TinyMCE 4切换工具栏按钮状态

Question

切换工具栏按钮状态的最简单方法是什么(就像使用默认的粗体按钮一样)？我无法"获得"那个将我从默认变为选中的按钮外观的Tinymce.这是我的插件代码(简化):

tinymce.PluginManager.add('myplugin', function (editor) {
    editor.addButton('mybutton', {
        text: false,
        image: 'someimage.png',

        onclick: function () {
            /* Toggle this toolbar button state to selected (like with the tinymce bold-button)*/
            /* and of course some other code goes here */
        }
    });
});

Answer 1

在TinyMCE 4中,您可以使用更简单的stateSelector设置:

editor.addButton('SomeButton', {
    text: 'My button',
    stateSelector: '.class-of-node' // or use an element (an id would probably work as well, but I haven't tried it myself)
});

或者您可以使用"nodechange"事件使用自定义逻辑

editor.addButton('SomeButton', {
    text: 'My button',
    onPostRender: function() {
        var ctrl = this;

        ed.on('NodeChange', function(e) {
            ctrl.active(e.element.nodeName == 'A');
        });
    }
});

参考:https: //www.tinymce.com/docs/advanced/migration-guide-from-3.x/#controlstates

Answer 2

如果其他人在这个问题上找到了这篇文章 - 我发现了一个更简单的方法,使用4.0.16中的onclick:

/* In the timymce > plugins, name your pluginfolder "my_crazy_plugin" and
plugin file as "plugin.min.js" */

/* Your plugin file: plugin.min.js */
tinymce.PluginManager.add('my_crazy_plugin', function(editor) {
   /* Actions to do on button click */
   function my_action() {
        this.active( !this.active() );
        var state = this.active();

        if (state){
            alert(state); /* Do your true-stuff here */
        }
        else {
            alert(state); /* Do your false-stuff here */
        }
    }

    editor.addButton('mybutton', {
        image: 'tinymce/plugins/my_crazy_plugin/img/some16x16icon.png',
        title: 'That Bubble Help text',
        onclick: my_action
    });
});


/* Your file with the tinymce init section: */
tinymce.init({
    plugins: [
        "my_crazy_plugin"
        ],
    toolbar: "mybutton"
});