Question

我遇到了以下脚本的问题，该脚本通过用匹配替换它们来添加指向文本块的链接。

它完美适用于它的设计目的，创建车辆ID的链接，但我想将其更改为车辆名称，问题是脚本重新读取被替换的文本并覆盖它。

例如：$ string =“blah blah ford mustang blah blah”
应该成为：“blah blah ford mustang”
相反，它变成：“福特＆模型=野马”＆gt;福特野马“

是否可以忽略超链接中的文本？

// The following script was written by Walkerneo

function replaceLinks($replacements, $string){
    foreach($replacements as $key=>$val){
        $key=strtolower((string)$key);
        $newReplacements[$key]=array();
        $newReplacements[$key]['id']=$val;
        $newReplacements[$key]['b']=array();
        $newReplacements[$key]['a']=array();
        foreach($replacements as $key2=>$val2){
            $key2=(string)$key2;
            $b = ($key=='11 22'&&$key2=='11 22 33');
            if($b){
                l('strlen $key2: '.strlen($key2));
                l('strlen $key: '.strlen($key));
                l('strpos: '.(strpos($key2,$key)));

            }
            if(strlen($key2)>strlen($key) && ($pos=strpos($key2,$key))!==false){
                if($pos!=0){
                    $newReplacements[$key]['b'][]=substr($key2,0,$pos);
                }
                if(($end=$pos+strlen($key))!=strlen($key2)){
                    $newReplacements[$key]['a'][]=substr($key2,$end);
                }
            }
        }
    }
    foreach($newReplacements as $key=>$item){
        $tmp="/(?<![\w>])";
        foreach($item['b'] as $b){
            $tmp.="(?<!$b)";
        }
        $tmp.="($key)";
        foreach($item['a'] as $a){
            $tmp.="(?!$a)";
        }
        $tmp.='/ie';
        $replacementMatches[]=$tmp;
    }
    return preg_replace($replacementMatches,'"<a href=\"".$newReplacements[strtolower("$1")]["id"]."\">$1</a>"' ,$string);

}

//$replaceWith = array('ford mustang'=>123,'ford'=>42,'honda'=>324);  <-- this is what the script was written for
$replaceWith = array('ford mustang'=> 'make=Ford&model=Mustang','ford'=> 'make=Ford','honda'=> 'make=Honda');
$string = "That is a very nice ford mustang, if only every other ford was quite as nice as this honda";

echo replaceLinks($replaceWith,$string);

将文本替换为不存在的超链接 - 现有脚本需要调整

0 个答案: