mixrouter
  1. 豆包
mixrouter
  • mixrouter接口说明
  • 已支持的模型列表
  • 大模型官方文档
  • gemini
    • openai格式
      • openai格式
      • gemini-开启thinking
    • google原生格式
      • google格式
  • anthropic
    • openai格式
      POST
    • anthropic格式
      POST
  • openai
    • chat/completions
      POST
    • 流式/chat/completions
      POST
    • responses
      POST
  • deepseek
    • chat/completions
      POST
  • qwen
    • chat/completions(开启thinking)
      POST
    • chat/completions
      POST
  • 豆包
    • chat/completions
      POST
    • 提交视频生成任务
      POST
    • 获取视频生成任务结果
      GET
  1. 豆包

chat/completions

POST
https://api.mixrouter.com/v1/chat/completions

Request

Authorization
Provide your bearer token in the
Authorization
header when making requests to protected resources.
Example:
Authorization: Bearer ********************
Body Params application/jsonRequired

Example
{
    "model": "doubao-seed-1-6-thinking-250715",
    "messages": [
        {
            "role": "user",
            "content": [
                {
                    "type": "text",
                    "text": "1+2+3+。。。。+100=?"
                }
            ]
        }
    ]
}

Request Code Samples

Shell
JavaScript
Java
Swift
Go
PHP
Python
HTTP
C
C#
Objective-C
Ruby
OCaml
Dart
R
Request Request Example
Shell
JavaScript
Java
Swift
curl --location --request POST 'https://api.mixrouter.com/v1/chat/completions' \
--header 'Authorization: Bearer <token>' \
--header 'Content-Type: application/json' \
--data-raw '{
    "model": "doubao-seed-1-6-thinking-250715",
    "messages": [
        {
            "role": "user",
            "content": [
                {
                    "type": "text",
                    "text": "1+2+3+。。。。+100=?"
                }
            ]
        }
    ]
}'

Responses

🟢200OK
application/json
Body

Example
{
    "choices": [
        {
            "finish_reason": "stop",
            "index": 0,
            "logprobs": null,
            "message": {
                "content": "要计算从1到100的连续整数和,可以用**等差数列求和公式**快速解决:\n\n\n### 公式推导  \n等差数列的求和公式为:  \n\\[ S_n = \\frac{n(a_1 + a_n)}{2} \\]  \n其中:  \n- \\( n \\) 是项数(这里从1到100共100项,\\( n=100 \\)),  \n- \\( a_1 \\) 是首项(\\( a_1=1 \\)),  \n- \\( a_n \\) 是末项(\\( a_n=100 \\))。  \n\n\n### 代入计算  \n\\[ S_{100} = \\frac{100 \\times (1 + 100)}{2} = \\frac{100 \\times 101}{2} = 50 \\times 101 = 5050 \\]  \n\n\n**结论**:1+2+3+…+100 = 5050。",
                "reasoning_content": "\n用户现在问的是1加到100的和是多少,首先我得回忆一下高斯求和的公式对吧。等差数列求和公式是(首项+末项)×项数÷2,这里首项是1,末项是100,项数是100,所以代入的话就是(1+100)×100÷2,算一下的话,101乘100是10100,再除以2就是5050。对,没错,高斯小时候就是用这个方法算出来的,不用一个个加。得确认一下公式有没有记错,等差数列求和确实是这样的,因为把数列倒过来加,比如1+100=101,2+99=101,一直到50+51=101,总共有50对,所以50乘101就是5050。没错,结果是对的。",
                "role": "assistant"
            }
        }
    ],
    "created": 1765547023,
    "id": "021765547013146aafb70253f86f133b2b849e439c9606481f7d7",
    "model": "doubao-seed-1-6-thinking-250715",
    "service_tier": "default",
    "object": "chat.completion",
    "usage": {
        "completion_tokens": 439,
        "prompt_tokens": 25,
        "total_tokens": 464,
        "prompt_tokens_details": {
            "cached_tokens": 0
        },
        "completion_tokens_details": {
            "reasoning_tokens": 209
        }
    }
}
Modified at 2025-12-12 13:46:09
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