this is the route to handle the login POST request:
POST /login/submit controllers.Users.loginSubmit(user : String, password : String)
this is the login.scala.html:
<form method="post" action="???">
<input type="text" name="username" /><br/>
<input type="password" name="password" /><br/>
<input type="submit" value="Login" />
</form>
I got two questions:
- what should be the value of action? is it "login/submit"?
- how do you pass this form to be handled in the loginSubmit function?
thanks
If it's
POST
form, you don't need to declare params in theroute
:Template:
Import:
Controller:
Template form helpers
There are also form template helpers available for creating forms in Play's template so the same can be done as:
They are especially useful when working with large and/or
pre-filled
formsIn Play Framework version 2.5.x Form.form() is deprecated and you should use inject a FormFactory
Here you can find example: The method form(Class) from Form class is deprecated in Play! Framework
Import:
Inject:
Controller: