擅长:python、mysql、java
<p>这对您提供的示例有效:</p>
<pre><code>def flatten_fields(fields_list):
c = {}
for item in fields_list:
for key in item:
if key == "fields":
c[item["name"]] = flatten_fields(item["fields"])
elif key != "name":
c[key] = item[key]
break
return [c]
</code></pre>
<p>但是它在字典列表上工作,所以您应该像<code>flatten_fields([data])[0]</code>那样调用它。你知道吗</p>
<p>输出为:</p>
<pre><code>{
"": [{
"keys": [{
"node-name": "0/0/CP0",
"interface-name": "TenGigE0/0/0/47",
"device-id": "ASR9K-H1902.corp.cisco.com"
}],
"content": [{
"lldp-neighbor": [{
"chassis-id": "78ba.f975.a64f",
"receiving-parent-interface-name": "Bndle-Ether403",
"enabled-capabilities": "R",
"device-id": "ASR9K-H1902.corp.cisco.com",
"hold-time": 120,
"receiving-interface-name": "TenGigE0/0/0/47",
"platform": "",
"header-version": 0,
"port-id-detail": "Te0/1/0/4/0"
}]
}]
}]
}
</code></pre>