i具有以下数据库结果(使用sequelize(,结果:
来自数据库的JSON:
[{"codSparePart":"SP001","name":"NAME","description":"DESCRIPTION","codProject":1,"available":1,"codManufacturer":1,"stock":4,"manufacturer.name":"MAN1","manufacturer.description":"DES1","manufacturer.codManufacturer":1}]
在控制器中我执行下一个:
vm.spareParts = data.data;
console.log('-------- ' + JSON.stringify(vm.spareParts));
for (var i = 0; i < vm.spareParts.length; i++) {
vm.items.push({
codSparePart: vm.spareParts[i].codSparePart,
name: vm.spareParts[i].name,
description: vm.spareParts[i].description,
available: vm.spareParts[i].available,
codManufacturer: vm.spareParts[i].codManufacturer,
nameManufacturer: vm.sparePart[i]['manufacturer.description'],
stock: vm.spareParts[i].stock,
codProject: vm.spareParts[i].codProject
});
除了用点读取变量外,所有人都可以正常工作。 nameManufacturer: vm.sparePart[i]['manufacturer.description']
nameManufacturer: vm.sparePart[i]['manufacturer.name']
我如何阅读这些变量,我正在寻找解决方案,但我认为太简单了,没有任何信息。预先感谢您的帮助。
您有错字,将vm.sparePart
更改为vm.spareParts
此行:
nameManufacturer: vm.sparePart[i]['manufacturer.description'],
您的代码似乎正常工作:
const array =[{"codSparePart":"SP001","name":"NAME","description":"DESCRIPTION","codProject":1,"available":1,"codManufacturer":1,"stock":4,"manufacturer.name":"MAN1","manufacturer.description":"DES1","manufacturer.codManufacturer":1}]
console.log(array[0]['manufacturer.description']);
更多信息:如果其名称包含点,如何获得JSON对象值?