使用SurrealDB构建CRUD API时创建记录失败的问题求助

问题场景 #

用SurrealDB构建CRUD API时，添加新记录触发反序列化异常。

请求JSON：

{"zipcode":"00-001","city":"Warsaw","country":"Poland","street":"Some street"}

错误响应：

Failed to convert `{ city: 'Warsaw', country: 'Poland', id: address:936jcs9jtx23qejgn5uc, street: 'Some street', zipcode: '00-001' }` to `T`: invalid type: map, expected a string

相关代码如下：

Address 模型

#[derive(Serialize, Deserialize, Debug)]
pub struct Address {
    #[serde(skip_serializing_if = "Option::is_none")]
    pub id: Option<String>,
    pub zipcode: String,
    pub city: String,
    pub country: String,
    pub street: String,
}

impl Address {
    pub fn new(zipcode: String, city:String, country:String, street: String) -> Address{
        Address{ id: None, zipcode, city, country, street, }
    }
}

Create 函数

pub type SurrealClient = Surreal<Client>;

pub async fn create(
    client: Arc<SurrealClient>,
    zipcode: String, city: String, country: String, street: String,
) -> Result<Address, SurrealError> {
     let address: Address = Address::new(zipcode, city, country, street);

    match client.create("address").content(address).await{
        Ok(address) => Ok(address),
        Err(e) => Err(e),
    }
}

问题原因 #

错误根源在id字段的类型定义：SurrealDB返回的id是复合结构（格式表名:UUID），默认会被序列化为map类型，但代码中将id声明为Option<String>，serde无法完成类型转换，因此抛出异常。

解决方案 #

使用SurrealDB官方提供的Thing（或新版本的RecordId）类型替代String来定义id字段：

1. 确保依赖中包含surrealdb及相关serde特性
2. 修改Address模型：

use surrealdb::sql::Thing;

#[derive(Serialize, Deserialize, Debug)]
pub struct Address {
    #[serde(skip_serializing_if = "Option::is_none")]
    pub id: Option<Thing>,
    pub zipcode: String,
    pub city: String,
    pub country: String,
    pub street: String,
}

如果使用的是较新版本的SurrealDB，将Thing替换为RecordId即可，两者用法一致。

需要将id转为字符串时，直接调用id.to_string()，会得到address:936jcs9jtx23qejgn5uc格式的字符串，符合预期。

修改后无需改动create函数，客户端会自动完成正确的序列化与反序列化。

内容的提问来源于stack exchange，提问作者mmich