Java applet中传递值

Question

我在java applet中传递值，但是值作为null到达那里

我的用于applet部署的javascript：

<script src="javascript/deployJava.js"></script>
<script>
    var attributes = { id:'dynamictreeapplet', code:'jstojava.com.vaannila.utility.dynamicTreeApplet.class',width:1000,height:1000} ;
    var parameters = {jnlp_href:'./appletjars/a.jnlp'} ;
    deployJava.runApplet(attributes, parameters, '1.6');
</script>

我的jnlp

<?xml version="1.0" encoding="UTF-8"?>
<jnlp spec="1.0+" codebase="http://localhost:8080/docRuleToolProtocol/appletjars/" href="a.jnlp">

    <information>
        <title>dynamictree</title>
        <vendor>Ashutosh</vendor>
    </information>

    <resources>
        <j2se version="1.5+" href="http://java.sun.com/products/autodl/j2se" />
        <jar href="dynamictree.jar" main="true" />
        <jar href="prefuse.jar" main="true" />
    </resources>

javascript传递值：

function showSelected(value){
alert("the value given from"+value);
var diseasename=value;
alert(diseasename);
dynamictreeapplet.dieasenameencode=diseasename; 
}

    <applet-desc name="dynamictree" main-class="com.vaannila.utility.dynamicTreeApplet" width="500" height="500">
    </applet-desc>
</jnlp>

Answer 1

更改

dynamictreeapplet.dieasenameencode=diseasename;

至

document.dynamictreeapplet.dieasenameencode=diseasename;

并且可能将dieasenameencode编码为diseasenameencode。