Question

webService中，数组如何直接传递？

Answer 1

import java.io.*;
import java.util.*;
import java.net.*;
import org.w3c.dom.*;
import org.apache.soap.util.xml.*;
import org.apache.soap.*;
import org.apache.soap.encoding.*;
import org.apache.soap.encoding.soapenc.*;
import org.apache.soap.rpc.*;
import org.apache.soap.transport.http.SOAPHTTPConnection;

public class testClient {

public static void main(String[] args) throws Exception {

URL url = new URL ("");
//改成你的地址
SOAPMappingRegistry smr = new SOAPMappingRegistry ();
StringDeserializer sd = new StringDeserializer ();
smr.mapTypes (Constants.NS_URI_SOAP_ENC, new QName ("", "Result"), null, null, sd);

// 创建传输路径和参数
SOAPHTTPConnection st = new SOAPHTTPConnection();

// 创建调用
Call call = new Call ();
call.setSOAPTransport(st);
call.setSOAPMappingRegistry (smr);

call.setTargetObjectURI ("");
call.setMethodName("addNumbers");
call.setEncodingStyleURI ("");

Vector params = new Vector();
params.addElement(new Parameter("NumberOne", Double.class, "10", null));
params.addElement(new Parameter("NumberTwo", Double.class, "25", null));
call.setParams(params);

Response resp = null;

try {
resp = call.invoke (url, "");
}
catch (SOAPException e) {
System.err.println("Caught SOAPException (" + e.getFaultCode () + "): " + e.getMessage ());
return;
}

// 检查返回值
if (resp != null && !resp.generatedFault()) {
Parameter ret = resp.getReturnValue();
Object value = ret.getValue();

System.out.println ("Answer--> " + value);
}
else {
Fault fault = resp.getFault ();
System.err.println ("Generated fault: ");
System.out.println (" Fault Code = " + fault.getFaultCode());
System.out.println (" Fault String = " + fault.getFaultString());
}
}
}

你可以参考一下这个.
URL就是axis的地址

Answer 2

你说的是什么WEBSERVER 是JAVA的还是PHP的 还是C#.NET的
C#.NET 有用ARRAY 的办法 真接传输解析
其他的全是自己想办法..
自己在后台造数组的 XML 然后 前台用JS解析XML
比如 你可以在后台这样解释XML
<array>
<0>abc</0>
<1>bcd</0>
</array>