通过表格在Flask上发布数据会给出400个错误的请求 [英] Posting Data on Flask via form is giving 400 Bad Request

问题描述

我试图通过我的前端发布数据，烧瓶应用程序正在抛出400个不好的请求。但是，如果我正在使用卷曲调用相同，似乎工作正常。

以下是我的表单代码

 <脚本> 
 function sub（）{
 console.log（'sub function'）; 
 $（＃fquery）。submit（）; 
} 
< / script> 
< form id =form1action =/ finalmethod =post> 
< input id ='query'type =text> 
< button type =submitonClick ='sub（）'>提交& raquo;< / button> 
< / form> 
  

在服务器端：

  @ app.route（'/ final'，methods = ['POST'，'GET']）
 def message（）：$ b $ if request.method =='POST '：
 app.logger.debug（输入消息函数+ request.form ['query']）
q = request.form ['query'] 
 return render_template（'final。这个服务器端对我来说似乎很好，但是我觉得这个服务器端对我来说似乎很不错。因为我得到了curl请求的响应
  curl http：// localhost：8000 / final -dquery = select starct st blah blah-X POST -v 
 * Trying 127.0.0.1 ... connected 
> POST / gc HTTP / 1.1 
> User-Agent：curl / 7.22.0（i686-pc-linux-gnu）libcurl / 7.22.0 OpenSSL / 1.0.1 zlib / 1.2.3.4 libidn / 1.23 librtmp / 2.3 
>主机：localhost：8000 
>接受：* / * 
>内容长度：41 
> Content-Type：application / x-www-form-urlencoded 
> 
 *上传完全发送：41个字节的41个
 * HTTP 1.0，假设在
之后关闭< b< HTTP / 1.0 200 OK 
< Content-Type：text / html; charset = utf-8 
<内容长度：1961 
 <服务器：Werkzeug / 0.9.4 Python / 2.7.3 
< Date：Thu，24 Oct 2013 23:33:12 GMT 
  
 
 
解决方案
 <呃，我想我明白了：你只为<$ c>设置 id ，而不是名称 $ c> input 元素。然而，名称用于发送到服务器的表单数据。这会导致 KeyError 在 request.form ['query'] 中，导致400错误。 b $ b
I am trying to post data via my front end and the flask app is throwing 400 bad request. However If I am doing the same using Curl call it seems to work fine. I dont know what I am missing in the form.

The following is my form code
<script>
function sub() {
    console.log('sub function');
    $("#fquery").submit();
}
</script>
<form id="form1" action="/final" method="post">
 <input id='query' type="text">
  <button type="submit" onClick='sub()'>Submit &raquo;</button>
</form>
At server side:
@app.route('/final',methods=['POST','GET'])
def message():
    if request.method == 'POST':
        app.logger.debug(" entered message function"+ request.form['query'])
        q = request.form['query']
    return render_template('final.html',query=q,result="Core_Table Output")
The server side seems fine to me. Since I am getting response for curl request
curl http://localhost:8000/final -d "query=select starct st blah blah" -X POST -v
*   Trying 127.0.0.1... connected
> POST /gc HTTP/1.1
> User-Agent: curl/7.22.0 (i686-pc-linux-gnu) libcurl/7.22.0 OpenSSL/1.0.1 zlib/1.2.3.4 libidn/1.23 librtmp/2.3
> Host: localhost:8000
> Accept: */*
> Content-Length: 41
> Content-Type: application/x-www-form-urlencoded
> 
* upload completely sent off: 41out of 41 bytes
* HTTP 1.0, assume close after body
< HTTP/1.0 200 OK
< Content-Type: text/html; charset=utf-8
< Content-Length: 1961
< Server: Werkzeug/0.9.4 Python/2.7.3
< Date: Thu, 24 Oct 2013 23:33:12 GMT

解决方案
Ah, I think I see it:  You only set the id but not the name for the input element.  Yet the name is used in the form data that is sent to the server.  This causes a KeyError at request.form['query'] which causes the 400 error.

                        
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