Javascript Creates An Element With Href

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An HTML link takes you to another page. An tag defines an anchor point. The "href" attribute indicates where a user will be directed when they click on the link. Between the and the closing tag is the anchor text that will be displayed to the user.

When you create a web page, you can reference another web page or website. For example, you can start a blog and want to link to an article that you feel is relevant.

Hyperlinks, or links, are used to link one web page to another and allow users to quickly navigate between multiple web pages. hyperlinks can be used to link to other pages on your site or pages on different sites.

In this guide, we’ll go over the basics of hyperlink in HTML. We are going to talk about how the link goals together, how to create bookmark anchors and links to email addresses and web elements.

HTML hyperlink

A hyperlink HTML points to another web resource. A hyperlink is defined between a tag and a closing tag. The text between these two tags takes the user to the linked web resource when the link is clicked.

Hyperlinks are created using label. Here is the syntax for link in an HTML file:

the text between our URL link tags we specify in our opening tag . We use the href attribute to choose the destination URL. This could be an absolute URL or an absolute link like https://python.engineering or a URL to part of our site

Here is an example of a link pointing to the web address of the Karma Career home page:

HTML link tag types

There are three types of links that can be viewed in a browser. These are:

Some websites replace these colors, which mean s they appear differently, but these are the main types of links that you can define in HTML HTML hyperlink

.: targets

You can change the way a link opens . For example, let’s say you want to open a link in a new tab in the user’s web browser. This will ensure that the user does not lose their place on the

web page they are currently viewing. This is where the targetHTML link attribute comes in. Using the target link, you can specify where the browser should open the resource you connected to

>

There are four types of targets you can use, which are:.

Here is an example of some of these links in action:

Hyperlink HTM L: bookmark anchors

Hyperlinks HTML can also refer to a specific part of a HTML document. This is useful if you have a long webpage and want to direct a user to a particular place in the text.

Before you start using anchors For bookmarks, you must first set an ID attribute on the element you want the user to go to. Here is an example of declaring an ID attribute on a title on a web page:

we assigned the HTML header tag has the id subtitle3, which we can use as a reference in an link.

Here is an example of link that references this header:

instead of specifying a url in our href tag, we specify the ID of the item we want to link to. Let’s bring hash before id (#) so that the browser knows that the linked document is on our web page.

HTML anchor tag

An HTML anchor tag refers to the label. This tag links a user to another web resource. The "href" attribute defines the resource a user will be directed to when they click on the link.

Examples of HTML anchor tags

regard Let ’s some examples of using the HTML tag.

Relative URLs

If you are linking to a resource on the same HTML document that you are developing, you can use a relative URL. Relative URLs are also called local links and do not use " https: //www.sitename .com < / a> " syntax. instead of this, the relative URLs pointing to a specific file path Web resource on the local server.

Let’s say we create a link on the website of career Karma which should point to our blog. Instead of using an absolute URL, we can use a local URL, because our site is the Karma Career site. Here is the code we would use to create this link:

When we clicked on the Karma career blog text, we were directed to the / blog resource on our site

email address

The anchor tag can also be used to link to an email address . When you click on a link to an email address , the user’s default email program will open. the . user will be asked if they want to send a message to the specified email address

to connect to an email address , you must start the link with the mailto protocol: Here is an example of a link to an email address in HTML:

When we click on our link, our email program opens and asks us to send an email to " [email protected] ¬ª The. anchor tag can link to an element on a web page. You can create this link by specifying the ID of the element to which the anchor should point. Suppose we want a link that takes our user to the header Topic test on a web page. We could create this link using the following code:

When clicking on our link our webpage scrolls to the item with id test_heading.

conclusion

hyperlinks can be used in HTML to associate a web page or other resource. You can use hyperlinks to connect to resources on your site or any other website. In this guide, we have explained how to use hyperlinks in an HTML document.

With all this information, you are ready to create hyperlinks HTML like a master!

To learn more about coding in HTML, read our How to Learn HTML Guide .

👻 Read also: what is the best laptop for engineering students?

Javascript Creates An Element With Href absolute: Questions

How to get an absolute file path in Python

3 answers

izb By izb

Given a path such as "mydir/myfile.txt", how do I find the file"s absolute path relative to the current working directory in Python? E.g. on Windows, I might end up with:

"C:/example/cwd/mydir/myfile.txt"
877

Answer #1

>>> import os
>>> os.path.abspath("mydir/myfile.txt")
"C:/example/cwd/mydir/myfile.txt"

Also works if it is already an absolute path:

>>> import os
>>> os.path.abspath("C:/example/cwd/mydir/myfile.txt")
"C:/example/cwd/mydir/myfile.txt"

Javascript Creates An Element With Href absolute: Questions

How to check if a path is absolute path or relative path in a cross-platform way with Python?

3 answers

UNIX absolute path starts with "/", whereas Windows starts with alphabet "C:" or "". Does python have a standard function to check if a path is absolute or relative?

175

Answer #1

os.path.isabs returns True if the path is absolute, False if not. The documentation says it works in windows (I can confirm it works in Linux personally).

os.path.isabs(my_path)

Javascript Creates An Element With Href absolute: Questions

How to join absolute and relative urls?

3 answers

I have two urls:

url1 = "http://127.0.0.1/test1/test2/test3/test5.xml"
url2 = "../../test4/test6.xml"

How can I get an absolute url for url2?

139

Answer #1

You should use urlparse.urljoin :

>>> import urlparse
>>> urlparse.urljoin(url1, url2)
"http://127.0.0.1/test1/test4/test6.xml"

With Python 3 (where urlparse is renamed to urllib.parse) you could use it as follow:

>>> import urllib.parse
>>> urllib.parse.urljoin(url1, url2)
"http://127.0.0.1/test1/test4/test6.xml"

exp

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)
None
>>> 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.)

5839

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)
    

Explanation

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()
    z.update(y)
    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:
        result.update(dictionary)
    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.

and

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

Usage:

>>> 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()
    z.update(y)
    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}))
1.0804965235292912
>>> min(repeat(lambda: merge_two_dicts(x, y)))
1.636518670246005
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
3.1779992282390594
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
2.740647904574871
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
4.266070580109954
$ 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

5839

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
print(z)
{"a": 1, "c": 11, "b": 10}

5839

Answer #3

An alternative:

z = x.copy()
z.update(y)

We hope this article has helped you to resolve the problem. Apart from Javascript Creates An Element With Href, check other absolute-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:



Chen Emmerson

London | 2022-11-30

sin is always a bit confusing 😭 Javascript Creates An Element With Href is not the only problem I encountered. I just hope that will not emerge anymore

Olivia Richtgofen

California | 2022-11-30

Maybe there are another answers? What Javascript Creates An Element With Href exactly means?. I am just not quite sure it is the best method

Chen Wu

Abu Dhabi | 2022-11-30

Thanks for explaining! I was stuck with Javascript Creates An Element With Href for some hours, finally got it done 🤗. I am just not quite sure it is the best method

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