网址无效的URI使用？ 在python请求模块中

Question

I was trying to extract data using:

urlall = url+'/'+i+'.json'+'\\?'+'page='+str(page)

r = requests.get(urlall)

I got an error 400 Client Error: Invalid URI for url: the '\?' turned out to be '%5C?' 

if I use:

urlall = url+'/'+i+'.json'+'?'+'page='+str(page)

Then I got another error: can only concatenate str (not "_io.TextIOWrapper") to str

How can I set '?' as a string and get only the '?' in the url instead of %5C?

Answer 1

You can use string formatting to create your resultant url ( string.format or f-strings )

In [4]: url = 'http.example.com'                                                                                                          
In [5]: i = 1                                                                                                                             
In [8]: page = 1                                                                                                                          
#f-strings for python>=3.6
In [10]: f'{url}/{i}.json?page={page}'                                                                                                    
Out[10]: 'http.example.com/1.json?page=1'
#String formatting
In [11]: '{}/{}.json?page={}'.format(url, i, page)                                                                                        
Out[11]: 'http.example.com/1.json?page=1'

Or you can use urllib.parse.urlunsplit library to create your url, for example

In [1]: from urllib.parse import urlunsplit                                                                                               

In [2]: urlunsplit(['http','example.com','1.json','page=1',''])                                                                           
Out[2]: 'http://example.com/1.json?page=1'