我正在创建一个jnlp文件,但我需要接收一些URL参数。
我有一个从jsp文件捕获URL的方法:
String getParameter (HttpServletRequest request, String param)问题是如何向jnlp文件中添加参数:
<?xml version="1.0" encoding="utf-8"?>
<jnlp spec="1.0" codebase="https://localhost:8443/java-web-start/test/" href="start.jnlp">
<information>
<title>TestApp</title>
<vendor>Oracle</vendor>
<offline-allowed/>
</information>
<security>
<all-permissions/>
</security>
<resources>
<java version="1.5+"/>
<jar href="start.jar" main="true"/>
</resources>
<application-desc main-class="com.Main"/>
</jnlp>下面是index.jsp文件:
<%!
String getParameter(HttpServletRequest request, String param) {
String result = request.getParameter(param);
return result.replace("&", "&").replace("\"", """).replace("<", "<").replace(">", ">").replace("'","$#039;");
}
%>
<%=getParameter(request, "requestURL")%> 我希望它在我的jnlp文件中,然后下载并执行:
<?xml version="1.0" encoding="utf-8"?>
<jnlp spec="1.0" codebase="https://localhost:8443/java-web-start/test/" href="start.jnlp">
<information>
<title>TestApp</title>
<vendor>Oracle</vendor>
<offline-allowed/>
</information>
<security>
<all-permissions/>
</security>
<resources>
<java version="1.5+"/>
<jar href="start.jar" main="true"/>
</resources>
<application-desc main-class="com.Main">
<argument><%= clientCount %></argument>
<argument><%=getParameter(request, "requestURL")%></argument>
</<application-desc>
</jnlp>发布于 2016-02-15 20:51:51
加载jnlp并追加所需的参数,将jnlp文件视为简单的XML。
看看如何在XML上追加节点的this示例。
https://stackoverflow.com/questions/35415727
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