Python request loop, retry if status code not 200

Question

I do a loop of requests and unfortunately sometimes there happens a server timeout. Thats why I want to check the status code first and if it is not 200 I want to go back and repeat the last request until the status code is 200. An example of the code looks like this:

for i in range(0, len(table)):
        var = table.iloc[i]
        url = 'http://example.request.com/var/'
        response = requests.get(url)
        if response.status_code == 200:
            data = response.json()
        else:
            "go back to response"

I am appending the response data of every i, so I would like to go back as long as the response code is 200 and then go on with the next i in the loop. Is there any easy solution?

Also: url = "http://example.request.com/{0:}/".format(var). — CristiFati
– CristiFati, Commented Sep 14, 2019 at 14:37
You might want to check that you actually get a timeout, either locally or remotely, before hammering the server... — thebjorn
– thebjorn, Commented Sep 14, 2019 at 14:50

Calslock · Accepted Answer · 2019-09-14 14:31:54Z

8

I believe you want to do something like this:

for i in range(0, len(table)):
        var = table.iloc[i]
        url = 'http://example.request.com/var/'
        response = requests.get(url)
        while response.status_code != 200:
            response = requests.get(url)     
        data = response.json()

answered Sep 14, 2019 at 14:31

Calslock

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1 Comment

leifericf Over a year ago

A more robust solution to the same problem can be found here: stackoverflow.com/a/35636367/195964

Sandro Meireles · Accepted Answer · 2019-09-14 14:44:46Z

2

I made a small example, used an infinite loop and used break to demonstrate when the status code is = 200

while True:
    url = 'https://stackoverflow.com/'

    response = requests.get(url)

    if response.status_code == 200:
        # found code
        print('found exemple')
        break

answered Sep 14, 2019 at 14:44

Sandro Meireles

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