Thanks so far.
. but that link does not point to a solution for my problem.
Here a bit more definition of my problem:
<html>
<head>
<title>test</title>
</head>
<body>
<table border="0" align="left" cellpadding="0" cellspacing="0">
<tr>
<td width="755" align="left" valign="top">
<span jwcid="kopf">KOPF</span>
</td>
</tr>
</table>
</body>
</html>
...
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE page-specification PUBLIC
"-//Apache Software Foundation//Tapestry Specification 3.0//EN"
"http://jakarta.apache.org/tapestry/dtd/Tapestry_3_0.dtd";>
<page-specification class="Home">
<component id="kopf" type="Insert">
<binding name="value" expression="kopfDaten"/>
<binding name="raw" expression="true"/>
</component>
</page-specification>
...
import org.apache.tapestry.html.BasePage;
public class Home extends BasePage {
public String getKopfDaten() {
return readURL("http://url/sript.php?blatt=??";);
}
private String readURL(String url) {
String urlDaten = "";
try{
URL urlCon = new URL(url);
BufferedReader in = new BufferedReader(
new InputStreamReader(
urlCon.openStream()));
String inputLine;
while ((inputLine = in.readLine()) != null)
urlDaten += inputLine;
in.close();
}
catch (MalformedURLException ex){
}
catch (IOException ex){
}
return urlDaten;
}
}
..
Please find above the Tapestry .html, .page and .java
Now I can start it with http://url:8080/name/app. But I want to start it
with a
value for 'blatt'. I already tried:
http://url:8080/name/app?service=page/Home
<http://url:8080/name/app?service=page/Home&blatt=xy> &blatt=xy.
Maybe this is the right way, but how can I read the the value for blatt in
Home.java?
It is not satisfactory for me to put the parameter 'blatt' into the web.xml.
I have to start
The Tapestry App. from a non-Tapestry Service!
Hopefully this description is better now and someone can help me.
Thanks.
