Get the number of images on a webpage

Question

Recreational languages win at code golf too much. This challenge is simple, when using higher level functions.

The task is as follows:

Given any valid url, which is guaranteed to result in a valid HTML document (for example https://en.wikipedia.org/wiki/Ruby_on_Rails, which has 9 images), count the number of images displayed from the resultant webpage.

Only images visible in the HTML code are counted, marked by the HTML <img> tag.

Websites that can continually load images (Google Images) are not valid tests.

Answer 1

JavaScript, 61 bytes

As per this consensus, the code needs to run under the same domain as the page being requested, to avoid CORS issues. Returns a Promise conatining the count.

u=>fetch(u).then(r=>r.text()).then(t=>t.split`<img`.length-1)

Thanks to tsh for pointing out some code I forgot to update before posting, saving 3 bytes.

---

Try it

(f=
u=>fetch(`https://crossorigin.me/`+u).then(r=>r.text()).then(t=>t.split`<img`.length-1)
)(i.value=`https://codegolf.stackexchange.com/questions/161669/`).then(x=>o.innerText=x);b.addEventListener(`click`,_=>f(i.value).then(x=>o.innerText=x),0)

<input id=i><button id=b>Count</button><pre id=o></pre>

Answer 2

CJam, 10 bytes

lg"<img"e=

Does not work online.

Description:

l          # read line
 g         # open url and get html
  "<img"   # this string
        e= # count arg2 in arg1

Answer 3

Python 3 + Selenium + Firefox, 124 bytes

from selenium.webdriver import*
d=Firefox()
d.get(input())
c=len(d.find_elements_by_xpath('//img'))
d.quit()
print(c)

Notice that https://en.wikipedia.org/wiki/Ruby_on_Rails report 8 instead of 9. This is due to one of the <img> tag is contained by <noscript>, which is regarded as comment by a browser who enabled script.

Changing webdriver.Firefox to webdriver.Ie may save 5 bytes... But I just dislike IE.

How to run:

• First install Python 3, and Firefox
• install Selenium by pip install selenium
• Download gecko driver and put it to your path
• Run this script
• A Firefox will be started, ignore it and input the url to stdin
• The result will be shown on stdout

Answer 4

Python 2 + BeautifulSoup, 103 bytes

import bs4 as B
import urllib as U
print len(B.BeautifulSoup(U.urlopen(input()).read()).findAll('img'))

Can't do a TIO as BeautifulSoup is not available there. This reports 9 for https://en.wikipedia.org/wiki/Ruby_on_Rails.

Note: A warning is displayed before the number as a parser is not specified and lxmlis taken as the default. Adding the lxml parser explicitly suppresses the warning but costs another 7 bytes.

import bs4 as B
import urllib as U
print len(B.BeautifulSoup(U.urlopen(input()).read(),'lxml').findAll('img'))

Answer 5

Bash, grep, and wc, 45 28 bytes

curl $1|grep -o "<img"|wc -l

-17 bytes thanks to tsh

Explanation:

curl $1                      # download html from first parametre
       |grep -o "<img"       # find all <img strings
                      |wc -l # count lines