Question

如何正确地将propertyName传递给isValid（）函数，以便能够检查它是否不为空？当我直接检查“ searchInside.attendeeList”时，它起作用了！

function isValid(searchInside, propertyName) {
  if(searchInside.propertyName)
    console.log("this doesnt work");
  if(searchInside.attendeeList)
    console.log("this works");
}
var requestBody = {
    "meetingType": "Company",
    "emailSendingReason": "CREATED",
    "attendeeList": [
        {
            "employeeId": "12345",
            "employeeDisplayName": "abc, xyz",
            "callInFlag": false,
            "infoPackRequiredFlag": true,
            "inviteForInfoOnly": true
        },
        {
            "employeeId": "374684678",
            "employeeDisplayName": "xyz, poi",
            "callInFlag": true,
            "infoPackRequiredFlag": true,
            "inviteForInfoOnly": false
        }
    ],
    "thirdPartyAttendee": {}
};
isValid(requestBody, 'attendeeList');

Answer 1

尝试searchInside.hasOwnProperty(propertyName)。对象 hasOwnProperty（）检查对象密钥并返回布尔结果。

function isValid(searchInside, propertyName) {
  if(searchInside.hasOwnProperty( propertyName)){
    console.log("this works");
  } else console.log("this doesn't work");
}

var requestBody = {
    "meetingType": "Company",
    "emailSendingReason": "CREATED",
    "attendeeList": [
        {
            "employeeId": "12345",
            "employeeDisplayName": "abc, xyz",
            "callInFlag": false,
            "infoPackRequiredFlag": true,
            "inviteForInfoOnly": true
        },
        {
            "employeeId": "374684678",
            "employeeDisplayName": "xyz, poi",
            "callInFlag": true,
            "infoPackRequiredFlag": true,
            "inviteForInfoOnly": false
        }
    ],
    "thirdPartyAttendee": {}
};
isValid(requestBody, 'attendeeList');

Answer 2

您已经正确将它作为字符串传递了。在您的 isValid（）函数中，您可以像这样引用它：

function isValid(searchInside, propertyName) {
  if(searchInside[propertyName]){
    console.log("this works");
  }
}

Answer 3

function isValid(searchInside, propertyName) {
  if(typeof searchInside[propertyName] !== 'undefined')
    console.log("this doesnt work");
  if(searchInside.attendeeList)
    console.log("this works");
}

动态检查json属性

3 个答案: