首页 文章

Angular2在组件中迭代JSON

提问于
浏览
0

我正在尝试使用不同的数据迭代JSON .

我正在使用Angular2,Ag-grid Master详细信息,详细介绍了CellRendererParams中的模板:

当我尝试迭代我的JSON获取数据时,我遇到了问题 .

ngOnInit() {
    this.gridOptions = <GridOptions>{};
    this.detailCellRendererParams = {
      detailGridOptions: {
        onGridReady: function(params) {
          params.api.sizeColumnsToFit();
        }
      },

      getDetailRowData: function(params) {},

  template: function(params) {
    console.log('before');
    console.log('datas ', params.data.masterdetail);
    for (const valor of params.data.masterdetail) {
      console.log('eyy');
      if (params.data.masterdetail.hasOwnProperty(valor)) {
        console.log('este valor es : ', valor);
      }
    }
    console.log('after');
    // console.log('template ', params.data.masterdetail);
    return '<div class="aggrid--full-row-container">' + '<div ref="eDetailGrid" </div>' + 'asereje ' + '</div>';
  }
 };
}

永远不要加入...

enter image description here

你可以在 console.log('datas ', params.data.masterdetail); 看到我如何得到 4 数据

新代码更新:

console.log('before');
 console.log('datas ', params.data.masterdetail);
 const objects = JSON.parse(params.data.masterdetail);
    for (const obj of objects) {
      const keys = Object.keys(obj);
      keys.forEach(key => {
        console.log(obj[key]);
      });
    }
    console.log('after');

错误:

ERROR SyntaxError: Unexpected token o in JSON at position 1
at JSON.parse (<anonymous>)
at template (projectunits.component.ts:43)
at DetailCellRenderer.webpackJsonp../node_modules/ag-grid-enterprise/dist/lib/rendering/detail/detailCellRenderer.js.DetailCellRenderer.selectAndSetTemplate (detailCellRenderer.js:96)
at DetailCellRenderer.webpackJsonp../node_modules/ag-grid-enterprise/dist/lib/rendering/detail/detailCellRenderer.js.DetailCellRenderer.init (detailCellRenderer.js:33)
at ComponentResolver.webpackJsonp../node_modules/ag-

enter image description here

我得到这个数据是un file.json . 构建我的Json:

{

 "headers": [ 
   "headerName": "asdasd",
 ],   //end headers

  "datas": [

  "idaam": "11",
  "idorigen": "11",

   "masterdetail": {
        "child1": {
          "name": "I AM",
          "age": "1"

        },

        "child2": {
          "name": "YOU ARE",
          "age": "2"
        },

        "child3": {
        "name": "HE IS",
        "age": "3"
        },
    } //end masterdetail
  ]//end datas

}//end JSON

2 回答

  • 0

    这是我的回答 . 您可以在解析之前检查数据是Object还是简单String . 此代码假定您确定要获取JSON而不是Array

    const objects = (typeof params.data.masterdetail == 'object') ?  JSON.parse(params.data.masterdetail) : params.data.masterdetail;
    for (const obj in objects) {
      const data = objects[obj];
      //Bellow if each attribute data
    }
    
  • 0

    看起来像你不是json数据,你可以尝试

    params.data.masterdetail.forEach((data)=>{
    });
    

    作为它的json,你应该把json分成对象,然后再去做

    const objects = JSON.prase(JSON.stringify(params.data.masterdetail));
     for(const obj of objects){
       const keys = Object.keys(obj);
       keys.forEach((key)=> {
        console.log(obj[key]);
      });
     }
    

相关问题