Is there a problem with
response.sendRedirect(myUrl);
Also, javascript may be disabled, so I'd use
out.println("<HTML><HEAD>\n");
out.println("<meta http-equiv=\"Refresh\" content=\"0, url= " + myUrl +
"\">\n");
out.println("<body>\n");
out.println("<a href=\"" + myUrl + "\"> Next</a>\n");
out.println("</body>\n</html>");
or something to that effect.
The operative line is the
<meta http-equiv="Refresh" content="0, url=someURL">
and it's always nice to leave an achor "just in case" but YMMV
/bill
Please respond to "A mailing list for discussion about Sun Microsystem's Java
Servlet API Technology." <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED]
cc: (bcc: Bill Fellows/MO/americancentury)
Subject: must be a better way
Below is a portion of code that I use to exit a servlet and redirect the web
browser to a different HTML page after writing data to a db.
out.println("<HTML><HEAD>");
out.println("<SCRIPT language=\"javascript\">");
out.println("var home=\"http://xxx.xxx.xxx.xx/next.html\";");
out.println("setTimeout(\"location.href=home;\", 1);");
out.println("</SCRIPT><HEAD><BODY>");
out.println("</BODY></HTML>");
There must be a better way than this to jump to a new HTML page!
Please, any suggestions?
Thanks.
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