Javascript Change The Background Image

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The CSS property of the background image transforms the background of a website into an image. A background image is set using the background-image: url (url_of_image) property. You can specify multiple overlapping background images using the background-image property.

Background images are a common feature on modern sites that create an aesthetic user experience. The web designers customize this feature based on unique predetermined themes via CSS background-image property.

This tutorial and examples will familiarize you with this CSS property. By the end of this article, you will be an expert in customizing background images.

CSS background image

The background-image property sets an image as the background of an element on a web page. It is often used to set the background of an entire page or section of a page.

When designing a web page, you need to consider whether the final product will have a background image.

For example, you might want a group photo of the team members in the background if you are designing an " About Us " page. If you are designing a website for a cafe, consider adding a header on a page that shows an image of a cup of coffee.

The background-image property adds images as a background to a HTML element . You can add a background image to an element using the following syntax:

background-image: url (imageUrl);

imageUrl refers to the location of the image you want to display.

example CSS background image

We’re working on a website design project for a cafe called ’The Coffee Grind.’ The cafe asked us to create a banner at the top of the page, displaying an image of a cup of coffee. This banner should have "Welcome to The Coffee Grind" in the center.

We could create the banner using the following code:

style.css .image {background- image: URL ( photos / 683039 / pexels-photo-683039.jpeg? auto = compress & cs = tinysrgb & h = 750 & w = 1260); height: 250px; } .header {padding-top: 50px ; White color; font size: 20px; text alignment: center; }

Our code returns:

Capture d ’screen 2020 10 08 to 19.33.45

Let’s analyze our code In our HTML file, we have defines a style.css .image {background-image: url ( ), url ( -photo-683039.jpeg? auto = compress & cs = tinysrgb & h = 750 & w = 1260); height: 250px; } .header {padding-top: 50px; White color; font size: 20px; text alignment: center; }

Our code returns:

Screen Shot 2020 10 08 to 19.34.02

As you can see, our coffee cup icon has been added in front of our background image. Our icon only appears in the foreground because we ’ve specified it as the first background image in our list of background images .

size of the background image

When working with the background-image property, you can customize the size of the background image in its container. The background-size property is used for this purpose.

The background-size property accepts four possible values, which are:

Change background size

Suppose we want the cafe banner image before r to be contained in our element. This means that the entire image should be visible, even if it does not fill the container. We could do it using this code:

style.css .image {background-image: url (; background size: contains ; height: 250px; }

Our code returns:

Screen Shot 2020 10 08 to 19.34.24

As you can see, unlike our first example, the entire background image is visible. This is because we specified the -size background in our code and set its value to contain. Alternatively, we could have used the cover value if we wanted our image to cover the dimensions of the entire container.

You could also specify a pixel size for the image. that we want our style image to be 200px high by 200px wide. We could do this using the following code:

style.css .image {background-image: url (; background size: contains ; height: 200px; width: 200px; }

Our code returns:

Screen 2020 10 08 at 19.35.22 = 850 =" 374 "src: // 19.35.22.png? Lossy = 1 & strip = 1 & webp = 1 "alt =" Screen Shot 2020 10 08 at 19.35.22 "class =" wp-image-23919 "srcset =" https: //744025.smushcdn .com / 1245953 / wp -content / uploads / 2020/10 / Screen-Shot-2020-10-08-at-19.35.22-20x9.png? Lossy = 1 & strip = 1 & webp = 1 20w, https: //744025.smushcdn. Com / 1245953 / wp-content / uploads / 2020/10 / Screen -Shot-2020-10-08-at-19.35.22.png? Size = 234x103 & lossy = 1 & strip = 1 & webp = 1,234w, https: // / uploads / 2020/10 / Screen-Shot-2020-10-08-at-19.35.22-385x169.png? lossy = 1 & strip = 1 & webp = 1 385w, / wp- content / uploads / 2020/10 / Screen-Shot-2020-10-08-at-19.35.22.png? size = 468x206 & lossy = 1 & band = 1 & webp = 1468w, https: // -content / uploads / 2020/10 / Screen-Shot-2020-10-08-at-19.35.22.png? size = 702x309 & lossy = 1 & strip = 1 & webp = 1 702w, wp-content / uploads / 2020/10 / Screen-Shot-2020-10-08- at -19.35.22-768x338.png? lossy = 1 & strip = 1 & webp = 1,768w, / wp-content / up loads / 2020/10 / Screen-Shot-2020-10-08-at-19.35. 22-770x339.png? lossy = 1 & strip = 1 & webp = 1770w, / wp-content / uploads / 2020/10 / Screen-Shot-2020-10-08-at-19.35.22 .png? lossy = 1 & strip = 1 & webp = 1850w "size =" (max width: 850px) 100vw, 850px ">

