[英]java.net.MalformedURLException: no protocol:
如果我的網址路徑中有http,為什么我沒有協議?
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>
postURL參數的值是可以無例外地解析的URL,因此問題出在其他地方。 你可以做什么:
以下程序僅顯示postURL值對於Java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));
}
}
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