当我访问“ dept_name”时,JSON返回“ undefined”。 如何返回正确的输出?
API中的JSON对象(内容)
"departments": [
{
"dept_no": "d005",
"dept_name": "Development",
"from_date": "1994-07-03",
"to_date": "9999-01-01",
"dept_manager": [
{
"emp_no": 110511,
"first_name": "DeForest",
"last_name": "Hagimont",
"email": "110511@cloud-spartan.com",
"from_date": "1985-01-01",
"to_date": "1992-04-25"
},
{
"emp_no": 110567,
"first_name": "Leon",
"last_name": "DasSarma",
"email": "110567@cloud-spartan.com",
"from_date": "1992-04-25",
"to_date": "9999-01-01"
}
]
}
],
当我访问数据['部门'] .dept_no返回'未定义'
var content_depart = content['departments'];
console.log(content_depart);
var department = content_depart.dept_name;
console.log(department);
console.log(content_depart)
[ { dept_no: 'd005',
dept_name: 'Development',
from_date: '1994-07-03',
to_date: '9999-01-01',
dept_manager: [ [Object], [Object] ] } ]
console.log(部门)
undefined
答案 0 :(得分:0)
在访问数据之前添加索引
var department = content_depart[0].dept_name;
console.log(department);
答案 1 :(得分:0)
data['departments']
是一个数组。
let data =
{ // v--------- array !
"departments": [
{ // ...
您不能直接访问其成员,但必须先指定索引。例如:
let data = {"departments":[{"dept_no":"d005","dept_name":"Development","from_date":"1994-07-03","to_date":"9999-01-01","dept_manager":[{"emp_no":110511,"first_name":"DeForest","last_name":"Hagimont","email":"110511@cloud-spartan.com","from_date":"1985-01-01","to_date":"1992-04-25"},{"emp_no":110567,"first_name":"Leon","last_name":"DasSarma","email":"110567@cloud-spartan.com","from_date":"1992-04-25","to_date":"9999-01-01"}]}]};
console.log(data.departments[0].dept_no);
答案 2 :(得分:0)
$ seq 20 | awk -v h=5 -v r=3 'NR<=h || ((NR-h)%r)==1'
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是数组而不是对象
因此,请使用content_depart