In this example, our image is 200px wide and 200px long.

Background image of the position

The background-origin property allows you to place the background image according to content, ai borders or padding of a web element.

The background-origin property accepts tro

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Javascript Change The Background Image exp: Questions


How do I merge two dictionaries in a single expression (taking union of dictionaries)?

5 answers

Carl Meyer By Carl Meyer

I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged (i.e. taking the union). The update() method would be what I need, if it returned its result instead of modifying a dictionary in-place.

>>> x = {"a": 1, "b": 2}
>>> y = {"b": 10, "c": 11}
>>> z = x.update(y)
>>> print(z)
>>> x
{"a": 1, "b": 10, "c": 11}

How can I get that final merged dictionary in z, not x?

(To be extra-clear, the last-one-wins conflict-handling of dict.update() is what I"m looking for as well.)


Answer #1

How can I merge two Python dictionaries in a single expression?

For dictionaries x and y, z becomes a shallowly-merged dictionary with values from y replacing those from x.

  • In Python 3.9.0 or greater (released 17 October 2020): PEP-584, discussed here, was implemented and provides the simplest method:

    z = x | y          # NOTE: 3.9+ ONLY
  • In Python 3.5 or greater:

    z = {**x, **y}
  • In Python 2, (or 3.4 or lower) write a function:

    def merge_two_dicts(x, y):
        z = x.copy()   # start with keys and values of x
        z.update(y)    # modifies z with keys and values of y
        return z

    and now:

    z = merge_two_dicts(x, y)


Say you have two dictionaries and you want to merge them into a new dictionary without altering the original dictionaries:

x = {"a": 1, "b": 2}
y = {"b": 3, "c": 4}

The desired result is to get a new dictionary (z) with the values merged, and the second dictionary"s values overwriting those from the first.

>>> z
{"a": 1, "b": 3, "c": 4}

A new syntax for this, proposed in PEP 448 and available as of Python 3.5, is

z = {**x, **y}

And it is indeed a single expression.

Note that we can merge in with literal notation as well:

z = {**x, "foo": 1, "bar": 2, **y}

and now:

>>> z
{"a": 1, "b": 3, "foo": 1, "bar": 2, "c": 4}

It is now showing as implemented in the release schedule for 3.5, PEP 478, and it has now made its way into the What"s New in Python 3.5 document.

However, since many organizations are still on Python 2, you may wish to do this in a backward-compatible way. The classically Pythonic way, available in Python 2 and Python 3.0-3.4, is to do this as a two-step process:

z = x.copy()
z.update(y) # which returns None since it mutates z

In both approaches, y will come second and its values will replace x"s values, thus b will point to 3 in our final result.

Not yet on Python 3.5, but want a single expression

If you are not yet on Python 3.5 or need to write backward-compatible code, and you want this in a single expression, the most performant while the correct approach is to put it in a function:

def merge_two_dicts(x, y):
    """Given two dictionaries, merge them into a new dict as a shallow copy."""
    z = x.copy()
    return z

and then you have a single expression:

z = merge_two_dicts(x, y)

You can also make a function to merge an arbitrary number of dictionaries, from zero to a very large number:

def merge_dicts(*dict_args):
    Given any number of dictionaries, shallow copy and merge into a new dict,
    precedence goes to key-value pairs in latter dictionaries.
    result = {}
    for dictionary in dict_args:
    return result

This function will work in Python 2 and 3 for all dictionaries. e.g. given dictionaries a to g:

z = merge_dicts(a, b, c, d, e, f, g) 

and key-value pairs in g will take precedence over dictionaries a to f, and so on.

