如何在Django表2中为每个记录放置提交按钮？ [英] How to put submit buttons for every record in django tables 2?

问题描述

我在表单中有一个django_tables2表，这是必要的，因为在顶部有一个删除选定的按钮。在每一行中，我都需要一个带有某些操作的提交按钮，例如接受，拒绝等。

I have a django_tables2 table inside a form, this is necessary because there's a delete selected button on top. In every row, I need a submit button with some action like 'accept', 'deny', etc.

但是我不知道如何识别按下了哪个按钮。我该怎么做呢？

But I don't know how to recognize the row of which the button was pressed. How can I accomplish this?

推荐答案

我发现这种方法可以向每行添加一个按钮，其ID指向您选择的行：

I have found this way to add a button to each row with an id that refers to the row you select :

在app / tables.py

in app/tables.py

import django_tables2 as tables
from app.models import MyModel
from django.utils.safestring import mark_safe
from django.utils.html import escape

class MyColumn(tables.Column): 
    empty_values = list() 
    def render(self, value, record): 
        return mark_safe('<button id="%s" class="btn btn-info">Submit</button>' % escape(record.id))

class MyTable(tables.Table):
    submit = MyColumn()
    class Meta:
        model = MyModel
        ...

在app / views.py中，只是像往常一样：

in app/views.py, just like usual :

def mypage(request) :
    table = MyTable(MyModel.objects.all())
    RequestConfig(request,paginate=False).configure(table)
    return render(request,'template.html',locals())

然后我使用Jquery触发我想通过获取每个按钮的id来获取每个按钮的内容...

Then I use Jquery to trigger what I want for each button by retrieving their id ...

也许对于您想做的事情来说有点太简单了，也许我看过其他解决方案使用LinkColumns和Django URL调度程序可以提高效率： Django-Tables2 LinkColumn链接无效

Maybe it is a little bit too simple for what you want to do, I've seen other solutions maybe more efficient with LinkColumns and Django URL dispatcher : Django-Tables2 LinkColumn link doesn't work

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