How To Get The Value Of The P Tag In Javascript

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The HTML tag allows you to apply styles to part of a web page or paragraph. It is often used to stylize text in a paragraph, for example by changing the color of a word. tag has no styles by default.

When programming in HTML, you may decide that you want to style something specific on your web page. For example, you might want to emphasize a certain word in a sentence or a specific link in a link list.

is where the HTML tag comes from. is a generic inline container that allows you to formulate content in a web document. is commonly used for styling - especially for text styling

This tutorial will discuss, with examples, how to use HTML ;. to apply a style to the elements of a web page. By the time you finish reading this tutorial, you will be an expert in using the HTML tag.

HTML tag

HTML part of a web page tag. It is often used in a

element to apply styles to a specific part of a paragraph. For example, you can use to change the color of a word in a paragraph.

By itself, tag has no default or renderer. This means that if you use the tag without attributes, it has no effect on the way your web page is displayed.

is often used with CSS to apply a certain style to one or more specific elements on a web page. lets you easily apply a certain style to multiple elements at once on a web page

>

HTML syntax

The HTML Span tag has an opening () and a closing () We can use the following syntax to use the tag:

The HTML tag has no specific tag-attributes. For example, the button tag has "height" and "width" as specific attributes of the tag. does not have any of these types of attributes because it relies on CSS for styling. tag supports all global HTML attributes .

You can think of the tag similar to tags are used to define sections and blocks of text because

Our code returns:

In our example, we have underlined the terms " delicious baked goods ‚" and " thousands‚". Mr. Pringle told us that these terms are important, so we directed the program to display them in red.

We have defined a text string enclosed in

HTML tag. This text traces the history of Pringle and Bakery Sons. Quindi, abbiamo racchiuso i termini che vogliamo appaiano in rosso nei tag .

Nel nostro file CSS, abbiamo specificato che il colore di ogni elemento assegnato alla class redText dovrebbe essere Red. This allows us to change the color of our text to red when we use the tag

;. is useful in this case because we just wanted to emphasize certain words in our text. We could have style text in red on the

tag itself. But that would have given rise to a whole red paragraph.

HTML in a list

Assume Pringle and Sons Bakery is asking us to add a list to their website with links to their social media pages. The bakery noted that the links to their Twitter and Facebook pages are particularly important because the team is more active on these platforms . . So they want us to highlight the links in blue

We could use the following code to do this:

Our code returns:

 HfpJCIX4ycBUcStlx4EzsAWQzvWNJ 2CLshhieSA BtyOWUe420vh 0KC45uZlVS SsXrpfIP0XBg7turyR5LmnqyarvOug2yL6bfrpS7 KERmAAONXuh5Za9I6x4uL8r FjvWz3

in our HTML, we have created an unordered list by using em> & lt <; ul & gt; tag. This list has three elements, each of which is contained in

  • tags

    List of articles for Twitter and Facebook are included in HTML span tags ( and ). The opening span tag for each item in the list specifies a class attribute. the value of the class attribute is "blue".

    We have defined a style called .blue which changes the background color of whatever element it is applied to to light blue. the background color of all links with CSS class .blue in our HTML to change from light blue.

  • The background color associated with our Instagram link has not changed because we have not included the text of this link in the tags.

    Conclusion

    HTML code is a generic container used to apply styles to specific elements on an HTML page. . By default, the tag does nothing, but it can be used in conjunction with CSS to style elements

    As a challenge, define a ; tag that makes a word in a sentence appear in all uppercase letters. You will need to use a CSS rule to draw text with your tag.

    This tutorial illustrates, with examples, how to use HTML << em> span> in your code. You are now ready to start using the tag like a professional web developer!

    For help and support with HTML learning, read our HTML Learning Guide .

    How To Get The Value Of The P Tag In Javascript __del__: Questions

    How can I make a time delay in Python?

    5 answers

    I would like to know how to put a time delay in a Python script.

    2973

    Answer #1

    import time
    time.sleep(5)   # Delays for 5 seconds. You can also use a float value.
    

    Here is another example where something is run approximately once a minute:

    import time
    while True:
        print("This prints once a minute.")
        time.sleep(60) # Delay for 1 minute (60 seconds).
    

    2973

    Answer #2

    You can use the sleep() function in the time module. It can take a float argument for sub-second resolution.

    from time import sleep
    sleep(0.1) # Time in seconds
    

    How to delete a file or folder in Python?

    5 answers

    How do I delete a file or folder in Python?

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    Path objects from the Python 3.4+ pathlib module also expose these instance methods:

    How To Get The Value Of The P Tag In Javascript exp: Questions

    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)
    

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