2个JSON对象合并问题~

求教怎么将columnDefs1 和columnDefs2合并变成columnDefs3谢谢~
    var columnDefs1 = [
                    { headerName: "AirBookingItemId", field: "AirBookingItemId", hide: true },
                    { headerName: "AirBookingHeadId", field: "AirBookingHeadId", hide: true },
                    { headerName: "S.Marks", field: "ShippingMarks", width: 120 },
{ headerName: "Item SEQ", field: "ItemSequence", width: 130 }
];

    var columnDefs2 = [
                     {
                         field: "ShippingMarks",
                         cellEditor: 'agLargeTextCellEditor',
                         editable: true,
                         width: 120,
                         cellEditorParams: {
                             maxLength: 2000,
                             rows: 15,
                             cols: 61
                         }
                     }
 ];
 
    var columnDefs3 = [
{ headerName: "AirBookingItemId", field: "AirBookingItemId", hide: true },
                    { headerName: "AirBookingHeadId", field: "AirBookingHeadId", hide: true },
                     {
 headerName: "S.Marks",
                         field: "ShippingMarks",
 width: 120,
                         cellEditor: 'agLargeTextCellEditor',
                         editable: true,
                         cellEditorParams: {
                             maxLength: 2000,
                             rows: 15,
                             cols: 61
                         }
                     },
 { headerName: "Item SEQ", field: "ItemSequence", width: 130 }
 ];    

解决方案 »

  1.   

    假设 columnDefs1和 columnDefs2数据有出入时以 columnDefs1为主，并且每个元素以field作为唯一标记columnDefs2.forEach(e=>{
    let isExist=false;//表示columnDefs2中的这个元素在columnDefs1中不存在
    for(let i=0;i<columnDefs1.length;i++){
       if(e.field===columnDefs1[i].field){//如果相同表示存在
          isExist=true;
          for(let key in e){
             columnDefs1[i][key]=columnDefs1[i][key]?columnDefs1[i][key]:e[key]
          };
          break;
       }
    }
    if(!isExist){//如果不存在，这把元素push到columnDefs1
    columnDefs1.push(e)
    }
    })
    console.log(columnDefs1)
      

  2.   

    var columnDefs1 = [
        { headerName: "AirBookingItemId", field: "AirBookingItemId", hide: true },
        { headerName: "AirBookingHeadId", field: "AirBookingHeadId", hide: true },
        { headerName: "S.Marks", field: "ShippingMarks", width: 120 },
        { headerName: "Item SEQ", field: "ItemSequence", width: 130 }
    ];
    var columnDefs2 = [
        {
            field: "ShippingMarks",
            cellEditor: "agLargeTextCellEditor",
            editable: true,
            width: 120,
            cellEditorParams: {
                maxLength: 2000,
                rows: 15,
                cols: 61
            }
        }
    ]
    var columnDefs3 =[];
    function ConnectJson(arr1, arr2,arr3) {
        for(var i in arr1){
            arr3.push({});
            for(var j in arr2){
                if(arr1[i].hasOwnProperty('field')&&arr2[j].hasOwnProperty('field')){
                    if(arr1[i].field===arr2[j].field){
                      Object.assign(arr3[i],arr1[i],arr2[j]);
                    }else {
                        Object.assign(arr3[i],arr1[i]);
                    }
                } 
            }
        }
    }
    ConnectJson(columnDefs1,columnDefs2,columnDefs3);合并对象用Object.assign方法。
      

  3.   

    用jQuery的extend方法就简单：
    var options = $.extend({}, defaults, options);
    参考
      

  4.   

    headerName没有了。
      

  5.   

    加一个深拷贝选项：
    var cc=$.extend(true,columnDefs1,columnDefs2);