我有以下對象數組：const formulas =[ { "formulaID": "1", "versionID": 1, "formulaClass": 3, "formulaType": "34", "outputName": "Chocolate Milk 2%" }, { "formulaID": "4", "versionID": 1, "formulaClass": 3, "formulaType": "17", "outputName": "Hazelnut Creamer" }, { "formulaID": "6", "versionID": 1, "formulaClass": 3, "formulaType": "23", "outputName": "White Milk 2%" }];const yields =[ { "formulaID": "4", "versionID": 1, "yieldFactor": 0.93 }, { "formulaID": "4", "versionID": 2, "yieldFactor": 0.98 }, { "formulaID": "6", "versionID": 1, "yieldFactor": 0.95 }, { "formulaID": "7", "versionID": 1, "yieldFactor": 0.85 }];并嘗試以編程方式創建此輸出：const result =[ { "formulaID": "7", "versionID": 1, "yieldFactor": 0.85, "outputName": "" }, { "formulaID": "4", "versionID": 1, "yieldFactor": 0.93, "outputName": "Hazelnut Creamer" }, { "formulaID": "4", "versionID": 2, "yieldFactor": 0.98, "outputName": "" }, { "formulaID": "6", "versionID": 1, "yieldFactor": 0.95, "outputName": "White Milk 2%" }];在這篇文章的幫助下，我編寫了以下代碼：const result = yields.map(yld => ({ formulaID: yld.formulaID, versionID: yld.versionID, yieldFactor: yld.yieldFactor, outputName: formulas.filter(f => (f.formulaID + '-' + f.versionID).includes(yld.formulaID + '-' + yld.versionID))}));console.log(result);它接近預期的結果，但我不確定如何僅隔離outputName. 在其當前狀態下，它給出了找到匹配項的整個數組。對于原始源數據數組之間不匹配的位置，如何僅顯示outputName匹配項和空字符串？outputName編輯 - 當前輸出如下所示：
從兩個不同的對象數組中選擇數據以創建一個新的對象數組
蠱毒傳說
2022-07-21 20:58:19
