使用Mechanize使用Python发送POST参数 [英] Sending POST parameters with Python using Mechanize

问题描述

我想使用Python填写此表单:

I want to fill out this form using Python:

    <form method="post" enctype="multipart/form-data" id="uploadimage">
  <input type="file" name="image" id="image" />
  <input type="submit" name="button" id="button" value="Upload File" class="inputbuttons" />
  <input name="newimage" type="hidden" id="image" value="1" />
  <input name="path" type="hidden" id="imagepath" value="/var/www/httpdocs/images/" />
</form>

如您所见，有两个名称完全相同的参数，因此，当我使用Mechanize进行操作时，将如下所示:

As you can see, there are two Parameters that are named exactly the same, so when I'm using Mechanize to do it, what would look like this:

    import mechanize
    br = mechanize.Browser()
    br.open('www.site.tld/upload.php')
    br.select_form(nr=0)

    br.form['image'] = '/home/user/Desktop/image.jpg'
    br.submit()

我收到错误消息:

mechanize._form.AmbiguityError: more than one control matching name 'image'

我在Internet(包括此站点)上找到的所有解决方案均无效.有其他方法吗? 遗憾的是，无法重命名HTML表单中的输入.

Every solution I found in the Internet (including this site) didn't work. Is there a different approach? Renaming the input in the HTML form is sadly not an option.

预先感谢.

推荐答案

您应该改用find_control；如果存在歧义，可以添加nr关键字以选择特定的控件.对于您来说，name和type关键字应该可以.

You should use find_control instead; you can add a nr keyword to select a specific control if there is ambiguity. In your case, the name and type keywords should do.

还请注意，文件控件不使用value；改用add_file并传入一个打开的文件对象:

Also note that a file control doesn't take a value; use add_file instead and pass in an open file object:

br.form.find_control(name='image', type='file').add_file(
    open('/home/user/Desktop/image.jpg', 'rb'), 'image/jpg', 'image.jpg')

请参见有关机械化表单的文档.

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