Critiques of Other Answers

Don"t use what you see in the formerly accepted answer:

z = dict(x.items() + y.items())

In Python 2, you create two lists in memory for each dict, create a third list in memory with length equal to the length of the first two put together, and then discard all three lists to create the dict. In Python 3, this will fail because you"re adding two dict_items objects together, not two lists -

>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: "dict_items" and "dict_items"

and you would have to explicitly create them as lists, e.g. z = dict(list(x.items()) + list(y.items())). This is a waste of resources and computation power.

Similarly, taking the union of items() in Python 3 (viewitems() in Python 2.7) will also fail when values are unhashable objects (like lists, for example). Even if your values are hashable, since sets are semantically unordered, the behavior is undefined in regards to precedence. So don"t do this:

>>> c = dict(a.items() | b.items())

This example demonstrates what happens when values are unhashable:

>>> x = {"a": []}
>>> y = {"b": []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: "list"

Here"s an example where y should have precedence, but instead the value from x is retained due to the arbitrary order of sets:

>>> x = {"a": 2}
>>> y = {"a": 1}
>>> dict(x.items() | y.items())
{"a": 2}

Another hack you should not use:

z = dict(x, **y)

This uses the dict constructor and is very fast and memory-efficient (even slightly more so than our two-step process) but unless you know precisely what is happening here (that is, the second dict is being passed as keyword arguments to the dict constructor), it"s difficult to read, it"s not the intended usage, and so it is not Pythonic.

Here"s an example of the usage being remediated in django.

Dictionaries are intended to take hashable keys (e.g. frozensets or tuples), but this method fails in Python 3 when keys are not strings.

>>> c = dict(a, **b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings

From the mailing list, Guido van Rossum, the creator of the language, wrote:

I am fine with declaring dict({}, **{1:3}) illegal, since after all it is abuse of the ** mechanism.


Apparently dict(x, **y) is going around as "cool hack" for "call x.update(y) and return x". Personally, I find it more despicable than cool.

It is my understanding (as well as the understanding of the creator of the language) that the intended usage for dict(**y) is for creating dictionaries for readability purposes, e.g.:

dict(a=1, b=10, c=11)

instead of

{"a": 1, "b": 10, "c": 11}

Response to comments

Despite what Guido says, dict(x, **y) is in line with the dict specification, which btw. works for both Python 2 and 3. The fact that this only works for string keys is a direct consequence of how keyword parameters work and not a short-coming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact, ** was designed precisely to pass dictionaries as keywords.

Again, it doesn"t work for 3 when keys are not strings. The implicit calling contract is that namespaces take ordinary dictionaries, while users must only pass keyword arguments that are strings. All other callables enforced it. dict broke this consistency in Python 2:

>>> foo(**{("a", "b"): None})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() keywords must be strings
>>> dict(**{("a", "b"): None})
{("a", "b"): None}

This inconsistency was bad given other implementations of Python (PyPy, Jython, IronPython). Thus it was fixed in Python 3, as this usage could be a breaking change.

I submit to you that it is malicious incompetence to intentionally write code that only works in one version of a language or that only works given certain arbitrary constraints.

More comments:

dict(x.items() + y.items()) is still the most readable solution for Python 2. Readability counts.

My response: merge_two_dicts(x, y) actually seems much clearer to me, if we"re actually concerned about readability. And it is not forward compatible, as Python 2 is increasingly deprecated.

{**x, **y} does not seem to handle nested dictionaries. the contents of nested keys are simply overwritten, not merged [...] I ended up being burnt by these answers that do not merge recursively and I was surprised no one mentioned it. In my interpretation of the word "merging" these answers describe "updating one dict with another", and not merging.

