grid.js php 用法,如何在Javascript函数中添加php变量/数组(在jsGrid({controller})中)

我正在为我的PHP应用程序使用http://js-grid.com。我选择这个库进行内联编辑/更新/添加/删除。现在我想查看来自数据库的$variables或$array数据。这是脚本

$(function() {

$("#jsGrid").jsGrid({

pageButtonCount: 5,

deleteConfirm: "Do you really want to delete the client?",

controller: db,

fields: [

{ name: "Name", type: "text", width: 150 },

{ name: "Age", type: "number", width: 50 },

{ type: "control" }

]

});

});

在上面的代码中，我得到了一个名为controller的属性，它正在渲染db (data)。 db来自文件db.js。该文件如下所示：

(function() {

var db = {

loadData: function(filter) {

return $.grep(this.clients, function(client) {

return (!filter.Name || client.Name.indexOf(filter.Name) > -1)

&& (!filter.Age || client.Age === filter.Age)

&& (!filter.Address || client.Address.indexOf(filter.Address) > -1)

&& (!filter.Country || client.Country === filter.Country)

&& (filter.Married === undefined || client.Married === filter.Married);

});

},

insertItem: function(insertingClient) {

this.clients.push(insertingClient);

},

updateItem: function(updatingClient) { },

deleteItem: function(deletingClient) {

var clientIndex = $.inArray(deletingClient, this.clients);

this.clients.splice(clientIndex, 1);

}

};

window.db = db;

db.countries = [

{ Name: "", Id: 0 },

{ Name: "United States", Id: 1 },

];

db.clients = [

{

"Name": "Otto Clay",

"Age": 61,

},

{

"Name": "Connor Johnston",

"Age": 73,

}

];

}());我也跟着github文档https://github.com/tabalinas/jsgrid-php。但我无法弄清楚如何将我的php $variable or $array放入我的视图以及javaScripts中。

我想要的是：

我想将$array作为controller : db调用到javaScripts部分。

错误：

当我使用controller: <?php echo $array; ?>', its returning null cause I can not call as they called as default fromdb.js`

请帮我解释一下，如何在javaScript中调用php $variable or $array而不是controller: db

先谢谢了。