PHP Cookbook

20.5. Displaying a GUI Widget in a Window

Problem

You want to display a window with a GUI widget, such as a button, in it.

Solution

Create the window, create the widget, and then add the widget to the window:

// create the window
$window = &new GtkWindow();

// create the button and add it to the window
$button = &new GTKButton('Click Me, Alice');
$window->add($button);

// display the window
$window->show_all();

// necessary so that the program exits properly
function shutdown() { gtk::main_quit(); }
$window->connect('destroy','shutdown');

// start GTK's signal handling loop
gtk::main();

Discussion

First, you create a window by instantiating a new GtkWindow object. GTK objects must be created as references: &new GtkWindow( ), not new GtkWindow( ). You then create a new GtkButton object with a label “Click Me, Alice”. Passing $button to the window’s add( ) method adds the button to the window. The show_all( ) method displays the window and any widgets inside of it. The only widget inside the window in this example is the button. The next two lines ensure that the program quits when the window is closed. The shutdown( ) function is a callback, as is explained later in Recipe 20.8.

The last line is necessary in all PHP-GTK programs. Calling gtk::main( ) starts the signal-handling loop. This means that the program waits for signals emitted by its GUI widgets and then responds to the signals as they occur. These signals are activities like clicking on buttons, resizing windows, ...