如何在现代和经典应用程序ExtJs之间自动切换
我有一个方法,点击按钮后切换到现代版本:如何在现代和经典应用程序ExtJs之间自动切换,extjs,extjs6-classic,extjs6-modern,Extjs,Extjs6 Classic,Extjs6 Modern,我有一个方法,点击按钮后切换到现代版本: Ext.beforeLoad = function (tags) { var s = location.search, // the query string (ex "?foo=1&bar") profile; if (s.match(/\bclassic\b/)) { profile = 'classic'; } else i
Ext.beforeLoad = function (tags) {
var s = location.search, // the query string (ex "?foo=1&bar")
profile;
if (s.match(/\bclassic\b/)) {
profile = 'classic';
}
else if (s.match(/\bmodern\b/)) {
profile = 'modern';
}
else {
profile = tags.phone ? 'modern' : 'classic';
}
Ext.manifest = profile; // this name must match a build profile name
};
该代码在加载页面之前设置配置文件。我使用以下方法添加参数:
onSwitchToClassicConfirmed: function (choice) {
if (choice === 'yes') {
var s = location.search;
// Strip "?modern" or "&modern" with optionally more "&foo" tokens following
// and ensure we don't start with "?".
s = s.replace(/(^\?|&)modern($|&)/, '').replace(/^\?/, '');
// Add "?classic&" before the remaining tokens and strip & if there are none.
location.search = ('?classic&' + s).replace(/&$/, '');
}
}
但如果设备是手机或平板电脑,我想在现代和经典屏幕之间自动切换。查看属性,它包含当前设备的信息。文档中的描述: 当前设备的常规类型 可能值:
- 电话
- 平板电脑
- 桌面
var profile = Ext.os.deviceType === "Phone" ? "modern" : "classic";