如何在不使用幻数的情况下将文件表示为SVG? [英] How can I say a file is SVG without using a magic number?

问题描述

SVG文件基本上是一个XML文件，因此我可以使用字符串<?xml(或十六进制表示形式:'3c 3f 78 6d 6c')作为幻数，但是有一些相反的原因不这样做:例如，有多余的空格可能会破坏此检查.

An SVG file is basically an XML file so I could use the string <?xml (or the hex representation: '3c 3f 78 6d 6c') as a magic number but there are a few opposing reason not to do that if for example there are extra white-spaces it could break this check.

我需要/期望检查的其他图像都是二进制文件，并且具有幻数.如何快速检查文件是否为SVG格式，而最终不使用Python来使用扩展名?

The other images I need/expect to check are all binaries and have magic numbers. How can I fast check if the file is an SVG format without using the extension eventually using Python?

推荐答案

XML不需要以<?xml开头开头，因此测试该前缀不是一种好的检测技术-更不用说它可以识别每个前缀XML作为SVG.一个不错的检测方法而且真的很容易实现，它是使用一个真正的XML解析器来测试该文件是否为格式正确的XML，其中包含svg顶级元素:

XML is not required to start with the <?xml preamble, so testing for that prefix is not a good detection technique — not to mention that it would identify every XML as SVG. A decent detection, and really easy to implement, is to use a real XML parser to test that the file is well-formed XML that contains the svg top-level element:

import xml.etree.cElementTree as et

def is_svg(filename):
    tag = None
    with open(filename, "r") as f:
        try:
            for event, el in et.iterparse(f, ('start',)):
                tag = el.tag
                break
        except et.ParseError:
            pass
    return tag == '{http://www.w3.org/2000/svg}svg'

使用cElementTree可确保通过使用expat进行检测； timeit显示在大约200μs内检测到SVG文件，在35μs内检测到非SVG文件. iterparse API使解析器可以放弃创建整个元素树(尽管具有模块名称)，并且仅读取文档的初始部分，而不管文件的总大小如何.

Using cElementTree ensures that the detection is efficient through the use of expat; timeit shows that an SVG file was detected as such in ~200μs, and a non-SVG in 35μs. The iterparse API enables the parser to forego creating the whole element tree (module name notwithstanding) and only read the initial portion of the document, regardless of total file size.

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