DJango REST框架 - 如何将参数传递给序列化器__init__

时间:2016-07-12 13:22:02

标签: python django django-rest-framework

我一直在阅读django rest framework doc,但我找不到如何将参数传递给序列化程序的 init 方法。在常规视图中我喜欢这样:

<div class="row-text" style="width: calc(100% - 84px);">
   <span class="row-item-name">
    <span class="row-item-link">
      <a class="grid-row-element-name">Eclipse111</a>
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   <span class="row-item-actions hover-child">
      <a>Share</a><span> | </span><a watchdox-rename="name" watchdox-save-func="rename(element, name)" class="rename-link"><span translate="">Rename</span></a>
   </span>
   </span>
   <br>
   <span class="row-meta-data">
    <span class="creation-date-formatted">Today at  10:30 | </span>
   <span class="row-email">orgadmin@mailinator.com</span>
   </span>
</div>
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   <div class="row-tools">
      <div class="btn-group dropdown" uib-dropdown="">
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您是否知道我如何使用CreateAPIView / UpdateAPIView?

来做到这一点

感谢您的帮助

1 个答案:

答案 0 :(得分:0)

DRF有解决方案。

您需要delivery_id来保存您的实例,在这种情况下,您会将其直接传递到the serializer's save并将其放入sudo apt-get install sumo sumo-tools sumo-doc / create update

您需要进行验证,在这种情况下,您可以将其传递给serializer's context,您可以在序列化程序/字段中将其用作validated_data。您需要覆盖视图的self.context['request']

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