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How do i select an image xpath without a classname. HTML code is like this

<img alt="" class src="https://images.craigslist.org/00J0J_i9BI6mN6rKP_300x300.jpg">

If I right click and copy xpath it gives me this //*[@id="sortable-results"]/ul/li[1]/a/img but when I use it in my code it has some error

In my code i use like this

 src = driver.find_elements_by_xpath('/li[@class="result-row"]/a[@class="result-image gallery"]/img[@class=""]/@src')

but it returns me an [] when i print(src)

Full div

<li class="result-row" data-pid="7017735595">

        <a href="https://vancouver.craigslist.org/van/ele/d/vancouver-sealed-brand-new-in-box/7017735595.html" class="result-image gallery" data-ids="1:00J0J_i9BI6mN6rKP"><img alt="" class="" src="https://images.craigslist.org/00J0J_i9BI6mN6rKP_300x300.jpg">
                <span class="result-price">$35</span>
        </a>

    <p class="result-info">
        <span class="icon icon-star" role="button" title="save this post in your favorites list">
            <span class="screen-reader-text">favorite this post</span>
        </span>

            <time class="result-date" datetime="2019-11-11 00:52" title="Mon 11 Nov 12:52:25 AM">Nov 11</time>


        <a href="https://vancouver.craigslist.org/van/ele/d/vancouver-sealed-brand-new-in-box/7017735595.html" data-id="7017735595" class="result-title hdrlnk">Sealed - Brand New in Box - Google Home Mini</a>


        <span class="result-meta">
                <span class="result-price">$35</span>


                <span class="result-hood"> (Vancouver)</span>

                <span class="result-tags">
                    <span class="pictag">pic</span>
                </span>

                <span class="banish icon icon-trash" role="button">
                    <span class="screen-reader-text">hide this posting</span>
                </span>

            <span class="unbanish icon icon-trash red" role="button" aria-hidden="true"></span>
            <a href="#" class="restore-link">
                <span class="restore-narrow-text">restore</span>
                <span class="restore-wide-text">restore this posting</span>
            </a>

        </span>
    </p>
</li>
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  • can you share the whole div where your image is located? you can create an absolute path using xpath. Commented Nov 11, 2019 at 9:20
  • ok sir ill post the whole html for that div Commented Nov 11, 2019 at 9:21
  • 1
    when I use it in my code it has some error What is it? and how do you use it? Commented Nov 11, 2019 at 9:21

2 Answers 2

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The xpath is close. You need to use // at the beginning of the path and remove the /@src

//li[@class="result-row"]/a[@class="result-image gallery"]/img[@class=""]

If you want to make sure the element has src attribute it's like that

//li[@class="result-row"]/a[@class="result-image gallery"]/img[@class=""][@src]

To get the src attribute use get_attribute('src)

src = driver.find_elements_by_xpath('//li[@class="result-row"]/a[@class="result-image gallery"]/img[@class=""]')[0].get_attribute('src')

Note that find_elements return list, use index to get the first element.

If you want to use class="result-info" to locate the element you can do

elements = driver.find_elements_by_xpath('//p[@class="result-info"]/../a[@class="result-image gallery"]/img[@class=""]')
for element in elements:
    src = element.get_attribute('src')
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11 Comments

how do I get the image src the one that ends with jpg file in order for me to save the image or the url in it? I tried to print(src) it returns me ` <selenium.webdriver.remote.webelement.WebElement (session="a1858a5fdf3e236394be883fc8f6bc5a", element="0ab23f0f-5420-498c-bcfd-22af33872247")>`
i want to get the image attribute src
oh i see, so i can add another function like get_attribute i didint know that. Thanks for the answer sir. Can you explain it to me deeper? why you used index 0 array and other things
@Vince get_attribute is part of Selenium, you don't need to add anything. The index 0 is already explained in the last line.
ya sir. i already figured it out. Thanks a lot for your answers !!
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2

Actually the xpath has been copied correctly, You have used it in a wrong way in the fetch code.

If you want the specific image, use

image = driver.find_element_by_xpath('//*[@id="sortable-results"]/ul/li[1]/a/img')

Or, if you want a list of all images of same xpath type, use:

images = driver.find_elements_by_xpath('//*[@id="sortable-results"]/ul/li/a/img')

(i.e. remove the specific number of 'li' div or any other div that you want to generalise and use find_elements; you need to use find_element for fetching a specific single element)


To get the attribute 'src', use get_attribute method:

For case 1:

website = image.get_attribute('src')

For case 2:

website = images[0].get_attribute('src')
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10 Comments

but i want the image src how could i achieve that?
got me an error 'list' object has no attribute 'get_attribute'
Updated the answer. If you fetch using find_elements, it fetches a list. So you need to select the element from list, i.e. list[0] in this case. If you fetched using find_element though, you would need to select from a list and you can directly get attribute
i dont get the output i want sir. It just gives me one image url but in a different form. like every letter there is a new line. its like this h t t p s : / / i m a g e s . c r a i g s l i s t . o r g / 0 0 R 0 R _ x w 4 R v C R Y D B I just want to list all the image url in every result-info
I am sure you are printing it wrong. If you want 'src' for all the images fetched in the list, use a loop. Like 'for image in images: website = image.get_attribute('src'); print(website)'
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