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如果我在 url 路径中有 http，为什么我没有协议？

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"

例外：

    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>
4

1 回答 1

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postURL 参数的值是一个可以毫无例外地解析的 URL，因此问题出在其他地方。你可以做什么：

  • 试验各种小程序参数
  • 请求小程序开发者的支持
  • 如果您没有小程序的源代码，您可能仍然可以对其进行反编译，并了解它是如何工作的：我在哪里可以找到 Java 反编译器？

以下程序仅显示 postURL 值适用于 Java：

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));
    }
}
于 2013-01-11T12:40:13.207 回答