UrlParse | A simple JavaScript URL Parsing Utility | Frontend Framework library

by thybag JavaScript Version: Current License: No License

X-Ray Key Features Code Snippets Community Discussions(10)Vulnerabilities Install Support

kandi X-RAY | UrlParse Summary

UrlParse is a JavaScript library typically used in User Interface, Frontend Framework, React applications. UrlParse has no bugs, it has no vulnerabilities and it has low support. You can download it from GitHub.

UrlParse aims to make extracting the part of a URL your interested in a more straightforward proccess. To try it out use the demo.htm packaged alongside the urlParse.js code.

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Quality

Security

License

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Support

UrlParse has a low active ecosystem.

It has 5 star(s) with 3 fork(s). There are 2 watchers for this library.

It had no major release in the last 6 months.

UrlParse has no issues reported. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of UrlParse is current.

Quality

UrlParse has no bugs reported.

Security

UrlParse has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

License

UrlParse does not have a standard license declared.

Check the repository for any license declaration and review the terms closely.

Without a license, all rights are reserved, and you cannot use the library in your applications.

Reuse

UrlParse releases are not available. You will need to build from source code and install.

Installation instructions are not available. Examples and code snippets are available.

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UrlParse Key Features

No Key Features are available at this moment for UrlParse.

UrlParse Examples and Code Snippets

No Code Snippets are available at this moment for UrlParse.

Community Discussions

Trending Discussions on UrlParse

How to parse rtsp url with boost qi?

Trying to download files without starting scrapy project but from .py file. Created Custom pipeline within python file, This error comes as metioned

Square brackets not getting escaped in regex

I cant get my flask application authenticated through google and apis usable

Parsing urls from a dataframe

Django project how to find admin username and password

Python (requests) - incorrect encoding when fetching headers

Apify - How to Enqueue URL Variations Efficiently

TypeError: descriptor 'split' requires a 'str' object but received a 'bytes'

Python flask app which returns list of files

QUESTION

How to parse rtsp url with boost qi?

Asked 2021-Jun-08 at 06:04

I'm trying to parse RTSP-url like this: ...

ANSWER

Answered 2021-Jun-07 at 22:01

The relatively obvious workaround would be to URL-escape the @:

Live On Coliru

Source https://stackoverflow.com/questions/67873608

QUESTION

Trying to download files without starting scrapy project but from .py file. Created Custom pipeline within python file, This error comes as metioned

Asked 2021-Jun-05 at 18:16

import scrapy
from scrapy.crawler import CrawlerProcess
from scrapy.pipelines.files import FilesPipeline
from urllib.parse import urlparse
import os

class DatasetItem(scrapy.Item):
    file_urls = scrapy.Field()
    files = scrapy.Field()

class MyFilesPipeline(FilesPipeline):
    pass



class DatasetSpider(scrapy.Spider):
    name = 'Dataset_Scraper'
    url = 'https://kern.humdrum.org/cgi-bin/browse?l=essen/europa/deutschl/allerkbd'
    

    headers = {
        'user-agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/53       7.36 (KHTML, like Gecko) Chrome/79.0.3945.130 Safari/537.36'
    }
    
    custom_settings = {
            'FILES_STORE': 'Dataset',
            'ITEM_PIPELINES':{"/home/LaxmanMaharjan/dataset/MyFilesPipeline":1}

            }
    def start_requests(self):
        yield scrapy.Request(
                url = self.url,
                headers = self.headers,
                callback = self.parse
                )

    def parse(self, response):
        item = DatasetItem()
        links = response.xpath('.//body/center[3]/center/table/tr[1]/td/table/tr/td/a[4]/@href').getall()
        
        for link in links:
            item['file_urls'] = [link]
            yield item
            break
        

if __name__ == "__main__":
    #run spider from script
    process = CrawlerProcess()
    process.crawl(DatasetSpider)
    process.start()

...

ANSWER

Answered 2021-Jun-05 at 18:16

In case if pipeline code, spider code and process launcher stored in the same file
You can use __main__ in path to enable pipeline:

Source https://stackoverflow.com/questions/67737807

QUESTION

Square brackets not getting escaped in regex

Asked 2021-Jun-02 at 13:20

I am a Python beginner. Python versions 3.8 and 3.9

In an existing URL validation code, I ran into issues with the password containing '['. The password is PN-[.d.g5(R{bK}[5ZLx,4~K*hHrSy32=q+

URL:

...

ANSWER

Answered 2021-Jun-02 at 13:20

There also curly braces in problematic password, so you need to add them too.

r"(?::[-a-z0-9._\[\]\{\}~%!$&'()*+,;=:]*)?@)?"

