如何将上传的图像传递给Flask中的template.html

fra*_*xel 22 python flask

我正在使用flask,并尝试使用快速入门教程做一些非常简单的事情,只需在我的机器(本地服务器)上运行.我生成一个简单的上传表单,成功上传图像文件.然后,我想将此图像作为变量传递template.html给页面中的显示.该template.html文件显示细腻,但图像始终是一个broken link image symbol.我尝试了很多不同的路径,但我觉得我做的事情有点不对劲.

import os
from flask import Flask, request, redirect, url_for, send_from_directory, 
                  render_template

UPLOAD_FOLDER = '/home/me/Desktop/projects/flask/uploads'
ALLOWED_EXTENSIONS = set(['txt', 'pdf', 'png', 'jpg', 'jpeg', 'gif'])

app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER

def allowed_file(filename):
    return '.' in filename and \
           filename.rsplit('.', 1)[1] in ALLOWED_EXTENSIONS

@app.route('/', methods=['GET', 'POST'])
def upload_file():
    if request.method == 'POST':
        file = request.files['file']
        if file and allowed_file(file.filename):
            filename = secure_filename(file.filename)
            file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
            return redirect(url_for('uploaded_file', filename=filename))
    return '''
    <!doctype html>
    <title>Upload new File</title>
    <h1>Upload new File</h1>
    <form action="" method=post enctype=multipart/form-data>
      <p><input type=file name=file>
         <input type=submit value=Upload>
    </form>
    '''

@app.route('/uploads/<filename>')
def uploaded_file(filename):
    filename = 'http://127.0.0.1:5000/uploads/' + filename
    return render_template('template.html', filename = filename)

if __name__ == '__main__':
    app.run()
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这是template.html:

<!doctype html>
<title>Hello from Flask</title>
{% if filename %}
  <h1>some text<img src="{{filename}}"> more text!</h1>
{% else %}
  <h1>no image for whatever reason</h1>
{% endif %}
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如何将上传的图像文件传递给template.html它以便正确显示?

谢谢

Mii*_*kka 21

现在发生的是/uploads/foo.jpg返回template.html中的HTML.在那里你尝试/uploads/foo.jpg用作img标签的来源.无处为您提供实际图像.

让我们像这样修改它:/show/foo.jpg返回HTML页面并/uploads/foo.jpg返回图像.用这两个替换后一个路线你应该好好去:

@app.route('/show/<filename>')
def uploaded_file(filename):
    filename = 'http://127.0.0.1:5000/uploads/' + filename
    return render_template('template.html', filename=filename)

@app.route('/uploads/<filename>')
def send_file(filename):
    return send_from_directory(UPLOAD_FOLDER, filename)
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Tra*_*mer 11

从uploaded_file函数,我们前往template.html,然后将重定向返回<img src="{{ url_for('send_file', filename=filename) }}">,我们点击send_file函数,该函数将显示模板内的HTML内容,其中图像已上传并存储在指定的UPLOAD_FOLDER中.你也在from werkzeug import secure_filenamepy文件中遗漏了这个

@app.route('/show/<filename>')
def uploaded_file(filename):
    return render_template('template.html', filename=filename)

@app.route('/uploads/<filename>')
def send_file(filename):
    return send_from_directory(UPLOAD_FOLDER, filename)
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现在你的template.html看起来像这样..

<!doctype html>
<title>Hello from Flask</title>
{% if filename %}
  <h1>some text <img src="{{ url_for('send_file', filename=filename) }}">more text!</h1>
{% else %}
  <h1>no image for whatever reason</h1>
{% endif %}
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  • 我发现烧瓶手册无法遵循。我已经在 Python 中编码超过 10 年,在编程中已经 30 年了,但从未见过如此骇人听闻的文档。我已经尝试了 3-4 个不同页面的示例,但没有一个语法是正确的。例如,使用同名函数在 Flask 0.12.4 中不起作用。filename 实际上并不代表磁盘上的文件路径,如果使用绝对文件路径,访问文件路径需要第二个路径。这些都不在文档中。数小时的调试和猜测工作。 (2认同)