擅长:python、mysql、java
<p>我在这里看到了两个选择。</p>
<h2>一。手动过滤(相当难看)</h2>
<pre><code>diff = []
for all in alllists:
found = False
for sub in subscriptionlists:
if sub.id == all.id:
found = True
break
if not found:
diff.append(all)
</code></pre>
<h2>2。再问一次</h2>
<pre><code>diff = List.objects.filter(datamode = 'A').exclude(member__id=memberid, datamode='A')
</code></pre>