问题:如何在Flask中获取POST JSON?

我正在尝试使用Flask构建一个简单的API,现在我想在其中读取一些POSTed JSON。我使用Postman Chrome扩展程序执行POST,而我发布的JSON就是{"text":"lalala"}。我尝试使用以下方法读取JSON:

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content
    return uuid

在浏览器上,它可以正确返回我放入GET中的UUID,但是在控制台上,它只是打印出来None(我希望它可以在其中打印出来{"text":"lalala"}。有人知道我如何从Flask方法中获取发布的JSON吗?

I’m trying to build a simple API using Flask, in which I now want to read some POSTed JSON. I do the POST with the Postman Chrome extension, and the JSON I POST is simply {"text":"lalala"}. I try to read the JSON using the following method:

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content
    return uuid

On the browser it correctly returns the UUID I put in the GET, but on the console, it just prints out None (where I expect it to print out the {"text":"lalala"}. Does anybody know how I can get the posted JSON from within the Flask method?


回答 0

首先,该.json属性是一个委托给request.get_json()method的属性,该属性记录了为什么None在此处看到。

您需要将请求内容类型设置为,application/json以使.json属性和.get_json()方法(不带参数)起作用,None否则两者都会产生这种情况。请参阅Flask Request文档

如果mimetype表示JSON(包含application / json,请参见),则它将包含已解析的JSON数据,否则为None

您可以request.get_json()通过向其传递force=True关键字参数来告知跳过内容类型要求。

请注意,如果此时引发异常(可能导致400 Bad Request响应),则您的JSON 数据无效。它在某种程度上畸形;您可能需要使用JSON验证程序进行检查。

First of all, the .json attribute is a property that delegates to the request.get_json() method, which documents why you see None here.

You need to set the request content type to application/json for the .json property and .get_json() method (with no arguments) to work as either will produce None otherwise. See the Flask Request documentation:

This will contain the parsed JSON data if the mimetype indicates JSON (application/json, see ), otherwise it will be None.

You can tell request.get_json() to skip the content type requirement by passing it the force=True keyword argument.

Note that if an exception is raised at this point (possibly resulting in a 400 Bad Request response), your JSON data is invalid. It is in some way malformed; you may want to check it with a JSON validator.


回答 1

作为参考,以下是有关如何从Python客户端发送json的完整代码:

import requests
res = requests.post('http://localhost:5000/api/add_message/1234', json={"mytext":"lalala"})
if res.ok:
    print res.json()

“ json =“输入将自动设置内容类型,如下所述:使用Python请求发布JSON

以上客户端将使用此服务器端代码:

from flask import Flask, request, jsonify
app = Flask(__name__)

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content['mytext']
    return jsonify({"uuid":uuid})

if __name__ == '__main__':
    app.run(host= '0.0.0.0',debug=True)

For reference, here’s complete code for how to send json from a Python client:

import requests
res = requests.post('http://localhost:5000/api/add_message/1234', json={"mytext":"lalala"})
if res.ok:
    print res.json()

The “json=” input will automatically set the content-type, as discussed here: Post JSON using Python Requests

And the above client will work with this server-side code:

from flask import Flask, request, jsonify
app = Flask(__name__)

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content['mytext']
    return jsonify({"uuid":uuid})

if __name__ == '__main__':
    app.run(host= '0.0.0.0',debug=True)

回答 2

这就是我要做的方法,应该是

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.get_json(silent=True)
    # print(content) # Do your processing
    return uuid

随着silent=True集,该get_json功能将尝试检索JSON的身体的时候默默的失败。默认情况下,此设置为False。如果您始终希望使用json正文(而不是可选),请将其保留为silent=False

设置force=True将忽略request.headers.get('Content-Type') == 'application/json'烧瓶对您的 检查。默认情况下,它也设置为False

请参见烧瓶文档

我强烈建议您离开force=False并让客户端发送Content-Type标头以使其更加明确。

希望这可以帮助!

This is the way I would do it and it should be

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.get_json(silent=True)
    # print(content) # Do your processing
    return uuid

With silent=True set, the get_json function will fail silently when trying to retrieve the json body. By default this is set to False. If you are always expecting a json body (not optionally), leave it as silent=False.

Setting force=True will ignore the request.headers.get('Content-Type') == 'application/json' check that flask does for you. By default this is also set to False.

See flask documentation.

I would strongly recommend leaving force=False and make the client send the Content-Type header to make it more explicit.

Hope this helps!


回答 3

假设您已发布具有application/json内容类型的有效JSON ,request.json将具有已解析的JSON数据。

from flask import Flask, request, jsonify

app = Flask(__name__)


@app.route('/echo', methods=['POST'])
def hello():
   return jsonify(request.json)

Assuming you’ve posted valid JSON with the application/json content type, request.json will have the parsed JSON data.

from flask import Flask, request, jsonify

app = Flask(__name__)


@app.route('/echo', methods=['POST'])
def hello():
   return jsonify(request.json)

回答 4

对于所有问题都是来自ajax调用的人,这里有一个完整的示例:

Ajax调用:这里的关键是使用a dict然后JSON.stringify

    var dict = {username : "username" , password:"password"};

    $.ajax({
        type: "POST", 
        url: "http://127.0.0.1:5000/", //localhost Flask
        data : JSON.stringify(dict),
        contentType: "application/json",
    });

在服务器端:

from flask import Flask
from flask import request
import json

app = Flask(__name__)

@app.route("/",  methods = ['POST'])
def hello():
    print(request.get_json())
    return json.dumps({'success':True}), 200, {'ContentType':'application/json'} 

if __name__ == "__main__":
    app.run()

For all those whose issue was from the ajax call, here is a full example :

Ajax call : the key here is to use a dict and then JSON.stringify

    var dict = {username : "username" , password:"password"};

    $.ajax({
        type: "POST", 
        url: "http://127.0.0.1:5000/", //localhost Flask
        data : JSON.stringify(dict),
        contentType: "application/json",
    });

And on server side :

from flask import Flask
from flask import request
import json

app = Flask(__name__)

@app.route("/",  methods = ['POST'])
def hello():
    print(request.get_json())
    return json.dumps({'success':True}), 200, {'ContentType':'application/json'} 

if __name__ == "__main__":
    app.run()

回答 5

给出另一种方法。

from flask import Flask, jsonify, request
app = Flask(__name__)

@app.route('/service', methods=['POST'])
def service():
    data = json.loads(request.data)
    text = data.get("text",None)
    if text is None:
        return jsonify({"message":"text not found"})
    else:
        return jsonify(data)

if __name__ == '__main__':
    app.run(host= '0.0.0.0',debug=True)

To give another approach.

from flask import Flask, jsonify, request
app = Flask(__name__)

@app.route('/service', methods=['POST'])
def service():
    data = json.loads(request.data)
    text = data.get("text",None)
    if text is None:
        return jsonify({"message":"text not found"})
    else:
        return jsonify(data)

if __name__ == '__main__':
    app.run(host= '0.0.0.0',debug=True)

回答 6

假设您发布了有效的JSON,

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content['uuid']
    # Return data as JSON
    return jsonify(content)

Assuming that you have posted valid JSON,

@app.route('/api/add_message/<uuid>', methods=['GET', 'POST'])
def add_message(uuid):
    content = request.json
    print content['uuid']
    # Return data as JSON
    return jsonify(content)

回答 7

尝试使用力参数…

request.get_json(force = True)

Try to use force parameter…

request.get_json(force = True)


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