Find the xpath with get_attribute() in python selenium

Question

This is a somewhat backwards approach to web scraping. I need to locate the xpath of a web element AFTER I have already found it with a text()= identifier

Because the xpath values are different based on what information shows up, I need to use predictable labels inside the row for locating the span text next to found element. I found a simple and reliable way is locating the keyword label and then increasing td integer by one inside the xpath.

    def x_label(self, contains):
         mls_data_xpath = f"//span[text()='{contains}']"
         string = self.driver.find_element_by_xpath(mls_data_xpath).get_attribute("xpath")
         digits = string.split("td[")[1]
         num = int(re.findall(r'(\d+)', digits)[0]) + 1
         labeled_data = f'{string.split("td[")[0]}td[{num}]/span'
         print(labeled_data)
         labeled_text = self.driver.find_element_by_xpath(labeled_data).text
         return labeled_text

I cannot find too much information on .get_attribute() and get_property() so I am hoping there is something like .get_attribute("xpath") but I haven't been able to find it.

Basically, I am taking in a string like "ApprxTotalLivArea" which I can rely on and then increasing the integer after td[0] by 1 to find the span data from cell next door. I am hoping there is something like a get_attributes("xpath") to locate the xpath string from the element I locate through my text()='{contains}' search.

That is a great starting point but I have to make it work for python. — Richard Modad
– Richard Modad, Commented Mar 31, 2022 at 21:21

undetected Selenium · Accepted Answer · 2022-03-31 21:19:13Z

2

The Remote WebElement does includes the following methods:

But xpath isn't a valid property of a WebElement. So get_attribute("xpath") will always return NULL

answered Mar 31, 2022 at 21:19

undetected Selenium

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1 Comment

Richard Modad Over a year ago

That is pseudo code that obviously doesn't work but explains what I am looking for - I haven't found anything in get_attribute or get_property that is relevant for xpath values but it seems like I should be able to generate an xpath from the element somehow.

Tom Fuller · Accepted Answer · 2023-01-10 22:26:25Z

This function iteratively get's the parent until it hits the html element at the top

from selenium import webdriver
from selenium.webdriver.common.by import By


def get_xpath(elm):
    e = elm
    xpath = elm.tag_name
    while e.tag_name != "html":
        e = e.find_element(By.XPATH, "..")
        neighbours = e.find_elements(By.XPATH, "../" + e.tag_name)
        level = e.tag_name
        if len(neighbours) > 1:
            level += "[" + str(neighbours.index(e) + 1) + "]"
        xpath = level + "/" + xpath
    return "/" + xpath

driver = webdriver.Chrome()
driver.get("https://www.stackoverflow.com")
login = driver.find_element(By.XPATH, "//a[text() ='Log in']")
xpath = get_xpath(login)
print(xpath)

assert login == driver.find_element(By.XPATH, xpath)

Hope this helps!

Richard Modad · Accepted Answer · 2022-03-31 23:01:17Z

I was able to find a python version of the execute script from this post that was based off a JavaScript answer in another forum. I had to make a lot of .replace() calls on the string this function creates but I was able to universally find the label string I need and increment the td/span xpath by +1 to find the column data and retrieve it regardless of differences in xpath values on different page listings.

def x_label(self, contains):
    label_contains = f"//span[contains(text(), '{contains}')]"
    print(label_contains)
    labeled_element = self.driver.find_element_by_xpath(label_contains)
    print(labeled_element)
    element_label = labeled_element.text
    print(element_label)

    self.driver.execute_script("""
    window.getPathTo = function (element) {
        if (element.id!=='')
            return 'id("'+element.id+'")';
        if (element===document.body)
            return element.tagName;

        var ix= 0;
        var siblings= element.parentNode.childNodes;
        for (var i= 0; i<siblings.length; i++) {
            var sibling= siblings[i];
            if (sibling===element)
                return window.getPathTo(element.parentNode)+'/'+element.tagName+'['+(ix+1)+']';
            if (sibling.nodeType===1 && sibling.tagName===element.tagName)
                ix++;
        }
    }
    """)

    generated_xpath = self.driver.execute_script("return window.getPathTo(arguments[0]);", labeled_element)
    generated_xpath = f'//*[@{generated_xpath}'.lower().replace('tbody[1]', 'tbody')

    print(f'generated_xpath = {generated_xpath}')

    expected_path = r'//*[@id="wrapperTable"]/tbody/tr/td/table/tbody/tr[26]/td[6]/span'

    generated_xpath = generated_xpath.replace('[@id("wrappertable")', '[@id="wrapperTable"]').replace('tr[1]', 'tr')
    clean_path = generated_xpath.replace('td[1]', 'td').replace('table[1]', 'table').replace('span[1]', 'span')
    print(f'clean_path = {clean_path}')
    print(f'expected_path = {expected_path}')
    digits = generated_xpath.split("]/td[")[1]
    print(digits)
    num = int(re.findall(r'(\d+)', digits)[0]) + 1
    print(f'Number = {num}')
    labeled_data = f'{clean_path.split("td[")[0]}td[{num}]/span'
    print(f'labeled_data = {labeled_data}')
    print(f'expected_path = {expected_path}')

    if labeled_data == expected_path:
        print('Congrats')
    else:
        print('Rats')

    labeled_text = self.driver.find_element_by_xpath(labeled_data).text
    print(labeled_text)
    return labeled_text

L Tyrone · Accepted Answer · 2024-06-21 18:30:07Z

An upgrade of Tom Fuller's function. The following helps to find the correct xpath if there are elements with the same tag_name (and, for example, class) in the parent element:

def get_xpath(elm):
    e = elm
    xpath = elm.tag_name
    i=0 # Счетчик финального элемента
    while e.tag_name != "html":
        if i==0: # Сохраняем родительский элемент финального-искомого (только в первый цикл)
            parent_elm=e.find_element(By.XPATH, "..")
            i+=1
        e = e.find_element(By.XPATH, "..")
        neighbours = e.find_elements(By.XPATH, "../" + e.tag_name)
        level = e.tag_name
        if len(neighbours) > 1:
            level += "[" + str(neighbours.index(e) + 1) + "]"
        xpath = level + "/" + xpath
    
    
    elm_count=1
    other_elements=parent_elm.find_elements('xpath', elm.tag_name)
    for other_element in other_elements:
        if other_element==elm:
            final_element_count=elm_count
        else:
            elm_count+=1
    if final_element_count>1:
        final_xpath="/" + xpath+f'[{str(final_element_count)}]'
    else:
        final_xpath="/" + xpath
    return final_xpath

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