我试图在不知道对象的情况下通过value
找到key
,包含嵌套对象,因此该函数将获取一个键和一个对象并返回值或未定义。
这是我的功能:
/* Iterate over object and include sub objects */
function iterate (obj, key) {
for (var property in obj) {
if (obj.hasOwnProperty(property)) {
//in case it is an object
if (typeof obj[property] == "object") {
if (obj.hasOwnProperty(key)) {
return obj[key]; //return the value
}
}
else {
iterate(obj[property]);
}
}
}
return undefined;
}
我在循环中调用return
,因此效率更高(希望如此......)。
1.是否有人准备好此功能?这个不起作用。
2.有人知道要改变什么才能让它发挥作用?
任何帮助,包括angular.js
功能都会很棒。
感谢。
答案 0 :(得分:4)
你稍微扭转了一点,你忘了将第二个参数传递给递归函数调用。此外,在这种情况下,无需返回undefined
,因为这是默认值。
function iterate (obj, key) {
var result;
for (var property in obj) {
if (obj.hasOwnProperty(property)) {
// in case it is an object
if (typeof obj[property] === "object") {
result = iterate(obj[property], key);
if (typeof result !== "undefined") {
return result;
}
}
else if (property === key) {
return obj[key]; // returns the value
}
}
}
}
编辑:实际上,在我们要查找值为对象的键的情况下,我们应该检查属性是否等于键,然后检查值是否是可迭代对象。感谢@Catinodeh!
function iterate (obj, key) {
var result;
for (var property in obj) {
if (obj.hasOwnProperty(property)) {
if (property === key) {
return obj[key]; // returns the value
}
else if (typeof obj[property] === "object") {
// in case it is an object
result = iterate(obj[property], key);
if (typeof result !== "undefined") {
return result;
}
}
}
}
}
答案 1 :(得分:1)
首先使用解决此问题的函数取消它: Fastest way to flatten / un-flatten nested JSON objects
然后再次使用public class Cars {
private String carID;
private String plateNum;
private static String position;
private static Attendant assignedTo;
private long currTime;
static String[] tempArray2 = new String[Garage.getCarsCapacity()];
public Cars(String carID, String plateNum, String position, Attendant assignedTo, long currTime) {
this.carID = carID;
this.plateNum = plateNum;
Cars.position = position;
Cars.assignedTo = assignedTo;
this.currTime = currTime;
}
private static void createCarsID() {
for (int x = 0; x < Garage.getCarsCapacity(); x++) {
tempArray2[x] = ("CR" + (x + 1));
}
}
public static String getID() {
createCarsID();
String tempID = null;
String tempPos = null;
for (int x = 0; x < tempArray2.length; x++) {
if (tempArray2[x] != null) {
tempID = tempArray2[x];
tempPos = tempArray2[x];
getPos(tempPos);
tempArray2[x] = null;
break;
}
}
return tempID;
}
public static void getPos(String IdToPos) {
String strPos = IdToPos.substring(2);
int pos = Integer.parseInt(strPos);
position = "GR" + pos;
}
public String getPlateNum() {
return plateNum;
}
public String getCarID() {
return carID;
}
public static String getPosition() {
return position;
}
public long getCurrTime() {
return currTime;
}
public static Attendant getAssignedTo() {
return assignedTo;
}
public static String askCarID() {
boolean valid = false;
System.out.println("Please enter your car's plate number.");
Scanner scanCarID = new Scanner(System.in);
String scannedCarID = scanCarID.nextLine();
while (!valid) {
if (scannedCarID.matches("^[A-Za-z][A-Za-z] [0-9][0-9][0-9]$")) {
valid = true;
System.out.println(scannedCarID);
} else {
System.out.println("Please enter a valid plate number. Ex: AF 378");
askCarID();
}
}
return scannedCarID.toUpperCase();
}
public static String convert(long miliSeconds) {
int hrs = (int) TimeUnit.MILLISECONDS.toHours(miliSeconds) % 24;
int min = (int) TimeUnit.MILLISECONDS.toMinutes(miliSeconds) % 60;
int sec = (int) TimeUnit.MILLISECONDS.toSeconds(miliSeconds) % 60;
return String.format("%02d:%02d:%02d", hrs, min, sec);
}
}
答案 2 :(得分:0)
这是一个对我有用的代码段。对我来说,我只搜索没有数组的嵌套对象。
function searcObject(obj, key){
var found = false
if(typeof obj == 'object' && obj !== null){
Object.keys(obj).forEach(function(prop) {
if(prop == key){
value = obj[key];
return true;
} else{
searcObject(obj[prop], key);
}
});
} else {
return false;
};
};