Question

我遇到一个小问题，我想将selenium driver和PhantomJS拍摄的（截图）图像上传到我的网站并将其放入数据库。

我正在使用get_screenshot_as_file()函数将图像保存到当前目录中，如果成功完成了此过程，此函数只会给我True或False值，但是我需要的是文件名，所以我可以将它保存在数据库中。

以下是我的观点和模型摘要：

models.py

class Driver(PhantomJS):
    def __init__(self, *args, **kwargs):
        super(Driver, self).__init__(*args, **kwargs)
        self._log = StderrLog()

class ScreenShot(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    image = db.Column(db.String(255), default=None)
    title = db.Column(db.String(80))
    describe = db.Column(db.Text)

    def __init__(self, image, title, describe):
        self.image = image
        self.title = title
        self.describe = describe

    def __repr__(self):
        return "Title %r" %self.title

views.py

from blog.form import ScreenShotForm
from blog.models import ScreenShot, StderrLog, Driver

@app.route('/my_works/', methods=('GET','POST'))
@only_author
def my_works():
    blog = Blog.query.first()
    form = ScreenShotForm()

    if form.validate_on_submit():

        url = request.args.get(url_for('index'), form.title.data)
        width = int(request.args.get("w", 1200))
        min_height = int(request.args.get("h", 400))
        wait_time = float(request.args.get("t", 20)) / 1000  # ms

        driver = Driver()
        driver.set_window_position(0, 0)
        driver.set_window_size(width, min_height)

        driver.set_page_load_timeout(100)
        driver.implicitly_wait(100)
        driver.get(url)

        driver.set_window_size(width, min_height)
        time.sleep(wait_time)

        sys.stderr.write(driver.execute_script("return document.readyState") + "\n")

        image = driver.get_screenshot_as_file(os.urandom(3).encode('hex')+'.png')

        title = form.title.data
        describe = form.describe.data

        scrshot = ScreenShot(image, title, describe)
        db.session.add(scrshot)
        db.session.commit()
        driver.quit()
        flash('Screenshot added.')
        return redirect(url_for('index'))
        #driver.get_screenshot_as_file('foo.png')
        #return "<h1>Image saved</h1>"
        #return Response(png, mimetype="image/png")
    return render_template('blog/my_works.html', blog=blog, form=form)

form.py

class ScreenShotForm(Form):
    title = StringField('Site',[validators.Required(), validators.length(max=80)])
    describe = TextAreaField('Describtion', validators=[validators.Required()])

正如您所看到的，我正在使用os.urandom(3).encode('hex')+'.png'因此它可以为我的屏幕截图提供一个随机名称，这样我就不会感到困惑，在数据库中我得到1表示True，也在我的项目文件夹中我完全得到截图25de0a.png文件，听起来很棒!!!!，但我不希望这种情况发生，我想获取文件名而不是True或False表达。

这是我在数据库中获得的代码片段：

+----+-------+-------------------------+-------------------+
| id | image | title                   | describe          |
+----+-------+-------------------------+-------------------+
|  6 | 0     | http://www.youtube.com/ | testing testing . |
|  7 | 1     | http://www.youtube.com/ | asdasfasfwf324214 |
+----+-------+-------------------------+-------------------+

我尝试了get_screenshot_as_png()功能，我得到的是图像的源代码。

如果结果如下，那就太好了：

+----+-------+-------------------------+-------------------+
| id | image | title                   | describe          |
+----+-------+-------------------------+-------------------+
|  6 | 25de0a.png     | http://www.youtube.com/ | testing testing . |
|  7 | 25de0a.png     | http://www.youtube.com/ | asdasfasfwf324214 |
+----+-------+-------------------------+-------------------+

最终，请任何帮助吨吨吨赞赏:)。

Answer 1

您将所需的文件名传递给get_screenshot_as_file方法（正如您现在所做的那样）。如果文件成功写入，get_screenshot_as_file方法返回True，否则返回False。

您的代码应该是：

image_filename = os.urandom(3).encode('hex')+'.png' 

if driver.get_screenshot_as_file(image_filename):
    # file was successfully written
    title = form.title.data
    describe = form.describe.data

    scrshot = ScreenShot(image_filename, title, describe)
    db.session.add(scrshot)
    db.session.commit()
else:
    # file was not written
    pass

如何将png文件保存到数据库，FLASK？

1 个答案: