小编Ken*_*iro的帖子

dojo dgrid与来自url(json数据)的存储重新加载/刷新按钮

首先,我使用的是dojo 1.1.0和dgrid 0.4.0.我正在创建一个包含几个dgrids的页面.特别是一个网格需要从URL加载JSON数据并显示它.它目前使用RequestMemory存储工作正常.然而,这是一个"记忆"商店.虽然这对其他一些人来说可能是微不足道的,但我需要找到一种方法来加载现在正在完成的数据,然后在屏幕上添加一个刷新按钮,调用必要的函数/方法从网址重新加载数据.重新填充dgrid.

JSON数据来自如下格式的url:

[{"id":1,"name":"trans1","username":"trans1","status":"Available","inactive":"no","checkedout":"false","details":"","assignedNum":"0","completedNum":"0","avgTime":"00:00","aaOn":true},{"id":2,"name":"trans2","username":"trans2","status":"Available","inactive":"no","checkedout":"false","details":"","assignedNum":"0","completedNum":"0","avgTime":"00:00","aaOn":true},{"id":3,"name":"trans3","username":"trans3","status":"Available","inactive":"no","checkedout":"false","details":"","assignedNum":"0","completedNum":"0","avgTime":"00:00","aaOn":false},{"id":4,"name":"trans4","username":"trans4","status":"Available","inactive":"no","checkedout":"false","details":"","assignedNum":"0","completedNum":"0","avgTime":"00:00","aaOn":false},{"id":5,"name":"trans5","username":"trans5","status":"Available","inactive":"no","checkedout":"false","details":"","assignedNum":"0","completedNum":"0","avgTime":"00:00","aaOn":false},{"id":6,"name":"trans6","username":"trans6","status":"Available","inactive":"no","checkedout":"false","details":"","assignedNum":"0","completedNum":"0","avgTime":"00:00","aaOn":false}]
Run Code Online (Sandbox Code Playgroud)

要求你在下面看到实际上包含的不仅仅是这个网格...因此所有额外的.

我不需要更新本地商店,也不想监视更改的URL.我只是希望能够根据需要从URL重新加载数据.

这是我目前用于初始加载网格(工作)的代码和我需要开始工作的刷新按钮.

require([ 'dojo/store/Observable' ,'dijit/Dialog', 'dijit/form/Select', 'dijit/form/MultiSelect', 'dijit/form/TextBox', 'dijit/layout/TabContainer', 'dijit/layout/ContentPane','dojo/request', 'dojo/request/xhr', 'dijit/form/ToggleButton', 'dijit/form/Button', 'dojo/parser', 'dojo/_base/declare', 'dgrid/Grid', 'dgrid/Selection', 'dgrid/Editor', 'dgrid/extensions/ColumnHider', 'dgrid/extensions/Pagination', 'dstore/RequestMemory', 'dijit/form/Select', 'dijit/registry','dojox/data/XmlStore', 'dojo/domReady!'], function (Observable, Dialog, Select, MultiSelect, TextBox, TabContainer, ContentPane, request, xhr, ToggleButton, Button, parser, declare, Grid, Selection, Editor, ColumnHider, Pagination, RequestMemory, Select, registry, XmlStore) {


//workers dgrid
    var workersStore = new RequestMemory({ target: '/autoAssign/getWorkers.aa?TASKTYPE=transport&INACTIVE=FALSE' });        
    var workerGrid = new (declare([ Grid, Selection, Pagination, Editor, ColumnHider ]))({
        collection: workersStore,
        className: 'dgrid-autoheight',
        id: …
Run Code Online (Sandbox Code Playgroud)

javascript dojo store dgrid dstore

3
推荐指数
1
解决办法
2570
查看次数

StringBuilder和string.IsIntern方法

任何人都可以解释,为什么这段代码返回abc,True?

        var str1 = "abc";
        var str2 = "";

        var builder = new StringBuilder();
        builder.Append("a");
        builder.Append("b");
        builder.Append("c");


        str2 = builder.ToString();
        Console.WriteLine(string.IsInterned(str2));
        Console.WriteLine(str1.Equals(str2));
Run Code Online (Sandbox Code Playgroud)

我用反射器看StringBuilder.ToString()方法但是找不到任何使用string.Intern的方法.

我使用的是Net Framework 4.5

c# string stringbuilder

2
推荐指数
1
解决办法
146
查看次数

Kendoui条形图设置系列点击一个按钮

我正在使用KendoUI条形图.请检查这个小提琴.在这个小提琴中,我series在设计时设置配置.我想series在运行时单击按钮添加到图表.如何series在单击按钮时添加到kendo图表.

目前series是在设计时分配的,我希望在运行时点击按钮时也这样做.

码:

<!DOCTYPE html>
<html>
<head>
    <base href="http://demos.telerik.com/kendo-ui/bar-charts/grouped-stacked-bar">
    <style>html { font-size: 14px; font-family: Arial, Helvetica, sans-serif; }</style>
    <title></title>
    <link rel="stylesheet" href="//kendo.cdn.telerik.com/2016.1.112/styles/kendo.common-material.min.css" />
    <link rel="stylesheet" href="//kendo.cdn.telerik.com/2016.1.112/styles/kendo.material.min.css" />

    <script src="//kendo.cdn.telerik.com/2016.1.112/js/jquery.min.js"></script>
    <script src="//kendo.cdn.telerik.com/2016.1.112/js/kendo.all.min.js"></script>
</head>
<body>
<div id="example">
    <div class="demo-section k-content wide">
        <div id="chart"></div>
    </div>
    <script>
        function createChart() {
            $("#chart").kendoChart({
                title: {
                    text: "World population by age group and sex"
                },
                legend: {
                    visible: false
                },
                seriesDefaults: {
                    type: "column"
                },
                series: [{ …
Run Code Online (Sandbox Code Playgroud)

javascript kendo-ui kendo-chart

2
推荐指数
1
解决办法
1175
查看次数

如何在Dojo中创建对象

如何在Dojo中创建对象?如何在Dojo中继承该对象(例如:我想用字段a和方法ACK创建类A,然后创建类B和C,B继承A和B继承C)?怎么做 ?

dojo

1
推荐指数
1
解决办法
3667
查看次数

Dgrid 0.4和dstore:更新UI中的行而不放置请求

在Dgrid 0.3.16我使用的是Observable商店,当我在商店中的数据发生变化时,我调用了商店通知功能.(不是'put',因为我只需要一个UI更新,这是一个特定的情况)

store.notify(object, existingId);
Run Code Online (Sandbox Code Playgroud)

我现在已将Dgrid升级到版本0.4,我使用'dstore'作为商店.商店是这样创建的:

        var store = new declare([ Rest, SimpleQuery, Trackable, Cache, TreeStore ])(lang.mixin({
            target:"/ac/api?fetchview",
            idProperty:"$uniqueid", 
            useRangeHeaders: true
        }, config));

        store.getRootCollection = function (parent, options) {
            return this.root.filter({parent: parent.$position},options);
        };

        store.getChildren = function (parent, options) {
            return this.root.filter({parent: parent.$position},options);
        };

        store.mayHaveChildren = function (item) {
            return item.$iscategory;
        };

        this.collection = store;
Run Code Online (Sandbox Code Playgroud)

如何更改一行而不调用"put"时如何通知商店?我需要dGrid重新渲染行.

dojo dgrid dstore

1
推荐指数
1
解决办法
883
查看次数

标签 统计

dojo ×3

dgrid ×2

dstore ×2

javascript ×2

c# ×1

kendo-chart ×1

kendo-ui ×1

store ×1

string ×1

stringbuilder ×1