擅长:python、mysql、java
<p>我更新了<strong>获取URL()</strong>以返回请求的URL列表,否则将不返回。
我希望我能帮忙</p>
<pre><code>def get_urls(distribution):
urls = list()
tag1 = "source"
tag2 = "url"
result_list = []
# Open distribution.yaml file and collect urls in list
try:
for file in os.listdir("distribution/"):
if file.startswith(distribution):
dist_file = "distribution/" + file
stream = open(dist_file, 'r')
data = yaml.safe_load(stream)
except:
pass
for elt in data:
result_list.append(data[elt]['source']['url'])
return result_list
return None
</code></pre>
<p><strong>输出:</strong></p>
<pre><code>['https://github.com/ament/ament_tools.git', 'https://github.com/ros2/cartographer.git']
</code></pre>