Python 在 Flask 中获取数据 JSON
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Get Data JSON in Flask
提问by BrunoCX92
Even following many example here & there, i can't get my API work in POST Method. Here the code about it :
即使在这里和那里遵循许多示例，我也无法在 POST 方法中使用我的 API。这里是关于它的代码：
from flask import Flask, jsonify, request
@app.route('/api/v1/lists', methods=['POST'])
def add_entry():
print("p0")
content = request.get_json()
appname = content.get('title')
print(content)
print(appname)
When i query with curl (i'm running it on Windows):
当我使用 curl 查询时（我在 Windows 上运行它）：
curl.exe -i -H "Content-Type: application/json" -X POST -d '{"title":"titi"}' http://localhost:5000/api/v1/lists
curl.exe -i -H "Content-Type: application/json" -X POST -d '{"title":"titi"}' http://localhost:5000/api/v1/lists
curl.exe -i -H "Content-Type: application/json" -X POST -d "{"""title""":"""Read a book"""}" http://localhost:5000/api/v1/lists
curl.exe -i -H "Content-Type: application/json" -X POST -d "{"""title""":"""读一本书"""}" http://localhost:5000/ api/v1/列表
I have always a 400 error in return:
我总是有 400 错误作为回报：
HTTP/1.0 400 BAD REQUEST Content-Type: text/html Content-Length: 192 Server: Werkzeug/0.12.1 Python/3.6.0 Date: Tue, 04 Apr 2017 21:55:29 GMT 400 Bad Request Bad Request The browser (or proxy) sent a request that this server could not understand.
HTTP/1.0 400 BAD REQUEST 内容类型：text/html 内容长度：192 服务器：Werkzeug/0.12.1 Python/3.6.0 日期：2017 年 4 月 4 日星期二 21:55:29 GMT 400 Bad Request Bad Request 浏览器（或代理）发送了此服务器无法理解的请求。
I dont see where the error is.
我没有看到错误在哪里。
Thanks for your help.
谢谢你的帮助。
回答by Farhan Ahmed
Even with the "working" code, in case of your if-block failure (when value1and value2is None) will cause the same error, since you are not handling the elsepart of the if-block. The corrected should be:
即使使用“工作”代码，如果您的 if 块失败（whenvalue1和value2is None）也会导致相同的错误，因为您没有处理elseif 块的一部分。更正的应该是：
@app.route('/api/v1/list', methods=['POST'])
def add_entry():
print("p0")
request_json = request.get_json()
value1 = request_json.get('First_Name')
value2 = request_json.get('Last_Name')
response_content = None
if value1 is not None and value2 is not None:
print("p3")
cursor.execute("INSERT INTO person (first_name,last_name) VALUES (%s,%s)", (value1, value2))
response_content = conn.commit()
return jsonify(response_content)
Of course you may want something better than Noneas the response.
当然，您可能想要比None响应更好的东西。
回答by Allie Fitter
Your add_entryfunction isn't returning a response. You must return somethingeven if it's just return 'OK'.
您的add_entry函数没有返回响应。你必须返回一些东西，即使它只是return 'OK'.
EDIT:You're still not returning anything. In Flaskthe Python printstatement is not the equivalent of PHP's echo. All printdoes is print to the console. You still have to return a value. If what you need is to return contentand appnameJSON encoded, then add
编辑：你仍然没有返回任何东西。在FlaskPythonprint语句中并不等同于 PHP 的echo. 所有print做的是打印到控制台。您仍然必须返回一个值。如果你需要的是返回content和appnameJSON 编码，然后添加
return json.loads({'contents': contents, 'appname': appname})
return json.loads({'contents': contents, 'appname': appname})
to the end of your function.
到你的函数结束。
And to be clear in Flask views must return a value. That Python functions implicitly return None is inconsequential. The error that's occurring is that your function has no return statment.
并且要在 Flask 视图中明确必须返回一个值。Python 函数隐式返回 None 是无关紧要的。发生的错误是您的函数没有返回语句。
回答by BrunoCX92
Here a code working in my case :
这是在我的情况下工作的代码：
@app.route('/api/v1/list', methods=['POST'])
def add_entry():
print("p0")
request_json = request.get_json()
value1 = request_json.get('First_Name')
value2 = request_json.get('Last_Name')
if value1 is not None and value2 is not None:
print("p3")
cursor.execute("INSERT INTO person (first_name,last_name) VALUES (%s,%s)", (value1, value2))
data = conn.commit()
return jsonify(data)
