从PHP中删除HTML中的所有样式属性

Question

I have to load the body of an HTML page without any style attribute and no link images and everything that is not 'plain text. I would like to do it in PHP and tried very solution but I have not solved. I load the html page with an ajax call to my script and then with a regular expression I take the body that then I want this cleared. Can you help me? This is the ajax call:

$.ajax({
       type: "GET"
       url: "core/proxy.php?url="+cerca,              
       success: function(data){
       var body = data.replace(/^[\S\s]*<body[^>]*?>/i, "")
       .replace(/<\/body[\S\s]*$/i, "");
        $("div#risultato").html(body);
    },
      error: function(){
      alert("failed");
    }
    });
});

Answer 1

Jose Antonio Riaza Valverde I corrected but nothing changes:

$.ajax({
            //definisco il tipo della chiamata
            type: "GET",
            //url della risorsa da contattare
            url: "core/proxy.php?url="+cerca,
            //azione in caso di successo
            success: function(data)
            {
                var body = data.replace(/^[\S\s]*<body[^>]*?>/i, "")
                .replace(/<\/body[\S\s]*$/i, "");
                $("div#risultato").html(body);
                clearStyles(document.getElementById('risultato'));

            },
            //azione in caso di errore
            error: function()
            {
                alert("Chiamata fallita");
            }
    });
});

and the function:

function clearStyles(element) {
element.setAttribute('style', ' ');
element.setAttribute('img', ' ');
element.setAttribute('a', ' ');
for (var i = 0; i < element.children.length; i++) {
    clearStyles(element.children[i]);
}

}