Question

我需要从另一个dojo模板小部件调用dojo模板小部件中的函数。问题是我需要将此函数称为静态函数，而不是创建模板的新实例 - 无论如何要执行此操作？在我的应用程序中，模板1是数据网格，而模板2是通过单击数据网格中的行打开的表单。我需要根据表单中采取的操作刷新数据网格

由于

   Start Page:

<!DOCTYPE html>
<html >
<head>
<meta name="viewport" content="initial-scale=1, maximum-scale=1,user-scalable=no"/>
    <link rel="stylesheet" href="http://js.arcgis.com/3.10/js/dojo/dijit/themes/claro/claro.css">
   <link rel="stylesheet" href="http://js.arcgis.com/3.10/js/esri/css/esri.css">

  <script>
    var dojoConfig = {
    parseOnLoad:true,
    async: true,
    isDebug:true,
    packages: [
     {name: "Scripts", location: location.pathname.replace(/\/[^/]+$/, "") + "/Scripts"}
    ]

};
</script>
  <script src="http://js.arcgis.com/3.10/"></script>

    <script>
        require(["Scripts/Mod1", "Scripts/Mod2"],
        function (Mod1, Mod2) {
            M1 = new Mod1();
            M1.M1Method("call from main page");
           // Mod1.M1Method("call from main page");//any way to make so this could work like static function?
        });
    </script>
</head>
<body class="claro">
    <div>look here you</div>
</body>
</html>

模板1：

define(["Scripts/Mod2", "dojo/_base/declare", "dijit/_WidgetBase"],
 function (Mod2, declare, _WidgetBase) {
     return declare([_WidgetBase], {
         M1Method: function (msg) {
             alert(msg);
             M2 = new Mod2();
             M2.M2Method("call from Mod1"); //works great
             // Mod2.M2Method("call from Mod1"); //any way to make so this could work like static function?
         },
         M1Method2: function (msg) {
             alert(msg);
         }
     }
 )
 }
);

模板2：

define(["require", "dojo/_base/declare", "dijit/_WidgetBase"],
 function (require, declare, _WidgetBase) {
     return declare([_WidgetBase], {
         M2Method: function (msg) {
             alert(msg);
             try {
                 require(["Scripts/Mod1"], function (Mod1) {
                     M1 = new Mod1();
                     M1.M1Method2("call from Mod2");
                     //Mod1.M1Method2("call from Mod2");//any way to make so this could work like static function?
                 });
             } catch (dohObj) {
                 alert(dohObj.message);
             }
         }
     }
 )
 }
);

Answer 1

我不完全确定你的实际问题是什么，但如果你想创建静态方法＆＃34;，那么是的，它是可能的。只需在declare()语句之后添加它们，如下所示：

var M1 = declare([_WidgetBase], {
    /** Your code */
});
M1.M1Method = function(msg) {
    /** Your static method */
};
return M1;

但是，请注意，如果您创建静态方法，则在创建实例时它不可用，例如以下方法将起作用：

M1.M1Method("test");

但这不再有效：

var m = new M1();
m.M1Method("test");

如果你想创建一个既可以作为原型的一部分而又可以作为＆＃34;静态方法＆＃34;的方法，那么你可以写下这样的东西：

/** An object containing your "static methods" */
var statics = {
    static: function (msg) {
        console.log("Static:" + msg);
    }
};

/** Your actual prototype */
var M1 = declare([], lang.mixin({
    nonStatic: function (msg) {
        console.log("Non static:" + msg);
    }
}, statics));
/** Mixin the statics */
lang.mixin(M1, statics);
return M1;

因此，我们在此处使用dojo/_base/lang::mixin()两次混合静态对象，首先我们将其用于将其混合到原型中，因此您仍然可以在＃39时使用该方法;有一个实例，然后我们将它混合到结果中。

我在JSFiddle上设置了一个基本示例：http://jsfiddle.net/g00glen00b/FE9Qz/

来自另一个dojo模板小部件的dojo模板小部件中的call函数

1 个答案: