Javascript Object In String

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[object Object] is a string representation of an object. You can see this text if you use alert () to print an object to the screen, for example. You can view the contents of an object using console.log (), JSON.stringify (), or ... in the loop.

While developing the use of JavaScript, many of us encountered the output: [object Object]. When I saw him, I went to my mentor at the time and asked him, "What does that mean too?" ". I was confused.

This article is intended to tell you about this release and what it mean s. We are going to talk about how you can translate [object object] into human readable content that you can work with.

What is JavaScript [object object]?

> [object Object] is a string version of an object instance. This value is returned by a JavaScript program if you try to print an object without having to first put in the form of the object as a string

This is the syntax of the object [object Object ]:.

No wonder developers get confused about this: there are no error messages or warnings guard tell us what is happening. Look Let an example of this object in the action

[object Object] JavaScript Example

Take this example:.

When the alert declaration () is executed, our code returns [object Object]. Our program tries to return a string representation of what was passed to the alert () method. But, since our code sees it as an object, it tells us that it is an instance of an object instead

[Object object] message is not very descriptive. But that doesn’t mean that we can’t see the values ‚Äã‚Äãin our object. Let’s talk about ways that we can read the values ‚Äã‚Äãin an object

Inside object?

Knowing that [object Object] is an instance of an object is great, but we want to know is inside the object. There are three ways to do this:

  1. Log in to the console with console.log ()
  2. stringify with JSON.stringify ()
  3. Use the ... in loop and look at each property < / li>

    Register at the console

    Probably the best way to see what is inside an object is to register the object at the console . The console.log () statement allows you to view all of you values ‚Äã‚Äãin a JavaScript object

    Consider the following code:.

    We can see the values ‚Äã‚Äãin our object.

    use JSON.stringify ()

    method JSON.stringify () converts a JavaScript object to a string . We can then manipulate this chain.

    Then we can use JSON.stringify () to convert an object to a string. So we could use alert () to show the string value to the user:

    As in our last example, we have defined an object called objA. Next, we use the JSON.stringify () method to convert the object to a string. We then use alert to show the string value to the console.

    Our code opens a prompt window with the following content:.

    Use a for ... loop

    JavaScript ... in loop allows us to scroll through the contents of an object. We can use this loop to print each key value pair

Consider the following code:

We declared a JSON object called objA like we did in the last two examples. Then we use for ... in loop to iterate through the contents of that object. The "key‚" value represents each key.

We use the value "key" to access the key and objA [key] to access the value associated with that key. Our code returns:

We use string concatenation to add a colon (:) between each key and the value. This allows us to separate the keys and values ‚Äã‚Äãso that they are more readable in the output of our code.


The JavaScript code [Object object] is a string representation of an object. To see the contents of an object, you must print the object to the console using console.log () or convert the object to a string. Or you can use a for ... in loop to iterate through the object and see its contents.

Want to learn more about JavaScript? Check out our Complete Guide to Learning JavaScript for tips on the best learning resources and online courses .

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Javascript Object In String find: Questions

Finding the index of an item in a list

5 answers

Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?


Answer #1

>>> ["foo", "bar", "baz"].index("bar")

Reference: Data Structures > More on Lists

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

list.index(x[, start[, end]])

Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

>>> import timeit
>>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000)
>>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000)

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

>>> [1, 1].index(1)
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
>>> next(g)

Most places where I once would have used index, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item"s not present.

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If the item might not be present in the list, you should either

  1. Check for it first with item in my_list (clean, readable approach), or
  2. Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)


Answer #2

One thing that is really helpful in learning Python is to use the interactive help function:

>>> help(["foo", "bar", "baz"])
Help on list object:

class list(object)

 |  index(...)
 |      L.index(value, [start, [stop]]) -> integer -- return first index of value

which will often lead you to the method you are looking for.


Answer #3

The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use enumerate():

for i, j in enumerate(["foo", "bar", "baz"]):
    if j == "bar":

The index() function only returns the first occurrence, while enumerate() returns all occurrences.

As a list comprehension:

[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]

Here"s also another small solution with itertools.count() (which is pretty much the same approach as enumerate):

from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]

This is more efficient for larger lists than using enumerate():

$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 196 usec per loop


InsecurePlatformWarning: A true SSLContext object is not available. This prevents urllib3 from configuring SSL appropriately

3 answers

Tried to perform REST GET through python requests with the following code and I got error.

Code snip:

import requests
header = {"Authorization": "Bearer..."}
url = az_base_url + az_subscription_id + "/resourcegroups/Default-Networking/resources?" + az_api_version
r = requests.get(url, headers=header)


          InsecurePlatformWarning: A true SSLContext object is not available. 
          This prevents urllib3 from configuring SSL appropriately and may cause certain SSL connections to fail. 
          For more information, see

My python version is 2.7.3. I tried to install urllib3 and requests[security] as some other thread suggests, I still got the same error.

Wonder if anyone can provide some tips?


Answer #1

The docs give a fair indicator of what"s required., however requests allow us to skip a few steps:

You only need to install the security package extras (thanks @admdrew for pointing it out)

$ pip install requests[security]

or, install them directly:

$ pip install pyopenssl ndg-httpsclient pyasn1

Requests will then automatically inject pyopenssl into urllib3

If you"re on ubuntu, you may run into trouble installing pyopenssl, you"ll need these dependencies:

$ apt-get install libffi-dev libssl-dev


Dynamic instantiation from string name of a class in dynamically imported module?

3 answers

In python, I have to instantiate certain class, knowing its name in a string, but this class "lives" in a dynamically imported module. An example follows:

loader-class script:

import sys
class loader:
  def __init__(self, module_name, class_name): # both args are strings
      modul = sys.modules[module_name]
      instance = modul.class_name() # obviously this doesn"t works, here is my main problem!
    except ImportError:
       # manage import error

some-dynamically-loaded-module script:

class myName:
  # etc...

I use this arrangement to make any dynamically-loaded-module to be used by the loader-class following certain predefined behaviours in the dyn-loaded-modules...


Answer #1

You can use getattr

getattr(module, class_name)

to access the class. More complete code:

module = __import__(module_name)
class_ = getattr(module, class_name)
instance = class_()

As mentioned below, we may use importlib

import importlib
module = importlib.import_module(module_name)
class_ = getattr(module, class_name)
instance = class_()


How to get all of the immediate subdirectories in Python

3 answers

I"m trying to write a simple Python script that will copy a index.tpl to index.html in all of the subdirectories (with a few exceptions).

I"m getting bogged down by trying to get the list of subdirectories.


Answer #1

import os
def get_immediate_subdirectories(a_dir):
    return [name for name in os.listdir(a_dir)
            if os.path.isdir(os.path.join(a_dir, name))]

We hope this article has helped you to resolve the problem. Apart from Javascript Object In String, check other find-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:

Schneider Krasiko

Tallinn | 2022-11-30

median is always a bit confusing 😭 Javascript Object In String is not the only problem I encountered. Will get back tomorrow with feedback

Marie Emmerson

London | 2022-11-30

I was preparing for my coding interview, thanks for clarifying this - Javascript Object In String in Python is not the simplest one. Checked yesterday, it works!

Jan Jackson

New York | 2022-11-30

sin is always a bit confusing 😭 Javascript Object In String is not the only problem I encountered. Will use it in my bachelor thesis


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