Python+Django+Js并利用表单实现文件上传

最新推荐文章于 2023-08-16 11:40:59 发布

原创最新推荐文章于 2023-08-16 11:40:59 发布 · 1.7k 阅读

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该博客介绍了如何使用Python的Django框架结合JavaScript，通过表单实现文件上传功能。内容涉及前端JS处理、后端Django models.py和views.py的设置，特别强调了上传文件必须为txt格式，并在文件存在时进行替换处理。

这段代码适用于利用form表单进行文件的提交，并保证所提交的文件格式为txt。

判断文件是否存在，如果存在则删除旧文件保存新文件并执行A，如果不存在则执行B。

前端：

<form action="{% url 'searchform' %}  method="POST" enctype="multipart/form-data">
    {% csrf_token %}
    <input type="file" name="uploadfile" accept="text/plain" onchange="fileChange(this);">
</form>

JS:

<script>
function fileChange(target){
    var name = target.value;
    var filename = name.substring(name.lastIndexOf(".")+1).toLLowerCase();
    if( fileName != "txt"){
        alert("Please upload .txt file");
        target.value="";
        return
    }
}
</script>

models.py:

class UploadFile(models.Model):
    uploadpath = models.FileField('file',upload_to="/home/")

views.py：

class UploadForm(forms.Form):

    uploadfile = forms.FileField()

def upload():
    if request.method == "POST":
        uf = UploadForm(request.POST,request.FILES)
        if uf.is_valid():
            uploadfile = uf.cleaned_data['uploadfile']
            name = uploadfile.name
            path = '/home/'+name
            exist = os.path.exists(path)
            if exist == false:
                file = UploadFile()
                file.uploadpath = file
                file.save()
                //B
            else:
                noexists = "rm -f"+path
                cmd = os.system(noexists)
                file = UploadFile()
                file.uploadpath - file
                file.save()
                //A
    else:
        uf = UploadForm()
    return render(request,"xx.html",locals())