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javascript - 如何在 Angular ui-grid 中显示从 json 到 ui-grid 的数据

转载 作者:塔克拉玛干 更新时间:2023-11-02 21:15:06 24 4
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我从 django 获取这个 json,我想在 Angular ui-grid 中显示它,但我收到错误:

Error: colDef.name or colDef.field property is required
preprocessColDef@http://127.0.0.1:8000/static/buddy/js/ui-grid.js:3771:1
buildColumns/<@http://127.0.0.1:8000/static/buddy/js/ui-grid.js:3630:7
buildColumns@http://127.0.0.1:8000/static/buddy/js/ui-grid.js:3629:5
dataWatchFunction@http://127.0.0.1:8000/static/buddy/js/ui-grid.js:2749:27
$watchCollectionAction@http://127.0.0.1:8000/static/buddy/js/angular.js:15693:13
$RootScopeProvider/this.$get</Scope.prototype.$digest@http://127.0.0.1:8000/static/buddy/js/angular.js:15826:23
$RootScopeProvider/this.$get</Scope.prototype.$apply@http://127.0.0.1:8000/static/buddy/js/angular.js:16097:13
done@http://127.0.0.1:8000/static/buddy/js/angular.js:10546:36
completeRequest@http://127.0.0.1:8000/static/buddy/js/angular.js:10744:7
requestLoaded@http://127.0.0.1:8000/static/buddy/js/angular.js:10685:1

我只想显示来自“字段”的属性

json 是:

[{"fields": {"joiningtime": null, "boozprofileId": 1, "userId": 1, "likeStatus": true}, "model": "buddy.guestentry", "pk": 1}, {"fields": {"joiningtime": null, "boozprofileId": 1, "userId": 1, "likeStatus": true}, "model": "buddy.guestentry", "pk": 2}, {"fields": {"joiningtime": "2015-10-18T15:53:58.243Z", "boozprofileId": 12, "userId": 3, "likeStatus": true}, "model": "buddy.guestentry", "pk": 3}, {"fields": {"joiningtime": "2015-10-18T15:54:24.055Z", "boozprofileId": 8, "userId": 3, "likeStatus": true}, "model": "buddy.guestentry", "pk": 4}, {"fields": {"joiningtime": null, "boozprofileId": 3, "userId": 1, "likeStatus": true}, "model": "buddy.guestentry", "pk": 5}, {"fields": {"joiningtime": null, "boozprofileId": 3, "userId": 1, "likeStatus": true}, "model": "buddy.guestentry", "pk": 6}, {"fields": {"joiningtime": null, "boozprofileId": 3, "userId": 1, "likeStatus": true}, "model": "buddy.guestentry", "pk": 7}, {"fields": {"joiningtime": null, "boozprofileId": 3, "userId": 1, "likeStatus": true}, "model": "buddy.guestentry", "pk": 8}]

最佳答案

您收到的错误表明您没有为 UI 网格定义列定义,或者可能没有正确定义它们。简单引用嵌套的fields属性为 fields.<attributeName> :

//the JSON from above 
$scope.gridOptions.data = [{"fields": {"joiningtime": null, "boozprofileId": ....}];

$scope.gridOptions.columnDefs = [
{name: 'fields.joiningtime' },
{name: 'fields.boozprofileId' },
{name: 'fields.userId' },
{name: 'fields.likeStatus' }
];

演示 -> http://plnkr.co/edit/KXvES4G64RVwneFbZzV2?p=preview


请记住以正确的 Controller 为目标。你有两个 IndexCtrlajax :

<div ng-controller="ajax">
<div ui-grid="gridOptions" ui-grid-cellNav class="grid"></div>
</div>

关于javascript - 如何在 Angular ui-grid 中显示从 json 到 ui-grid 的数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33462167/

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