J2ME中的java URL编码？

Question

我在项目中工作时遇到了一个问题，当我试图将字符串解析为URL时，我可以发送我的HttpRequest。问题是无法转换字符串，而我

java。lang.IllegalArgumentException:no'：'在URL中

这就是我正在使用的方法

 public String getResultat()  throws Exception {


     /// CODE J2ME

     int ch;
        StringBuffer sb = new StringBuffer("");
        HttpConnection ht;
        DataInputStream ds;
        try {

       String chx="http:x.com";

       String ecodedString=EnvoyerSMS.urlEncode(chx);

            ht = (HttpConnection) Connector.open(ecodedString);
            ds = ht.openDataInputStream();
            while ((ch = ds.read()) != -1) {
                sb.append((char)ch);
            }
        } catch (IOException ex) {
            ex.printStackTrace();
        }
        String reponse = sb.toString().trim();
        System.out.println("REPONSE "+reponse);
        return  reponse;
        }

这是我用来转换字符串的静态方法

    public static String urlEncode(String s) {
    StringBuffer sbuf = new StringBuffer();
    int len = s.length();
    for (int i = 0; i < len; i++) {
        int ch = s.charAt(i);
        if ('A' <= ch && ch <= 'Z') { // 'A'..'Z'
            sbuf.append((char)ch);
        } else if ('a' <= ch && ch <= 'z') { // 'a'..'z'
            sbuf.append((char)ch);
        } else if ('0' <= ch && ch <= '9') { // '0'..'9'
            sbuf.append((char)ch);
        } else if (ch == ' ') { // space
            sbuf.append('+');
        } else if (ch == '-' || ch == '_'   //these characters don't need encoding
                || ch == '.' || ch == '*') {
            sbuf.append((char)ch);
        } else if (ch <= 0x007f) { // other ASCII
            sbuf.append(hex(ch));
        } else if (ch <= 0x07FF) { // non-ASCII <= 0x7FF
            sbuf.append(hex(0xc0 | (ch >> 6)));
            sbuf.append(hex(0x80 | (ch & 0x3F)));
        } else { // 0x7FF < ch <= 0xFFFF
            sbuf.append(hex(0xe0 | (ch >> 12)));
            sbuf.append(hex(0x80 | ((ch >> 6) & 0x3F)));
            sbuf.append(hex(0x80 | (ch & 0x3F)));
        }
    }
    return sbuf.toString();
}

    //get the encoded value of a single symbol, each return value is 3 characters long
static String hex(int sym)
 {
     return(hex.substring(sym*3, sym*3 + 3));
 }

// Hex constants concatenated into a string, messy but efficient
final static String hex =
"%00%01%02%03%04%05%06%07%08%09%0a%0b%0c%0d%0e%0f%10%11%12%13%14%15%16%17%18%19%1a%1b%1c%1d%1e%1f" +
"%20%21%22%23%24%25%26%27%28%29%2a%2b%2c%2d%2e%2f%30%31%32%33%34%35%36%37%38%39%3a%3b%3c%3d%3e%3f" +
"%40%41%42%43%44%45%46%47%48%49%4a%4b%4c%4d%4e%4f%50%51%52%53%54%55%56%57%58%59%5a%5b%5c%5d%5e%5f" +
"%60%61%62%63%64%65%66%67%68%69%6a%6b%6c%6d%6e%6f%70%71%72%73%74%75%76%77%78%79%7a%7b%7c%7d%7e%7f" +
"%80%81%82%83%84%85%86%87%88%89%8a%8b%8c%8d%8e%8f%90%91%92%93%94%95%96%97%98%99%9a%9b%9c%9d%9e%9f" +
"%a0%a1%a2%a3%a4%a5%a6%a7%a8%a9%aa%ab%ac%ad%ae%af%b0%b1%b2%b3%b4%b5%b6%b7%b8%b9%ba%bb%bc%bd%be%bf" +
"%c0%c1%c2%c3%c4%c5%c6%c7%c8%c9%ca%cb%cc%cd%ce%cf%d0%d1%d2%d3%d4%d5%d6%d7%d8%d9%da%db%dc%dd%de%df" +
"%e0%e1%e2%e3%e4%e5%e6%e7%e8%e9%ea%eb%ec%ed%ee%ef%f0%f1%f2%f3%f4%f5%f6%f7%f8%f9%fa%fb%fc%fd%fe%ff";
}

谢谢：D

Answer 1

URLEncoder与您编写的代码不同。与此相当的是

new URI(null, string, null).toASCIIString()

尽管名称不同，URLEncoder并不用于编码URL。它用于编码URL参数、和POST键值对

Answer 2

它不应该是http://x.com而不是http:x.com吗

Bu t如果你有这样一个非常简单的URL（即没有参数），你根本不需要对它进行编码

但是，假设您的URL确实包含参数-例如http://x.com?param1=value1&param2=value2，那么您将只对参数值进行如下编码：

String encodedString = "http://x.com?param1=" + EnvoyerSMS.urlEncode(value1) + "&param2=" + EnvoyerSMS.urlEncode(value2);

。。。但是使用StringBuffer来连接这些部分，而不是像我一样使用+操作符

注意-您得到no ':' in URL错误的原因是您对整个URL进行了错误编码，因此:字符也被编码（编码为任何等效的编码）

如果执行System.out.println(encodedString);，您将看到HttpConnection对象试图使用的URL