java.net.MalformedURLException : no protocol:

Question

如果 url 路径中有 http，为什么我得不到协议(protocol)？

日志:

network: Connecting http://xxx.ccc.local/upload/up.php?aa=0&bb=Ap%F3lice+de+Seguro&cc=1028&from=documentos with cookie "CLinkLanguage=en; __utma=232844939.1396040569.1356709687.1357294077.1357902500.12; __utmz=232844939.1356709687.1.1.utmcsr=(direct)|utmccn=(direct)|utmcmd=(none); __utmb=232844939.24.10.1357902500; symfony=e0lpbkrcu0bidkpiujd1if4pt4; __utmc=232844939; CLinkLanguage=en; PHPSESSID=uv31kr1vpojvqgnc9ae9nda921"

异常(exception):

    java.net.MalformedURLException: no protocol: 
        at java.net.URL.<init>(Unknown Source)
        at java.net.URL.<init>(Unknown Source)
        at java.net.URL.<init>(Unknown Source)

html

 <APPLET  CODE = "wjhk.jupload.JUploadApplet" ARCHIVE = "upload/wjhk.jupload.jar" WIDTH = "600" HEIGHT = "400" MAYSCRIPT></XMP>
        <PARAM NAME = CODE VALUE = "wjhk.jupload.JUploadApplet" >
        <PARAM NAME = ARCHIVE VALUE = "upload/wjhk.jupload.jar" >
        <PARAM NAME = "type" VALUE="application/x-java-applet;version=1.4">
        <PARAM NAME = "scriptable" VALUE="false">
        <PARAM NAME = "postURL" VALUE ="{$url}">
        <PARAM NAME = "anexosID" VALUE ="{$anexosID}">
        <PARAM NAME = "subanexosID" VALUE ="{$IdConsulta}">
        <PARAM NAME = "companyID" VALUE ="{$companyID}">
        <PARAM NAME = "resultURL" VALUE ="{$resultUrl}">
        <param name="debug" value="true">


    Java 1.4 or higher plugin required.

<APPLET  CODE = "wjhk.jupload.JUploadApplet" ARCHIVE = "upload/wjhk.jupload.jar" WIDTH = "600" HEIGHT = "400" MAYSCRIPT></XMP>
    <PARAM NAME = CODE VALUE = "wjhk.jupload.JUploadApplet" >
    <PARAM NAME = ARCHIVE VALUE = "upload/wjhk.jupload.jar" >
    <PARAM NAME = "type" VALUE="application/x-java-applet;version=1.4">
    <PARAM NAME = "scriptable" VALUE="false">
    <PARAM NAME = "postURL" VALUE ="http://xxx.ccc.local/upload/up.php?aa=0&bb=Alvar%E1%2Ffg&cc=1028&from=documentos">
    <PARAM NAME = "anexosID" VALUE ="">
    <PARAM NAME = "subanexosID" VALUE ="">
    <PARAM NAME = "companyID" VALUE ="">
    <PARAM NAME = "resultURL" VALUE ="">
    <param name="debug" value="true">


Java 1.4 or higher plugin required.
</APPLET>

Answer 1

postURL参数的值是一个可以正常解析的URL，因此问题出在其他地方。你可以做什么:

• 尝试各种小程序参数
• 请求小程序开发人员的支持
• 如果您没有该小程序的源代码，您仍然可以对其进行反编译，并了解其工作原理:Where can I find a Java decompiler?

以下程序仅显示 postURL 值对于 Java 来说是可以的:

public class A {
    public static void main(String[] args) throws MalformedURLException {
        String s = "http://xxx.ccc.local/upload/up.php?aa=0&bb=Alvar%E1%2Ffg&cc=1028&from=documentos";

        URL url = new URL(s);
        String protocol = url.getProtocol();
        System.out.println(String.format("A::main: protocol = '%s'", protocol));
    }
}