Yes. I must refer you back to the question, which is asking for a shallow merge of two dictionaries, with the first"s values being overwritten by the second"s - in a single expression.

Assuming two dictionaries of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dictionaries from either source, and the surest way to avoid that is to make a copy when assigning values. As keys must be hashable and are usually therefore immutable, it is pointless to copy them:

from copy import deepcopy

def dict_of_dicts_merge(x, y):
    z = {}
    overlapping_keys = x.keys() & y.keys()
    for key in overlapping_keys:
        z[key] = dict_of_dicts_merge(x[key], y[key])
    for key in x.keys() - overlapping_keys:
        z[key] = deepcopy(x[key])
    for key in y.keys() - overlapping_keys:
        z[key] = deepcopy(y[key])
    return z


>>> x = {"a":{1:{}}, "b": {2:{}}}
>>> y = {"b":{10:{}}, "c": {11:{}}}
>>> dict_of_dicts_merge(x, y)
{"b": {2: {}, 10: {}}, "a": {1: {}}, "c": {11: {}}}

Coming up with contingencies for other value types is far beyond the scope of this question, so I will point you at my answer to the canonical question on a "Dictionaries of dictionaries merge".

Less Performant But Correct Ad-hocs

These approaches are less performant, but they will provide correct behavior. They will be much less performant than copy and update or the new unpacking because they iterate through each key-value pair at a higher level of abstraction, but they do respect the order of precedence (latter dictionaries have precedence)

You can also chain the dictionaries manually inside a dict comprehension:

{k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7

or in Python 2.6 (and perhaps as early as 2.4 when generator expressions were introduced):

dict((k, v) for d in dicts for k, v in d.items()) # iteritems in Python 2

itertools.chain will chain the iterators over the key-value pairs in the correct order:

from itertools import chain
z = dict(chain(x.items(), y.items())) # iteritems in Python 2

Performance Analysis

I"m only going to do the performance analysis of the usages known to behave correctly. (Self-contained so you can copy and paste yourself.)

from timeit import repeat
from itertools import chain

x = dict.fromkeys("abcdefg")
y = dict.fromkeys("efghijk")

def merge_two_dicts(x, y):
    z = x.copy()
    return z

min(repeat(lambda: {**x, **y}))
min(repeat(lambda: merge_two_dicts(x, y)))
min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
min(repeat(lambda: dict(chain(x.items(), y.items()))))
min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))

In Python 3.8.1, NixOS:

>>> min(repeat(lambda: {**x, **y}))
>>> min(repeat(lambda: merge_two_dicts(x, y)))
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
$ uname -a
Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux

Resources on Dictionaries


Answer #2

In your case, what you can do is:

z = dict(list(x.items()) + list(y.items()))

This will, as you want it, put the final dict in z, and make the value for key b be properly overridden by the second (y) dict"s value:

>>> x = {"a":1, "b": 2}
>>> y = {"b":10, "c": 11}
>>> z = dict(list(x.items()) + list(y.items()))
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python 2, you can even remove the list() calls. To create z:

>>> z = dict(x.items() + y.items())
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python version 3.9.0a4 or greater, then you can directly use:

x = {"a":1, "b": 2}
y = {"b":10, "c": 11}
z = x | y
{"a": 1, "c": 11, "b": 10}


Answer #3

An alternative:

z = x.copy()

JSON datetime between Python and JavaScript

4 answers

kevin By kevin

I want to send a datetime.datetime object in serialized form from Python using JSON and de-serialize in JavaScript using JSON. What is the best way to do this?


Answer #1

You can add the "default" parameter to json.dumps to handle this:

date_handler = lambda obj: (
    if isinstance(obj, (datetime.datetime,
    else None
json.dumps(, default=date_handler)

Which is ISO 8601 format.

