1

テキストから img URL だけを引き出す単純な JavaScript 関数を作成しようとしています。

私はほとんどそれを機能させましたが、私はparamsに問題があり、どのようにそれらを削除し、完全なimg URLを返すことができますか?

document.write(getImagesInText("Whether the view should  hidden from (i.e., ignored by) the accessibility service. http://jsfiddle.net/ http://thickbox.net/images/plant4.jpg afhttp://thickbox.net/images/plant4.jpg?4t34t34t"));


function getImagesInText(text){

        var html = text;
        var urlRegex = /((ftp|http|https):\/\/(\w+:{0,1}\w*@)?(\S+)(:[0-9]+)?(\/|\/([\w#!:.?+=&%@!\-\/]))?)+\.(?:jpg|jpeg|gif|png)/gi;
        return html.replace(urlRegex, '<img src="$1" width="48" height="48"/>');

}

こちらの例をご覧ください。
http://jsfiddle.net/7dJCm/1/

アップデート

function getImagesInText( s ) {
    var html = s;
    var imgregex = /((http(s)?|ftp):\/\/[\S]*(\.jpg|.jpeg|\.gif|.png)[\S]*)/gi;
    var urlRegex = /((ftp|http|https):\/\/(\w+:{0,1}\w*@)?(\S+)(:[0-9]+)?(\/|\/([\w#!:.?+=&%@!\-\/]))?)/gi;

    html = html.replace(/(\.jpg|\.jpeg|\.gif|\.png)/gi, function( ext ) { return ext + " "; }); 

    html = html.replace(/(http|https)/gi, function( ext ) { return " " + ext; });

    html = html.replace(urlRegex, function( path ) {
           if(path.match(/jpg|png|gif|jpeg/g)){
             return "<img width='48' height='48' src='" + path + "' />";
           }else{
             return "<a href='" + path + "'>" + path + "</a>";
           }
    });

    return html;
}

http://jsfiddle.net/7dJCm/16/

4

1 に答える 1

1 
var s = "Whether the view should  hidden from (i.e., ignored by) the accessibility service. http://jsfiddle.net/ http://thickbox.net/images/plant4.jpg afhttp://thickbox.net/images/plant4.jpg?4t34t34t";

console.log( getImagesInText(s) ) // "Whether the view should  hidden from (i.e., ignored by) the accessibility service. http://jsfiddle.net/ <img src='http://thickbox.net/images/plant4.jpg' /> af<img src='http://thickbox.net/images/plant4.jpg?4t34t34t' />"

function getImagesInText( s ) {
    var regex = /((http(s)?|ftp):\/\/[\S]*(\.jpg|\.jpeg|\.gif|\.png)[\S]*)/gi;

    return s.replace(regex, function( path ) {
        return "<img src='" + path + "' />";
    });
}
于 2012-11-13T14:50:07.787 に答える