替“狗”，用“鸟”代替“猫”吗？如果是这样，那么“猴子”应该被替换成什么？或者逐步用新的动物替换，直到所有动物都被使用且不重复？请澄清。 - Josh

啊，那我建议使用 preg_replace 并迭代每个 $remove。负向先行断言应该可以解决问题。 - Josh

1个回答

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- mowwwalker · Accepted Answer

编辑：

我花了很长时间，但以下是我的成果：

function replaceLinks($replacements, $string){
    foreach($replacements as $key=>$val){
        $key=strtolower((string)$key);
        $newReplacements[$key]=array();
        $newReplacements[$key]['id']=$val;
        //strings to make sure the search isn't in front of
        $newReplacements[$key]['behinds']=array();
        //strings to make sure the search isn't behind
        $newReplacements[$key]['aheads']=array();
        //check for other searches this is a substring of
        foreach($replacements as $key2=>$val2){
            $key2=(string)$key2;
            /* 
            //debugging
            $b = ($key=='11 22'&&$key2=='11 22 33');
            if($b){
                l('strlen $key2: '.strlen($key2));
                l('strlen $key: '.strlen($key));
                l('strpos: '.(strpos($key2,$key)));

            }
            */
            //the second search is longer and the first is a substring of it
            if(strlen($key2)>strlen($key) && ($pos=strpos($key2,$key))!==false){
                //the first search isn't at the start of the second search ('the ford' and 'ford')
                if($pos!=0){
                    $newReplacements[$key]['behinds'][]=substr($key2,0,$pos);
                }
                //it's not at the end ('ford' and 'fords')
                if(($end=$pos+strlen($key))!=strlen($key2)){
                    $newReplacements[$key]['aheads'][]=substr($key2,$end);
                }
            }
        }
    }
    foreach($newReplacements as $key=>$item){
        //negative lookbehind for words or >
        $tmp="/(?<![\w>=])";
        //negative lookbehinds for the beginnings of other searches that this search is a subtring of
        foreach($item['behinds'] as $b){
            $tmp.="(?<!$b)";
        }
        //the actual search
        $tmp.="($key)";
        //negative lookaheads for ends of other searches that this is a substring of.
        foreach($item['aheads'] as $a){
            $tmp.="(?!$a)";
        }
        //case insensitive
        $tmp.='/ie';
        $replacementMatches[]=$tmp;
    }
    return preg_replace($replacementMatches,'"<a href=\"".$newReplacements[strtolower("$1")]["id"]."\">$1</a>"' ,$string);

}

将数组作为参数传递，就像你之前所说的那样：

$replaceWith = array('ford mustang'=>123,'ford'=>42,'honda'=>324);

以及一个字符串：

$string = "That is a very nice ford mustang, if only every other ford was quite as nice as this honda";

echo replaceLinks($replaceWith,$string);

它优先考虑更大的字符串键，因此如果您有ford和ford mustang，它将用链接替换ford mustang。

~~不是很实用，但也许会起作用。~~

$string = "That is a very nice ford mustang, if only every other ford was quite as nice as this honda";
$remove = array("/(?<![\w>])ford mustang(?![\w<])/",'/(?<![>\w])ford(?! mustang)(?![<\w])/',"/(?<![>\w])honda(?![<\w])/");
$replaceWith = array("<a href='fordmustangID'>ford mustang</a>","<a href='fordID'>ford</a>","<a href='hondaID'>honda</a>");
echo preg_replace($remove, $replaceWith,$string);

我使用带有负预测和正预测的正则表达式，以确保我们要替换的字符串部分不是字母数字序列（如12ford23或afford）的一部分，也不接触元素的开始或结束标签。