how do I select xpath image without a class name using selenium in python?

Question

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>

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                    <span class="screen-reader-text">hide this posting</span>
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</li>

Answer 1

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')

Answer 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')

(ie 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')