Source https://stackoverflow.com/questions/67805404

QUESTION

I cant get my flask application authenticated through google and apis usable

Asked 2021-May-27 at 15:59

im quite new to coding but this is barely documented so i need some help. Im building a flask application but I cant get the google auth flow working. Im using Pycharm and python version 3.9

My issues are : I cant find any beginner tutorial that explains how to go through the Auth flow.

I dont understand how to interact with google APis through flask. (i want to use the android-management-api)

I do understand that i need to create a service object but that only works when i can authenticate the google flow and this is where im already stuck for 5 days now.

I already followed the instructions from realpython and MattButton. when trying these instructions i keep getting errors. now im getting:

...

ANSWER

Answered 2021-May-27 at 15:59

Solved! I found out that there were 2 conflicting libraries that both used the requests name

Flask.requests
Requests

Source https://stackoverflow.com/questions/67585492

QUESTION

Parsing urls from a dataframe

Asked 2021-May-26 at 08:13

I am trying to parse urls from a dataframe to get the 'path'. My dataframe has 3 columns: ['url'], ['impressions'], ['clicks']. I want to replace all the urls by their Path. Here is my code:

...

ANSWER

Answered 2021-May-26 at 07:42

urlparse only takes one string at a time, not a series.

try:

Source https://stackoverflow.com/questions/67691891

QUESTION

Django project how to find admin username and password

Asked 2021-May-21 at 03:07

I found a Django project and failed to get it running in Docker container in the following way:

git clone https://github.com/hotdogee/django-blast.git
$ cat requirements.txt in this files the below dependencies had to be updated:
- kombu==3.0.30
- psycopg2==2.8.6

I have the following Dockerfile:

...

ANSWER

Answered 2021-May-21 at 02:26

The script code not have command line for create superuser, please try this in terminal and you have user

Source https://stackoverflow.com/questions/67629837

QUESTION

Python (requests) - incorrect encoding when fetching headers

Asked 2021-May-19 at 13:02

I am using requests library (python 3.9) to get filename from URL.^[1] For some reason a file name is incorrectly encoded. I should get "Ogłoszenie_0320.pdf" instead of "OgÅ\x82oszenie_0320.pdf".

My code looks something like this:

...

ANSWER

Answered 2021-May-19 at 12:48

Only characters from the ascii based latin-1 should be used as header values [rfc]. Here the file name has been escaped.

Source https://stackoverflow.com/questions/67600066

QUESTION

Apify - How to Enqueue URL Variations Efficiently

Asked 2021-May-10 at 14:40

I am creating a new actor in Apify with Cheerio to read an input file of URLs and return primarily two items: (1) the HTTP status code and (2) the HTML title. As part of our process, I would like to be able to try up to 4 variations of each input URL, such as:

HTTP://WWW.SOMEURL.COM
HTTPS://WWW.SOMEURL.COM
HTTP://SOMEURL.COM
HTTPS://SOMEURL.COM

If one of the 4 variations is successful, then the process should ignore the other variations and move to the next input URL.

I read the original input list into a RequestList, and then would like to create the variations in a RequestQueue. Is this the most efficient way to do it? Please see code below, and thank you!

...

ANSWER

Answered 2021-May-10 at 14:40

you should create your URL list beforehand. the handlePageFunction is only used for the actual scraping part, and you should only have the Apify.pushData there:

Source https://stackoverflow.com/questions/67471659

QUESTION

TypeError: descriptor 'split' requires a 'str' object but received a 'bytes'

Asked 2021-May-08 at 22:01

I am trying to scrape data from ESPN Cricinfo using a python script available on Github. The code is the following.

...

ANSWER

Answered 2021-May-07 at 04:56

Use

Source https://stackoverflow.com/questions/67428909

QUESTION

Python flask app which returns list of files

Asked 2021-May-04 at 11:52

I have a basic flask app which is used to get logs from pods. I want to use send_from_directory when the query is not complete. I do not want to create a zip file and list it rather I would like to list the logs files that are generated with the query.

...

ANSWER

Answered 2021-May-04 at 11:52

You can list all files using os.listdir, glob.glob or the pathlib library. The result can then be output within a template.
Then you add a new route which expects the file name as a parameter and delivers the file from the folder via send_from_directory. You can pass the name of the file as a path within a rule.
As far as I understand you correctly, the following example should meet your requirements.

Source https://stackoverflow.com/questions/67373819

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install UrlParse

You can download it from GitHub.

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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