我有这个脚本贯穿目录 images 的所有子目录,并打印出所有图像。
<?php
$dirname = 'images/*/';
$images = glob($dirname . "*");
foreach ($images as $image) {
echo '<img src="' . $image . '" class=image /><br>';
}
?>
我还想知道子目录的名称(&#34; *&#34;在$ dirname中),其中每个图像都是从中取出的,因此我可以将其打印出来。
所以在浏览器中它应该是这样的:
答案 0 :(得分:1)
获取目录名称,然后获取其中的尾随部分:
import csv
import requests
from bs4 import BeautifulSoup
import pprint
import sys
url = 'http://www.yellowpages.com/search?search_terms=restaurants&geo_location_terms=Charleston%2C%20SC'
response = requests.get(url)
html = response.content
soup = BeautifulSoup(html, "html.parser")
g_data = soup.find_all("div", {"class": "info"}) #this isolates the big chunks of data which houses our child tags
for item in g_data: #iterates through big chunks
try:
eateryName = (item.contents[0].find_all("a", {"class": "business-name"})[0].text)
except:
pass
print(eateryName)
with open('csvnametest.csv', "w") as csv_file:
writer = csv.writer(csv_file)
writer.writerow([eateryName])