Question

我在URLField上保留了图片的网址，这样我就可以在我的主页模型上显示图像，而不是URL的字符串，我无法转换那个方向，“www。 google.es/images/car“在汽车的图像中？

models.py

def home (request):
     photos = Photo.objects.all ()
     html = '<ul>'
     for photo in photos:
         html + = '<li>' + photo.url + '</ li>'
     html + = '</ ul>'
     return HttpResponse (html)

views.py

photo.url

如何将图像<template>转换为图像。

Answer 1

您无法将字符串转换为图片，但您可以使用<img>标记内的for photo in photos: html + = '<li><img src="' + photo.url + '"></ li>'属性中的网址

obj\Debug\main.o||In function `main':|
E:\project\test\main.cpp|15|undefined reference to `Stack<char>::Stack()'|
E:\project\test\main.cpp|23|undefined reference to `Stack<char>::shiftL()'|
E:\project\test\main.cpp|28|undefined reference to `Stack<char>::shiftR()'|
E:\project\test\main.cpp|33|undefined reference to `Stack<char>::Del()'|
E:\project\test\main.cpp|36|undefined reference to `Stack<char>::push(char)'|
||error: ld returned 1 exit status|
||=== Build failed: 6 error(s), 0 warning(s) (0 minute(s), 0 second(s)) ===|

请记住，您正在为网站生成HTML而不提供内容

Answer 2

尝试：

def home (request):
     photos = Photo.objects.all ()
     html = '<ul>'
     for photo in photos:
         html + = '<li><img src="{}"><li>'.format(photo.url)
     html + = '</ ul>'
     return HttpResponse (html)

也许更好，再次依赖于格式而不是字符串连接：

def home (request):
     photos = Photo.objects.all ()
     html = '<ul>{}</ul>'
     html_photos = []
     for photo in photos:
         html_photos.append('<li><img src="{}"><li>'.format(photo.url))
     html = html.format("\n".join(html_photos))
     return HttpResponse (html)

如何在Django上的图像中转换URLField

2 个答案: