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var jsonString = Ext.util.JSON.decode(response.responseText);
Ext.MessageBox.alert("result : ", jsonString.data);
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
String username = request.getParameter("username");
String password = request.getParameter("password");
System.out.println(username + "\t" + password);
PrintWriter out = response.getWriter();
out.println("{data : '11111'}");
out.flush();
out.close();
}
<?php
$result = "{data : '11111'}";
echo '('.$result.')';
?>
可以跳过php页面,只请求servlet就可以了
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
String username = request.getParameter("username");
String password = request.getParameter("password");
System.out.println(username + "\t" + password);
PrintWriter out = response.getWriter();
out.print("{data : '11111'}");
out.flush();
out.close();
}
servelt不实现跳转页面既可以了。
如果确实要定向php就在php页面输出{data : '11111'}当你请求的时候,页面只显示{data : '11111'}
说明是正确的,有其他字符就错误了。然后用ajax请求请求这个php页面就可以了。