Xpath 选择器 - 是否可以在 selenium 中按标题查找元素？

Question

I've been trying to build a bot that scrapes the name of followers liking a picture in instagram. The website opens a popup box with the accounts in side, and the box seemingly refreshes the account links as you scroll it. I've written a code that will open the box and then scroll it, but i can't get selenium to scrape the account names. My code looks like this to scroll the pop up box:

realscroll_box = browser.find_element_by_xpath('/html/body/div[4]/div/div[3]/div')
while last_ht != ht:
last_ht = ht
time.sleep(2)
ht = browser.execute_script('''
                            arguments[0].scrollTo(0, arguments[0].scrollHeight);
                            return arguments[0].scrollHeight;
                            ''', realscroll_box)
namelinkstemp1 = realscroll_box.find_elements_by_xpath('//*[contains(@href,"/")')

But it gives me this error:

selenium.common.exceptions.InvalidSelectorException: Message: invalid selector: Unable to locate an element with the xpath expression // [starrts-with(@href,"/") because of the following error: SyntaxError: Failed to execute 'evaluate' on 'Document': The string '// [starrts-with(@href,"/")' is not a valid XPath expression.

The text i'm trying to extract looks like this:

<a title="instagramusername" href="/instagramusername/">
<div class="                   Igw0E     IwRSH      eGOV_     ybXk5    _4EzTm                                                                                                              ">
<div class="_7UhW9   xLCgt       qyrsm KV-D4            fDxYl      rWtOq">
<div class="                   Igw0E   rBNOH        eGOV_     ybXk5    _4EzTm                                                                                                              ">instagramusername</div>
</div>
</div>
</a>

Any help is appreciated, xpath queries are very new to me.

Thanks:)

Answer 1

Starts-with function finds the element based on attribute value. In your code function is not correct. Also //*[contains(@href,"/") , here you missed ] Correct one: //a[contains(@href, '/')]

  //*[starts-with(@href,'/')]

or

//a[starts-with(@href,'/instagramuse')]