非常着名的IndexError。不幸的是，我真的没有找到解决方案。
最后一次访问最后一个网址时，我总是收到错误消息。网站是否为空。无论范围是2还是20，都会发生此错误。
text_file = open("Results-from-{}.txt".format(self.entry_get), "w")
### Iterator for end of the url
multiple_url = []
for iterator_page in range(15):
iterator_page = iterator_page + 1
multiple_url.append("".join([self.sub_url, str(iterator_page)]))
### loop for visit all 20 pages ###
parser = 0
while parser < len(multiple_url):
print(multiple_url[parser])
parser += 1
with urllib.request.urlopen(multiple_url[parser]) as url:
soup = BeautifulSoup(url, "html.parser")
### html tag parsing
names = [name.get_text().strip() for name in soup.findAll("div", {"class": "name m08_name"})]
street = [address.get_text().strip() for address in soup.findAll(itemprop="streetAddress")]
plz = [address.get_text().strip() for address in soup.findAll(itemprop="postalCode")]
city = [address.get_text().strip() for address in soup.findAll(itemprop="addressLocality")]
### zip and write
for line in zip(names, street, plz , city):
print("%s;%s;%s;%s;\n" % line)
text_file.write("%s;%s;%s;%s;\n" % line)
### output of the path main: cwd_out_final
cwd = os.getcwd()
cwd_out = "\{}".format(text_file.name)
cwd_out_final = cwd + cwd_out
text_file.close()
我的错误：
Exception in Tkinter callback
Traceback (most recent call last):
File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/tkinter/__init__.py", line 1699, in __call__
return self.func(*args)
File "/Users/x/PycharmProjects/hackday/parser.py", line 55, in search_complete_inner
with urllib.request.urlopen(multiple_url[parser]) as url:
IndexError: list index out of range
谢谢！
答案 0 :(得分：0)
在将parser用作with语句中的索引之前，您将其增加 <?php
$a='[
{
"user_id":"11",
"check_id":"38",
"pcode_id":"14",
"platform_id":"2",
"vin":"MA11340DP0DN09661",
"date":"2017-06-09-10-48-25",
"status":"completed",
"description":"Check for proper insert of the connector and loose",
"result":true
},
{
"user_id":"11",
"check_id":"39",
"pcode_id":"14",
"platform_id":"2",
"vin":"MA11340DP0DN09661",
"date":"2017-06-09-10-48-25",
"status":"completed",
"description":"Damaged\/Cracked",
"result":false
}]';
$json = json_decode($a);
for($i=0; $i<count($json); $i++)
{
echo "user_id :- ".$json[$i]->user_id."<br>";
echo "check_id :- ".$json[$i]->check_id."<br>";
echo "pcode_id :- ".$json[$i]->pcode_id."<br>";
echo "platform_id :- ".$json[$i]->platform_id."<br>";
echo "vin :- ".$json[$i]->vin."<br>";
echo "date :- ".$json[$i]->date."<br>";
echo "status :- ".$json[$i]->status."<br>";
echo "description :- ".$json[$i]->description."<br>";
echo "result :- ".$json[$i]->result."<br><hr>";
}
?>
;在最后一个元素上执行此操作将生成有问题的错误。此外，这意味着您永远不会使用列表中的 first 元素。