'Listen to shopify cart update quantity

I am writing a Shopify app and I would want to listen to the shopify cart update event. I know that when user clicks on remove or increase the item quantity. Shopify send a POST request to the backend to update the card. My app calculates the shipping value based on shopify cart item quantity. I add my app script to the shopify via script-tag.

I tried to listen to the quantity input but this event only fires one. I think shopify updates the input DOM after every cart update so it may remove my listener.

 $('body').on('change', '.input[name="updates[]"]', function() { console.log('HELLO')});

How can I listen to the cart update event? It seems to be a simple question but I really cannot find any answer regarding to Shopify!

shopifyshopping-cartshopify-appalgorithmrecursionmathbig-obacktracking


Solution 1:[1]

To update item quantity fields in the cart in sectioned theme

Step 1: You access Shopify admin, click Online Store and go to Themes.

Step 2: Find the themes that you want to edit, then press Actions then Edit Code.

Step 3: Look at Sections, you click on cart-template.liquid. In case that you are not using the theme that does not have the address, you can access Template directory and click cart.liquid.

Step 4: Find the phrase written as update[item.id] in input tag.

Step 5: Change the name into the value of name= “updates[]”.

Step 6: Reiterate these aforementioned steps whenever you see an name value of updates[] in input tag.

Step 7: Click Save and you update item quantity fields successfully..

Solution 2:[2]

Here's a simple way: You could calculate the time complexity by expanding each of the additive terms and observe the pattern:
T(n)
= T(n-1) + T(n-2) + T(n-3) + ... + T(1)
= (T(n-2) + T(n-3) + ... + T(1)) + T(n-2) + T(n-3) + ... + T(1)
= 2T(n-2) + 2T(n-3) + ... + 2T(1)
= 4T(n-3) + ... + 4T(1)
= 2^(k-1) T(n-k) + ... + 2^(k-1) T(1)
= 2^(n-2) T(1)
= 2^(n-2)

Solution 3:[3]

T(n) = (T(1) + T(2) + ... + T(n-2)) + T(n-1)
     = T(n-1) + T(n-1)
     = 2T(n-1)

Thus T(n) = c * 2^n, where c is T(1)/2.

Solution 4:[4]

You have the following:

T(1) = 1
T(2) = T(2-1) = T(1) = 1
T(3) = T(3-1) + T(3-2) = T(2) + T(1) = 2
T(4) = T(4-1) + T(4-2) + T(4-3) = T(3) + T(2) + T(1) = 2 + 1 + 1 = 4
T(5) = 4 + 2 + 1 + 1 = 8
T(6) = 8 + 4 + 2 + 1 + 1 = 16
...

From the above it follows by induction that T(n) = 2*T(n-1) with the base cases T(1) = T(2) = 1

We can further built on this saying that since T(n) = 2*T(n-1), T(n) = 2*2*T(n-2). It follows that T(n) = (2^i)*T(n-i) for i < n-2 and n > 2.

So in terms of time complexity we get O(2^n).

Note: Please correct me if I'm wrong. I'm not the very best with this stuff. My appologies in advance. I'm trying to learn myself.

Sources

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Source: Stack Overflow

Solution Source
Solution 1
Solution 2 Jingjin Wang
Solution 3 Paul Hankin
Solution 4

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