JsonFormat.printer().print()如何工作? [英] How does JsonFormat.printer().print() work?

问题描述

我有一个非常简单的原型:

I have a very simple proto:

message ChargeCardResponse
{
    bool success = 1;
}

当我尝试将其解析为String时，JsonFormat.printer().print()会得到非常奇怪的输出，具体取决于成功值.

When I'm trying to parse it to String I get a very strange output from the JsonFormat.printer().print() which depends on the success value.

ChargeCardResponse.Builder builder = ChargeCardResponse.newBuilder();
System.out.println(JsonFormat.printer().print(builder.setSuccess(true).build()));

输出: { 成功":正确 }

Output: { "success": true }

ChargeCardResponse.Builder builder2 = ChargeCardResponse.newBuilder();
System.out.println(JsonFormat.printer().print(builder2.setSuccess(false).build()));

输出:{}

因此，如果我使用成功属性为打印机赋予错误的值，则无法制作JSON格式. 我究竟做错了什么?有人可以确认吗?

So if I give false value to the printer with the success property it can't make the JSON format. What am I doing wrong? Is there anybody who can confirm it?

推荐答案

问题是将成功设置为"false"等同于保留默认的"false"值.

The thing is that setting success to "false" is equivalent to leaving the default "false" value intact.

JsonPrinter忽略默认的原型值，除非该选项已关闭:

JsonPrinter omits default proto values, unless that option it turned off:

https://developers.google.com/protocol-buffers/docs/reference/java/com/google/protobuf/util/JsonFormat.Printer#includeDefaultValueFields--

这应该有效，我认为:

ChargeCardResponse.Builder builder2 = ChargeCardResponse.newBuilder();
System.out.println(JsonFormat.printer().includingDefaultValueFields().print(builder2.setSuccess(false).build()));

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