改进Django中的表单的简单方法
首先，search()视图对于空字符串的处理相当薄弱——仅显示一条”Please submit a search term.”的提示信息。 若用户要重新填写表单必须自行点击“后退”按钮， 这种做法既糟糕又不专业。如果在现实的案例中，我们这样子编写，那么Django的优势将荡然无存。
在检测到空字符串时更好的解决方法是重新显示表单，并在表单上面给出错误提示以便用户立刻重新填写。 最简单的实现方法既是添加else分句重新显示表单，代码如下：
from django.http import HttpResponse
from django.shortcuts import render_to_response
from mysite.books.models import Book
def search_form(request):
return render_to_response('search_form.html')
def search(request):
if 'q' in request.GET and request.GET['q']:
q = request.GET['q']
books = Book.objects.filter(title__icontains=q)
return render_to_response('search_results.html',
{'books': books, 'query': q})
else:
**return render_to_response('search_form.html', {'error': True})**
（注意，将search_form()视图也包含进来以便查看）
这段代码里，我们改进来search()视图：在字符串为空时重新显示search_form.html。 并且给这个模板传递了一个变量error，记录着错误提示信息。 现在我们编辑一下search_form.html，检测变量error：
<html>
<head>
<title>Search</title>
</head>
<body>
**{% if error %}**
**<p style="color: red;">Please submit a search term.</p>**
**{% endif %}**
<form action="/search/" method="get">
<input type="text" name="q">
<input type="submit" value="Search">
</form>
</body>
</html>
我们修改了search_form()视图所使用的模板，因为search_form()视图没有传递error变量，所以在条用search_form视图时不会显示错误信息。
通过上面的一些修改，现在程序变的好多了，但是现在出现一个问题： 是否有必要专门编写search_form()来显示表单？ 按实际情况来说，当一个请求发送至/search/（未包含GET的数据）后将会显示一个空的表单（带有错误信息）。 所以，只要我们改变search()视图：当用户访问/search/并未提交任何数据时就隐藏错误信息，这样就移去search_form()视图以及对应的URLpattern。
def search(request):
error = False
if 'q' in request.GET:
q = request.GET['q']
if not q:
error = True
else:
books = Book.objects.filter(title__icontains=q)
return render_to_response('search_results.html',
{'books': books, 'query': q})
return render_to_response('search_form.html',
{'error': error})
在改进后的视图中，若用户访问/search/并且没有带有GET数据，那么他将看到一个没有错误信息的表单； 如果用户提交了一个空表单，那么它将看到错误提示信息，还有表单； 最后，若用户提交了一个非空的值，那么他将看到搜索结果。
最后，我们再稍微改进一下这个表单，去掉冗余的部分。 既然已经将两个视图与URLs合并起来，/search/视图管理着表单的显示以及结果的显示，那么在search_form.html里表单的action值就没有必要硬编码的指定URL。 原先的代码是这样：
<form action="/search/" method="get">
现在改成这样：
<form action="" method="get">
action=”“意味着表单将提交给与当前页面相同的URL。 这样修改之后，如果search()视图不指向其它页面的话，你将不必再修改action。
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