英文:
How to use `omitempty` with protobuf Timestamp in Golang
问题
我在我的结构体上有一个可选字段叫做ExpireTime。它的类型是time.Time，并且有一个json:"expire_time,omitempty"的标签，表示当它为空时不发送该字段。这部分工作得非常好。
当我想通过GRPC使用同样的字段时，将其转换为protobuf时间戳格式时遇到了问题。
type Timestamp struct {
// 表示自Unix纪元以来的UTC时间的秒数
// 1970-01-01T00:00:00Z。必须在0001-01-01T00:00:00Z到
// 9999-12-31T23:59:59Z之间（包括边界）。
Seconds int64 `protobuf:"varint,1,opt,name=seconds,proto3" json:"seconds,omitempty"`
// 非负的纳秒分辨率的秒的小数部分。具有负秒值的小数部分仍然必须具有非负的纳秒值，
// 以便按时间向前计数。必须在0到999,999,999之间（包括边界）。
Nanos int32 `protobuf:"varint,2,opt,name=nanos,proto3" json:"nanos,omitempty"`
// 包含已过滤或未导出的字段
}
ExpireTime *timestamppb.Timestamp `protobuf:"bytes,1,opt,name=expire_time,json=expireTime,proto3" json:"expire_time,omitempty"`
问题在于一个空的time.Time{}对象将被转换为一个负的seconds值，对应于0001-01-01T00:00:00Z。在这种情况下，omitEmpty标志将不会被应用，因为该值并没有被置零。我应该怎么做才能在字段实际为空时省略它呢？谢谢！
英文:
I have an optional field on my struct called ExpireTime. It has a time.Time type and a json:"expire_time,omitempty" tag to not send it, when it is empty. This part works perfectly fine.
When I want to use the same field via GRPC, I run into an issue when converting it to the protobuf timestamp format.
type Timestamp struct {
// Represents seconds of UTC time since Unix epoch
// 1970-01-01T00:00:00Z. Must be from 0001-01-01T00:00:00Z to
// 9999-12-31T23:59:59Z inclusive.
Seconds int64 `protobuf:"varint,1,opt,name=seconds,proto3" json:"seconds,omitempty"`
// Non-negative fractions of a second at nanosecond resolution. Negative
// second values with fractions must still have non-negative nanos values
// that count forward in time. Must be from 0 to 999,999,999
// inclusive.
Nanos int32 `protobuf:"varint,2,opt,name=nanos,proto3" json:"nanos,omitempty"`
// contains filtered or unexported fields
}
ExpireTime *timestamppb.Timestamp `protobuf:"bytes,1,opt,name=expire_time,json=expireTime,proto3" json:"expire_time,omitempty"`
The issue is that an empty time.Time{} object will be converted to a negative seconds value corresponding to 0001-01-01T00:00:00Z. Having the omitEmpty flag will not be applied in this case as the value is not zeroed out. What could I do to omit this field, when it is actually empty? Thanks!
答案1
得分: 2
根据你的要求，以下是翻译好的内容：
正如你所说，time.Time{} 转换为 0001-01-01T00:00:00Z；这是按预期工作的。请注意，在相反的方向转换时也需要小心（零值的 TimeStamp 将变为 1970-01-01T00:00:00Z）。
然而，通常 Timestamp 将作为消息的一部分，例如：
message MyMessage{
google.protobuf.Timestamp comm_time = 1;
}
通过 protoc 运行这段代码将得到类似以下的结果：
type MyMessage struct {
state protoimpl.MessageState
sizeCache protoimpl.SizeCache
unknownFields protoimpl.UnknownFields
CommTime *timestamppb.Timestamp `protobuf:"bytes, 1,opt,name=comm_time,json=commTime,proto3" json:"comm_time,omitempty"`
}
这意味着你可以通过 CommTime=nil 来实现你想要的结果，例如：
sourceTime := time.Time{} // 你想要编码的时间
var commTime *timestamp.Timestamp
if !sourceTime.IsZero() {
commTime = timestamppb.New(sourceTime)
}
msg := MyMessage{
CommTime: commTime,
}
英文:
As you say time.Time{} converts to 0001-01-01T00:00:00Z; this is working as intended. Note that you also need to be careful converting in the opposite direction (a zero TimeStamp will become 1970-01-01T00:00:00Z).
However generally the Timestamp will be part of a message, for example:
message MyMessage{
google.protobuf.Timestamp comm_time = 1;
}
Running this through protoc will result in something like:
type MyMessage struct {
state protoimpl.MessageState
sizeCache protoimpl.SizeCache
unknownFields protoimpl.UnknownFields
CommTime *timestamppb.Timestamp `protobuf:"bytes, 1,opt,name=comm_time,json=commTime,proto3" json:"comm_time,omitempty"`
}
This means you should be able to achieve the result you are looking for with CommTime=nil; e.g.
sourceTime := time.Time{} // Whatever time you want to encode
var commTime *timestamp.Timestamp
if !sourceTime.IsZero() {
commTime = timestamppb.New(sourceTime)
}
msg := MyMessage{
CommTime: commTime,
}
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