[FIXED] How to capture the first link in xpath?

Issue

I have this html code:

<ul id="main-menu">
    <li>
       <a href="/ru/products/pro_1" title="text_1">...</a>
       <ul class="sm-nowrap">...</ul>
    </li>
    <li>
       <a href="/ru/products/pro_2" title="text_2">...</a>
       <ul class="sm-nowrap">...</ul>
    </li>
    <li>
       <a href="/ru/products/pro_3" title="text_3">...</a>
       <ul class="sm-nowrap">...</ul>
    </li>
    <li>
       <a href="/ru/products/pro_4" title="text_4">...</a>
       <ul class="sm-nowrap">...</ul>
    </li>
</ul>

I need to collect all the links (which are above the <ul class="sm-nowrap"> tag). I'm trying to do this with the following loop:

for i in response.css('ul#main-menu li'):
    link = i.xpath('//ul[class="sm-nowrap"]/preceding::a[1]/@href').get()

but I only get None None None None.... where is my mistake, what is wrong?

Solution

!! I don't have your whole HTML code so considering the part you gave us: !!

from io import StringIO
from lxml import etree

f = StringIO('''\
<ul id="main-menu">
    <li>
       <a href="/ru/products/pro_1" title="text_1">...</a>
       <ul class="sm-nowrap">...</ul>
    </li>
    <li>
       <a href="/ru/products/pro_2" title="text_2">...</a>
       <ul class="sm-nowrap">...</ul>
    </li>
    <li>
       <a href="/ru/products/pro_3" title="text_3">...</a>
       <ul class="sm-nowrap">...</ul>
    </li>
    <li>
       <a href="/ru/products/pro_4" title="text_4">...</a>
       <ul class="sm-nowrap">...</ul>
    </li>
</ul>
''')

tree = etree.parse(f)

The easiest and fastest way:

links = [i for i in tree.xpath('//a/@href')]

Explanation:

// everything from root to first a tag(s) (same level) encountered, then get the href attribute .

Result:

['/ru/products/pro_1', '/ru/products/pro_2', '/ru/products/pro_3', '/ru/products/pro_4']

Answered By - Drakax