Revisions to Unity Json utility does not serialize DateTime

Added [Serializable] attribute to make the struct type usable in a complex serializable object.

Source Link

edit approved Nov 11, 2021 at 13:41

3
2

The problem with JsonUtility, is that it doesn't save properties, just public fields and fields with the [SerializeField] attribute.

You can solve this with another class and some conversion magic, like so:

Json Serializable Class

[Serializable]
struct JsonDateTime {
    public long value;
    public static implicit operator DateTime(JsonDateTime jdt) {
        Debug.Log("Converted to time");
        return DateTime.FromFileTimeUtc(jdt.value);
    }
    public static implicit operator JsonDateTime(DateTime dt) {
        Debug.Log("Converted to JDT");
        JsonDateTime jdt = new JsonDateTime();
        jdt.value = dt.ToFileTimeUtc();
        return jdt;
    }
}

Usage example

var time = DateTime.Now;
print(time);
var json = JsonUtility.ToJson((JsonDateTime) time);
print(json);
DateTime timeFromJson = JsonUtility.FromJson<JsonDateTime>(json);
print(timeFromJson);

Output

The problem with JsonUtility, is that it doesn't save properties, just public fields and fields with the [SerializeField] attribute.

You can solve this with another class and some conversion magic, like so:

Json Serializable Class

struct JsonDateTime {
    public long value;
    public static implicit operator DateTime(JsonDateTime jdt) {
        Debug.Log("Converted to time");
        return DateTime.FromFileTimeUtc(jdt.value);
    }
    public static implicit operator JsonDateTime(DateTime dt) {
        Debug.Log("Converted to JDT");
        JsonDateTime jdt = new JsonDateTime();
        jdt.value = dt.ToFileTimeUtc();
        return jdt;
    }
}

Usage example

var time = DateTime.Now;
print(time);
var json = JsonUtility.ToJson((JsonDateTime) time);
print(json);
DateTime timeFromJson = JsonUtility.FromJson<JsonDateTime>(json);
print(timeFromJson);

Output

The problem with JsonUtility, is that it doesn't save properties, just public fields and fields with the [SerializeField] attribute.

You can solve this with another class and some conversion magic, like so:

Json Serializable Class

[Serializable]
struct JsonDateTime {
    public long value;
    public static implicit operator DateTime(JsonDateTime jdt) {
        Debug.Log("Converted to time");
        return DateTime.FromFileTimeUtc(jdt.value);
    }
    public static implicit operator JsonDateTime(DateTime dt) {
        Debug.Log("Converted to JDT");
        JsonDateTime jdt = new JsonDateTime();
        jdt.value = dt.ToFileTimeUtc();
        return jdt;
    }
}

Usage example

var time = DateTime.Now;
print(time);
var json = JsonUtility.ToJson((JsonDateTime) time);
print(json);
DateTime timeFromJson = JsonUtility.FromJson<JsonDateTime>(json);
print(timeFromJson);

Output

Unity also serializes fields with the [SerializeField] attribute, which is almost always better than making them public

Source Link

edit approved May 28, 2019 at 19:52

Kevin

7k
1
12
32

The problem with JsonUtility, is that it doesn't save properties, just public fields and fields with the [SerializeField] attribute.

You can solve this with another class and some conversion magic, like so:

Json Serializable Class

struct JsonDateTime {
    public long value;
    public static implicit operator DateTime(JsonDateTime jdt) {
        Debug.Log("Converted to time");
        return DateTime.FromFileTimeUtc(jdt.value);
    }
    public static implicit operator JsonDateTime(DateTime dt) {
        Debug.Log("Converted to JDT");
        JsonDateTime jdt = new JsonDateTime();
        jdt.value = dt.ToFileTimeUtc();
        return jdt;
    }
}

Usage example

var time = DateTime.Now;
print(time);
var json = JsonUtility.ToJson((JsonDateTime) time);
print(json);
DateTime timeFromJson = JsonUtility.FromJson<JsonDateTime>(json);
print(timeFromJson);

Output

The problem with JsonUtility, is that it doesn't save properties, just public fields.

You can solve this with another class and some conversion magic, like so:

Json Serializable Class

struct JsonDateTime {
    public long value;
    public static implicit operator DateTime(JsonDateTime jdt) {
        Debug.Log("Converted to time");
        return DateTime.FromFileTimeUtc(jdt.value);
    }
    public static implicit operator JsonDateTime(DateTime dt) {
        Debug.Log("Converted to JDT");
        JsonDateTime jdt = new JsonDateTime();
        jdt.value = dt.ToFileTimeUtc();
        return jdt;
    }
}

Usage example

var time = DateTime.Now;
print(time);
var json = JsonUtility.ToJson((JsonDateTime) time);
print(json);
DateTime timeFromJson = JsonUtility.FromJson<JsonDateTime>(json);
print(timeFromJson);

Output

The problem with JsonUtility, is that it doesn't save properties, just public fields and fields with the [SerializeField] attribute.

You can solve this with another class and some conversion magic, like so:

Json Serializable Class

struct JsonDateTime {
    public long value;
    public static implicit operator DateTime(JsonDateTime jdt) {
        Debug.Log("Converted to time");
        return DateTime.FromFileTimeUtc(jdt.value);
    }
    public static implicit operator JsonDateTime(DateTime dt) {
        Debug.Log("Converted to JDT");
        JsonDateTime jdt = new JsonDateTime();
        jdt.value = dt.ToFileTimeUtc();
        return jdt;
    }
}

Usage example

var time = DateTime.Now;
print(time);
var json = JsonUtility.ToJson((JsonDateTime) time);
print(json);
DateTime timeFromJson = JsonUtility.FromJson<JsonDateTime>(json);
print(timeFromJson);

Output

Source Link

answered Feb 17, 2017 at 23:37

lilotop

624
5
15

The problem with JsonUtility, is that it doesn't save properties, just public fields.

You can solve this with another class and some conversion magic, like so:

Json Serializable Class

struct JsonDateTime {
    public long value;
    public static implicit operator DateTime(JsonDateTime jdt) {
        Debug.Log("Converted to time");
        return DateTime.FromFileTimeUtc(jdt.value);
    }
    public static implicit operator JsonDateTime(DateTime dt) {
        Debug.Log("Converted to JDT");
        JsonDateTime jdt = new JsonDateTime();
        jdt.value = dt.ToFileTimeUtc();
        return jdt;
    }
}

Usage example

var time = DateTime.Now;
print(time);
var json = JsonUtility.ToJson((JsonDateTime) time);
print(json);
DateTime timeFromJson = JsonUtility.FromJson<JsonDateTime>(json);
print(timeFromJson);

Output

Stack Exchange Network

Return to Answer