json.dumps() converts a dictionary to str object, not a json(dict) object! So you have to load your str into a dict to use it by using json.loads() method See json.dumps() as a save method and json.loads() as a retrieve method. This is the code sample which might help you understand it more:
Assuming you really mean easiest and are not necessarily looking for a way to do this programmatically, you can do this: Add, if not already there, a row of “column Musicians” to the spreadsheet. That is, if you have data in columns such as: Rory Gallagher Guitar Gerry McAvoy Bass Rod de’Ath Drums Lou Martin … Read more
The JSON you posted looks fine, however in your code, it is most likely not a JSON string anymore, but already a JavaScript object. This means, no more parsing is necessary. You can test this yourself, e.g. in Chrome’s console: JSON.parse() converts the input into a string. The toString() method of JavaScript objects by default … Read more
You need to set your content-type to application/json. But -d (or –data) sends the Content-Type application/x-www-form-urlencoded, which is not accepted on Spring’s side. Looking at the curl man page, I think you can use -H (or –header): Full example: (-H is short for –header, -d for –data) Note that -request POST is optional if you … Read more
NOTE: When I first spent time reading about REST, idempotence was a confusing concept to try to get right. I still didn’t get it quite right in my original answer, as further comments (and Jason Hoetger’s answer) have shown. For a while, I have resisted updating this answer extensively, to avoid effectively plagiarizing Jason, but … Read more
Your code produced an empty response body, you’d want to check for that or catch the exception raised. It is possible the server responded with a 204 No Content response, or a non-200-range status code was returned (404 Not Found, etc.). Check for this. Note: There is no need to use simplejson library, the same … Read more
I have the following code: The input in responsecontent is JSON, but it is not properly parsed into an object. How should I properly deserialize it?
You need to set your content-type to application/json. But -d (or –data) sends the Content-Type application/x-www-form-urlencoded, which is not accepted on Spring’s side. Looking at the curl man page, I think you can use -H (or –header): Full example: (-H is short for –header, -d for –data) Note that -request POST is optional if you use -d, as the -d flag implies a POST request. On Windows, things are slightly different. See the comment thread.
The wording of the error message corresponds to what you get from Google Chrome when you run JSON.parse(‘<…’). I know you said the server is setting Content-Type:application/json, but I am led to believe the response body is actually HTML. Feed.js:94 undefined “parsererror” “SyntaxError: Unexpected token < in JSON at position 0” with the line console.error(this.props.url, status, err.toString()) underlined. The err was actually thrown … Read more
UPDATE 2: YouTube has unfortunately dropped support for this method. Hopefully there will be a new solution in the future without using their API (since it has quote limits). Try this link: https://www.youtube.com/live_stats?v={videoid} Find a Live event on YouTube and replace the video id in the link. This should retrieve the number of concurrent views for that … Read more