burger | minimal hamburger menu with fullscreen navigation | Theme library

You need to wrap all your content except img in separate div and you need to add flex to your ".burger-item " , and you need to change some styles for your ".burgerimg "

But i suggest you change something , and experiment on your own

Working code :

I have tested the override works. please open in full page and see if this the result you are looking for?

you could add a jquery function that gets triggered when contact button is clicked.

$('#foo').trigger('click');

where foo is the id of the close button on the side nav.

hope it helps. Feel free to comment if you need any help or you need a vanilla javascript solution.

Just a little bit of CSS tweaking and consolidating. I removed the individual styles set for each of those four floating elements and added them to one .fixed_under_header class element.

This CSS should do the trick:

You can use a superclass and the __init_subclass__ hook to wire things up:

1. If I provide a space(anywhere in text field) it loads everything.

I don't observe that when trying your code. It will happen if your search string contains multiple consecutive spaces because String.split unfortunately does not collapse consecutive instances of the search pattern. For example, 'fooxxxbar'.split('x') will return ['foo', '', '', 'bar']. You later check if any of the tokens from recipeNamesList start with any of those elements, but ''.startsWith(string) is always true.

An easy way to fix that is to essentially collapse whitespace yourself by making the search pattern match one-or-more whitespace characters. That is, replace string.split(' ') with string.split(RegExp(r'\s+')).

Another easy way to fix it is to just continue if encountering an empty token to ignore it.

1. If I type Bengali Curry it returns both Bengali Lamb Curry & 'Chingri Malai Curry'.

Your current algorithm always adds the results list whenever it finds any match. It searches for 'bengali', finds one item, adds it to results, searches for 'curry', finds both of those results, and adds both of them.

Instead of iterating over search tokens in the outer loop and iterating over the recipeNamesList tokens in the inner loop, it would make more sense to iterate over recipeNamesList, and on each iteration, check if all of the search tokens match the tokens of the search string. If so, then add that recipe name to the list of results¹. That also would prevent the same recipe name from being added multiple times without needing to use a Set.

¹ Or if you want fuzzier matching, record a score for each recipe name (e.g. the number of search tokens that were matched), sort the results by the scores, and discard any below a certain threshold.

First, your TrieNode is using a List, which means you have to do a for-loop O(N) search for each char. It would be faster to use a Map.

(You could also use a 26D array, instead of a map, if you know that you'll only have English letters: [0] = 'A', [1] = 'B', ... [25] = 'Z')

Now I need to search using prefix so if the user searches for bur it'll show Burger. But if someone write Garlic Butter I need to Garlic Parmesan Butter. So, basically if the search query has multiple words I need to show the correct name.

bur is easy to do, as that's what a Trie data structure is for, but the garlic butter one is harder. You probably want to look into how to implement a fuzzy search or "did you mean...?" type algorithm:

• Fuzzy search algorithm (approximate string matching algorithm)

However, maybe this is more complex than you want.

Here are some alternatives I thought of:

Change your TrieNode to store a String variable stating the "standard" word. So the final TrieNode of garlic parmesan butter and garlic butter would both have an instance variable that stores garlic butter (the main standard/common word you want to use).

Your TrieNode would be like this:

Probably because your .card-about width is wider than your viewport.

Try using max-width instead so. I changed your height into auto so the content wont overflow when the height become bigger.

You can either use the index of the object in the array as its index or you can use Window functions [Django docs] and use the Rank or DenseRank (Reference [Django docs]) function to compute the ranks as per your need: