Flask跨域请求解决方案

Flask跨域请求解决方案

这个坑太坑了，记录一下。

我也不知道到底哪行代码起了作用，总之就是起了作用，问题解决了，那就这样吧！

跨域请求跟前端无关，可以通过后端代码更改：

1. 首先增加CROS:

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from flask import Flask, jsonify, request, make_response
from flask_cors import CORS

def after_request(resp):
    resp.headers['Access-Control-Allow-Origin'] = '*'
    return resp

app = Flask(__name__)
app.after_request(after_request)

CORS(app, resources={
    r"/api/*": {
        "origins": "*",  # Vite 默认开发服务器地址
        "methods": ["GET", "POST", "PUT", "DELETE", "OPTIONS"],
        "allow_headers": ["Content-Type"]
    }
})
api = Api(app)

2. 然后在接口调用的地方：

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# 如下：methods方法里加上 'OPTIONS',
@app.route('/api/analyze-analysis', methods=['POST', 'OPTIONS'])
def ai_analysis():
    # 再然后加上下面这几行代码
    if request.method == 'OPTIONS':
        response = make_response()
        response.headers['Access-Control-Allow-Origin'] = '*'
        response.headers['Access-Control-Allow-Methods'] = 'POST, OPTIONS'
        response.headers['Access-Control-Allow-Headers'] = 'Content-Type'
        return response 
    # 下面就是业务逻辑代码了。
    try:
        data = request.json
        code = data.get('code')
        prompt = data.get('prompt', '')
        model = data.get('model', 'gpt-4')

        if not code:
            return jsonify({'error': '代码不能为空'}), 400

        # Use the AIGrading instance
        result = ai_grader.analyze_code(code, prompt, model)
        
        # Convert the result keys to match frontend expectations
        return jsonify({
            'codeQualityScore': result['code_quality'],
            'functionalityScore': result['functionality'],
            'creativityScore': result['creativity'],
            'documentationScore': result['documentation'],
            'feedback': result['feedback']
        })

    except Exception as e:
        print(f"AI analysis failed: {str(e)}")
        return jsonify({'error': str(e)}), 500