Question

所以，这是我的功能：它应该从输入中提取链接。（从http或类似的东西开始，以“”结束）。问题是我无法调试这个函数的工作情况，因为它停留在循环中。我试图检查原因，仍然无法分辨。

 function linkify(input)
    {
        if (input === undefined || input == null)

        return input;

        var tinput = input;
        var urlextr = "";
        var url = [""];
        var num = 0;

        //search for "http://" first
        while(tinput.match(/http:\/\//gi) != null)
        {
            console.log("http");
            urlextr=tinput.substring(tinput.search("http://"));
            urlextr=urlextr.substring(0,urlextr.search(" "));
            url[num] = urlextr;
            url[num+1] = urlextr;
            num = num + 2;
            tinput = tinput.replace(urlextr,"");
        }

        //search for "https://"
        while(tinput.match(/https:\/\//gi) != null)
        {
            console.log("https");
            urlextr=tinput.substring(tinput.search("https://"));
            urlextr=urlextr.substring(0,urlextr.search(" "));
            url[num] = urlextr;
            url[num+1] = urlextr;
            num = num + 2;
            tinput = tinput.replace(urlextr,"");
        }

        //search for "ftp://" (Why not?)
        while(tinput.match(/ftp:\/\//gi) != null)
        {
            console.log("ftp");
            urlextr=tinput.substring(tinput.search("ftp://"));
            urlextr=urlextr.substring(0,urlextr.search(" "));
            url[num] = urlextr;
            url[num+1] = urlextr;
            num = num + 2;
            tinput = tinput.replace(urlextr,"");
        }

        //search for "www.*" (Must be last!)
        while(tinput.match(/www./gi) != null)
        {
            console.log("www");
            urlextr=tinput.substring(tinput.search("www."));
            urlextr=urlextr.substring(0,urlextr.search(" "));
            url[num] = urlextr;
            url[num+1] = urlextr;
            num = num + 2;
            tinput = tinput.replace(urlextr,"");
        }
        console.log(url);
        return output;
    }

Answer 1

我假设您有一个字符串，其中您有不同的空格分隔的网址。问题在于

urlextr=urlextr.substring(0,urlextr.search(" "));

如果您正在处理字符串中的最后一个URL，它将始终返回一个空字符串，从而使循环无限期地运行。要解决它，你可以像下面那样做

urlextr=tinput.substring(tinput.search("http://"));
urlextr=urlextr.substring(0,urlextr.search(" "));
if (!urlextr) {
    url[num] = tinput.trim();
    break;
}
url[num] = urlextr;
url[num+1] = urlextr;

JS函数陷入循环

1 个答案: