norris | 🔥 | Reactive Programming library

I think you could use a combination of TableView and Pagination like it is described in this posting: JavaFX TableView Paginator

Here is an example:

App.java:

it's pretty straightforward :

1. read all element and put them in a list:

As I already stated in my comment you'd need to count the cars you've already processed and print new page headers when you've hit a multiple of 30 cars.

First I'd suggest moving your header print statements to a separate method, e.g. printPageHeader(int pageNumber). Then change your loop like this:

To create a custom legend you need to get the plot objects and use them when defining your legend, as explicited on matplotlib's documentation:

1. Explicitly defining the elements in the legend

For full control of which artists have a legend entry, it is possible to pass an iterable of legend artists followed by an iterable of legend labels respectively:

ax.legend([line1, line2, line3], ['label1', 'label2', 'label3'])

Also, you may add the labels when creating the vertical bands.

From the docs for std::print:

Note that stdout is frequently line-buffered by default so it may be necessary to use io::stdout().flush() to ensure the output is emitted immediately.

So looks like you need to call io::stdout().flush().

Sounds like you are asking for help on how to get started. You are asking for help on learning to learn. Here is how I would approach your problem:

Apparently you are supposed to use a priority queue.

1. Write a tiny program that makes a priority queue and stores strings into it, then prints them out.
2. Define a class and store instances of that class into the priority queue instead of strings.
3. Modify the sort criteria on the priority queue and notice that the printed sequence changes according to the sort criteria.
4. Write a function that creates one class instance with random values.
5. Write a function that creates all 100 class instances.
6. Declare victory.

You can see the next page link in the response: "nextPageLink":"?q=&campus=col&academic-career=ugrd&p=2",

So you should use p instead of page.

You can simplify the Javascript a little and use a forEach loop on the value property within the response.

The error => {console.log(error + "")} piece is not quite right - that should be within a catch segment - ie .catch( err=>console.log(err) ) or similar

Let's learn how to read documentation:

1. Search the internet for Tweepy documentation -> https://docs.tweepy.org/en/latest/
2. Look up the search method you want to use -> https://docs.tweepy.org/en/latest/api.html#search-methods
3. It says 'Return type: SearchResults object. Find what that object is -> https://docs.tweepy.org/en/v3.10.0/search.html?q=SearchResults
4. Turns out the documentation for this is missing -> consider using something other than Tweepy.
5. You can now go check the Tweepy source or try to read the error... let's try the error
6. 'SearchResults' object has no attribute 'text' means that the var you call status is actually of the type SearchResults. When you've printed it looked like a list, so let's check if it behaves like one: