使用表单上传csv文件django
uploading csv file django using a form
我正在处理一个学校管理项目，我想上传一个 CSV 文件并通过覆盖当前数据或在数据库中更新它来保存它。但它没有被添加到数据库中。此外，我喜欢添加更多字段，因此如果可能的话使用动态（即循环），这样我以后就不必更改它了。
models.py
class Student(models.Model):
registration_number = models.CharField(max_length=200, unique=True)
firstname = models.CharField(max_length=200)
surname = models.CharField(max_length=200)
date_of_birth = models.DateField(default=timezone.now)
current_class = models.ForeignKey(StudentClass, on_delete=models.SET_NULL, blank=True, null=True)
date_of_admission = models.DateField(default=timezone.now)
parent_mobile_number = models.CharField(max_length=15)
address = models.TextField()
class StudentBulkUpload(models.Model):
date_uploaded = models.DateTimeField(auto_now=True)
csv_file = models.FileField(upload_to='students/bulkupload/')
forms.py
class StudentBulkUploadForm(forms.ModelForm):
class Meta:
model = StudentBulkUpload
fields = ("csv_file",)
views.py
def uploadcsv(request):
if request.method == 'GET':
form = StudentBulkUploadForm()
return render(request, 'students/students_upload.html', {'form':form})
# If not GET method then proceed
try:
form = StudentBulkUploadForm(data=request.POST, files=request.FILES)
if form.is_valid():
csv_file = form.cleaned_data['csv_file']
if not csv_file.name.endswith('.csv'):
messages.error(request, 'File is not CSV type')
return redirect('students:student-upload')
# If file is too large
if csv_file.multiple_chunks():
messages.error(request, 'Uploaded file is too big (%.2f MB)' %(csv_file.size(1000*1000),))
return redirect('students:student-upload')
file_data = csv_file.read().decode('utf-8')
lines = file_data.split('\n')
# loop over the lines and save them in db. If error, store as string and then display
for line in lines:
fields = line.split(',')
data_dict = {}
print(data_dict)
try:
form = StudentBulkUploadForm(data_dict)
if form.is_valid():
form.save()
else:
logging.getLogger('error_logger').error(form.errors.as_json())
except Exception as e:
logging.getLogger('error_logger').error(form.errors.as_json())
pass
except Exception as e:
logging.getLogger('error_logger').error('Unable to upload file. ' + repr(e))
messages.error(request, 'Unable to upload file. ' + repr(e))
return redirect('students:student-upload')
student_upload.html
<form method="POST" enctype="multipart/form-data">
{% csrf_token %}
{{ form.as_p }}
<input type="submit" class="btn btn-success" value="Submit">
</form>
在您的代码中，您正在对 CSV 文件进行所有解析和解码，而不是使用已编写的代码。我建议使用 Django 的 CSV import module 来解决这个问题。这样您就可以创建一个自动接收 CSV 数据并将其很好地转换为模型的模型：
from model import CsvModel
class MyCSvModel(CsvModel):
student_name = CharField()
foo = IntegerField()
然后，您将像往常一样创建一个可以指向该模型的 django 表单，这样您就可以处理用户上传。
但是，如果这不是一个选项，因为您已经有了某个模型，您希望保持不变（或任何原因），请查看 this Whosebug 问题，该问题解释了如何使用模块 Pandas 读取 csv 并将其转换为数据帧。
希望对您有所帮助！
是否有必要将数据通过表单传递并保存到db。否则你可以简单地创建一个模型对象并将字典传递给它并保存。
我认为你应该（简而言之）：
- 先使用
StudentBulkUpload() 上传并保存文件
- 获取文件路径并读取内容
- 如果模型字段和 csv 列的名称相同会更好
- 遍历每一行并创建一个字典，在迭代中只包含一个学生的详细信息
- 创建
Student()的实例并将字典传递给它并保存
- 对于外键，根据存储在 csv
中的值，使用 get() 从 StudentClass() 获取对象
您可以通过两种方式保存学生详细信息，我认为：
- 创建模型对象并通过正常方式读取每一行来赋值
new_student = Student()
new_student.registration_number = fields[0]
new_student.firstname = fields[1]
# like so for other fields
new_student.save()
- 创建一个模型对象，创建一个键值字典，其中键对应于模型的字段名称。
# create a dictionary `new_student_details` containing values of a student
new_student = Student()
new_student.__dict__.update(new_student_details)
new_student.save()
创建一个函数来读取 csv 文件并保存学生详细信息
import csv
def save_new_students_from_csv(file_path):
# do try catch accordingly
# open csv file, read lines
with open(file_path, 'r') as fp:
students = csv.reader(fp, delimiter=',')
row = 0
for student in students:
if row==0:
headers = student
row = row + 1
else:
# create a dictionary of student details
new_student_details = {}
for i in range(len(headers)):
new_student_details[headers[i]] = student[i]
# for the foreign key field current_class in Student you should get the object first and reassign the value to the key
new_student_details['current_class'] = StudentClass.objects.get() # get the record according to value which is stored in db and csv file
# create an instance of Student model
new_student = Student()
new_student.__dict__.update(new_student_details)
new_student.save()
row = row + 1
fp.close()
你的代码在之后应该看起来像这样：
def uploadcsv(request):
if request.method == 'GET':
form = StudentBulkUploadForm()
return render(request, 'students/students_upload.html', {'form':form})
# If not GET method then proceed
try:
form = StudentBulkUploadForm(data=request.POST, files=request.FILES)
if form.is_valid():
csv_file = form.cleaned_data['csv_file']
if not csv_file.name.endswith('.csv'):
messages.error(request, 'File is not CSV type')
return redirect('students:student-upload')
# If file is too large
if csv_file.multiple_chunks():
messages.error(request, 'Uploaded file is too big (%.2f MB)' %(csv_file.size(1000*1000),))
return redirect('students:student-upload')
# save and upload file
form.save()
# get the path of the file saved in the server
file_path = os.path.join(BASE_DIR, form.csv_file.url)
# a function to read the file contents and save the student details
save_new_students_from_csv(file_path)
# do try catch if necessary
except Exception as e:
logging.getLogger('error_logger').error('Unable to upload file. ' + repr(e))
messages.error(request, 'Unable to upload file. ' + repr(e))
return redirect('students:student-upload')
注意： csv 文件需要遵循正确的格式。您可以随心所欲地保存 Student()。无论如何，该文件需要上传和阅读。必须逐行 Student() 保存。这只是一个结构。由于我已经删除了您的大部分代码，请随时进行必要的更改
我正在处理一个学校管理项目，我想上传一个 CSV 文件并通过覆盖当前数据或在数据库中更新它来保存它。但它没有被添加到数据库中。此外，我喜欢添加更多字段，因此如果可能的话使用动态（即循环），这样我以后就不必更改它了。
models.py
class Student(models.Model):
registration_number = models.CharField(max_length=200, unique=True)
firstname = models.CharField(max_length=200)
surname = models.CharField(max_length=200)
date_of_birth = models.DateField(default=timezone.now)
current_class = models.ForeignKey(StudentClass, on_delete=models.SET_NULL, blank=True, null=True)
date_of_admission = models.DateField(default=timezone.now)
parent_mobile_number = models.CharField(max_length=15)
address = models.TextField()
class StudentBulkUpload(models.Model):
date_uploaded = models.DateTimeField(auto_now=True)
csv_file = models.FileField(upload_to='students/bulkupload/')
forms.py
class StudentBulkUploadForm(forms.ModelForm):
class Meta:
model = StudentBulkUpload
fields = ("csv_file",)
views.py
def uploadcsv(request):
if request.method == 'GET':
form = StudentBulkUploadForm()
return render(request, 'students/students_upload.html', {'form':form})
# If not GET method then proceed
try:
form = StudentBulkUploadForm(data=request.POST, files=request.FILES)
if form.is_valid():
csv_file = form.cleaned_data['csv_file']
if not csv_file.name.endswith('.csv'):
messages.error(request, 'File is not CSV type')
return redirect('students:student-upload')
# If file is too large
if csv_file.multiple_chunks():
messages.error(request, 'Uploaded file is too big (%.2f MB)' %(csv_file.size(1000*1000),))
return redirect('students:student-upload')
file_data = csv_file.read().decode('utf-8')
lines = file_data.split('\n')
# loop over the lines and save them in db. If error, store as string and then display
for line in lines:
fields = line.split(',')
data_dict = {}
print(data_dict)
try:
form = StudentBulkUploadForm(data_dict)
if form.is_valid():
form.save()
else:
logging.getLogger('error_logger').error(form.errors.as_json())
except Exception as e:
logging.getLogger('error_logger').error(form.errors.as_json())
pass
except Exception as e:
logging.getLogger('error_logger').error('Unable to upload file. ' + repr(e))
messages.error(request, 'Unable to upload file. ' + repr(e))
return redirect('students:student-upload')
student_upload.html
<form method="POST" enctype="multipart/form-data">
{% csrf_token %}
{{ form.as_p }}
<input type="submit" class="btn btn-success" value="Submit">
</form>
在您的代码中，您正在对 CSV 文件进行所有解析和解码，而不是使用已编写的代码。我建议使用 Django 的 CSV import module 来解决这个问题。这样您就可以创建一个自动接收 CSV 数据并将其很好地转换为模型的模型：
from model import CsvModel
class MyCSvModel(CsvModel):
student_name = CharField()
foo = IntegerField()
然后，您将像往常一样创建一个可以指向该模型的 django 表单，这样您就可以处理用户上传。
但是，如果这不是一个选项，因为您已经有了某个模型，您希望保持不变（或任何原因），请查看 this Whosebug 问题，该问题解释了如何使用模块 Pandas 读取 csv 并将其转换为数据帧。
希望对您有所帮助！
是否有必要将数据通过表单传递并保存到db。否则你可以简单地创建一个模型对象并将字典传递给它并保存。
我认为你应该（简而言之）：
- 先使用
StudentBulkUpload()上传并保存文件
- 获取文件路径并读取内容
- 如果模型字段和 csv 列的名称相同会更好
- 遍历每一行并创建一个字典，在迭代中只包含一个学生的详细信息
- 创建
Student()的实例并将字典传递给它并保存 - 对于外键，根据存储在 csv 中的值，使用
get() 从 StudentClass() 获取对象
您可以通过两种方式保存学生详细信息，我认为：
- 创建模型对象并通过正常方式读取每一行来赋值
new_student = Student()
new_student.registration_number = fields[0]
new_student.firstname = fields[1]
# like so for other fields
new_student.save()
- 创建一个模型对象，创建一个键值字典，其中键对应于模型的字段名称。
# create a dictionary `new_student_details` containing values of a student
new_student = Student()
new_student.__dict__.update(new_student_details)
new_student.save()
创建一个函数来读取 csv 文件并保存学生详细信息
import csv
def save_new_students_from_csv(file_path):
# do try catch accordingly
# open csv file, read lines
with open(file_path, 'r') as fp:
students = csv.reader(fp, delimiter=',')
row = 0
for student in students:
if row==0:
headers = student
row = row + 1
else:
# create a dictionary of student details
new_student_details = {}
for i in range(len(headers)):
new_student_details[headers[i]] = student[i]
# for the foreign key field current_class in Student you should get the object first and reassign the value to the key
new_student_details['current_class'] = StudentClass.objects.get() # get the record according to value which is stored in db and csv file
# create an instance of Student model
new_student = Student()
new_student.__dict__.update(new_student_details)
new_student.save()
row = row + 1
fp.close()
你的代码在之后应该看起来像这样：
def uploadcsv(request):
if request.method == 'GET':
form = StudentBulkUploadForm()
return render(request, 'students/students_upload.html', {'form':form})
# If not GET method then proceed
try:
form = StudentBulkUploadForm(data=request.POST, files=request.FILES)
if form.is_valid():
csv_file = form.cleaned_data['csv_file']
if not csv_file.name.endswith('.csv'):
messages.error(request, 'File is not CSV type')
return redirect('students:student-upload')
# If file is too large
if csv_file.multiple_chunks():
messages.error(request, 'Uploaded file is too big (%.2f MB)' %(csv_file.size(1000*1000),))
return redirect('students:student-upload')
# save and upload file
form.save()
# get the path of the file saved in the server
file_path = os.path.join(BASE_DIR, form.csv_file.url)
# a function to read the file contents and save the student details
save_new_students_from_csv(file_path)
# do try catch if necessary
except Exception as e:
logging.getLogger('error_logger').error('Unable to upload file. ' + repr(e))
messages.error(request, 'Unable to upload file. ' + repr(e))
return redirect('students:student-upload')
注意： csv 文件需要遵循正确的格式。您可以随心所欲地保存 Student()。无论如何，该文件需要上传和阅读。必须逐行 Student() 保存。这只是一个结构。由于我已经删除了您的大部分代码，请随时进行必要的更改