A more comprehensive default handler function:

def handler(obj):
    if hasattr(obj, "isoformat"):
        return obj.isoformat()
    elif isinstance(obj, ...):
        return ...
        raise TypeError, "Object of type %s with value of %s is not JSON serializable" % (type(obj), repr(obj))

Update: Added output of type as well as value.
Update: Also handle date

What blocks Ruby, Python to get Javascript V8 speed?

4 answers

Are there any Ruby / Python features that are blocking implementation of optimizations (e.g. inline caching) V8 engine has?

Python is co-developed by Google guys so it shouldn"t be blocked by software patents.

Or this is rather matter of resources put into the V8 project by Google.


Answer #1

What blocks Ruby, Python to get Javascript V8 speed?


Well, okay: money. (And time, people, resources, but if you have money, you can buy those.)

V8 has a team of brilliant, highly-specialized, highly-experienced (and thus highly-paid) engineers working on it, that have decades of experience (I"m talking individually – collectively it"s more like centuries) in creating high-performance execution engines for dynamic OO languages. They are basically the same people who also created the Sun HotSpot JVM (among many others).

Lars Bak, the lead developer, has been literally working on VMs for 25 years (and all of those VMs have lead up to V8), which is basically his entire (professional) life. Some of the people writing Ruby VMs aren"t even 25 years old.

Are there any Ruby / Python features that are blocking implementation of optimizations (e.g. inline caching) V8 engine has?

Given that at least IronRuby, JRuby, MagLev, MacRuby and Rubinius have either monomorphic (IronRuby) or polymorphic inline caching, the answer is obviously no.

Modern Ruby implementations already do a great deal of optimizations. For example, for certain operations, Rubinius"s Hash class is faster than YARV"s. Now, this doesn"t sound terribly exciting until you realize that Rubinius"s Hash class is implemented in 100% pure Ruby, while YARV"s is implemented in 100% hand-optimized C.

So, at least in some cases, Rubinius can generate better code than GCC!

Or this is rather matter of resources put into the V8 project by Google.

Yes. Not just Google. The lineage of V8"s source code is 25 years old now. The people who are working on V8 also created the Self VM (to this day one of the fastest dynamic OO language execution engines ever created), the Animorphic Smalltalk VM (to this day one of the fastest Smalltalk execution engines ever created), the HotSpot JVM (the fastest JVM ever created, probably the fastest VM period) and OOVM (one of the most efficient Smalltalk VMs ever created).

In fact, Lars Bak, the lead developer of V8, worked on every single one of those, plus a few others.

Django Template Variables and Javascript

4 answers

When I render a page using the Django template renderer, I can pass in a dictionary variable containing various values to manipulate them in the page using {{ myVar }}.

Is there a way to access the same variable in Javascript (perhaps using the DOM, I don"t know how Django makes the variables accessible)? I want to be able to lookup details using an AJAX lookup based on the values contained in the variables passed in.


Answer #1

The {{variable}} is substituted directly into the HTML. Do a view source; it isn"t a "variable" or anything like it. It"s just rendered text.

Having said that, you can put this kind of substitution into your JavaScript.

<script type="text/javascript"> 
   var a = "{{someDjangoVariable}}";

This gives you "dynamic" javascript.

We hope this article has helped you to resolve the problem. Apart from Javascript Change The Background Image, check other exp-related topics.

Want to excel in Python? See our review of the best Python online courses 2022. If you are interested in Data Science, check also how to learn programming in R.

By the way, this material is also available in other languages:

Marie Danburry

Massachussetts | 2022-11-30

Simply put and clear. Thank you for sharing. Javascript Change The Background Image and other issues with JavaScript was always my weak point 😁. I am just not quite sure it is the best method

Ken Danburry

Abu Dhabi | 2022-11-30

Thanks for explaining! I was stuck with Javascript Change The Background Image for some hours, finally got it done 🤗. I just hope that will not emerge anymore

Angelo Innsbruck

Abu Dhabi | 2022-11-30

Thanks for explaining! I was stuck with Javascript Change The Background Image for some hours, finally got it done 🤗. I am just not quite sure it is the best method


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