Question

I m trying to save files to a directory after scraping them from the web using scrapy. I m extracting a date from the file and using that as the file name. The problem I m running into, however, is that some files have the same date, i.e. there are two files that would take the name "June 2, 2009". So, what I m looking to do is somehow check whether there is already a file with the same name, and if so, name it something like "June 2, 2009.1" or some such.

使用Im的代码如下:

def parse_item(self, response):
    self.log( Hi, this is an item page! %s  % response.url) 

    response = response.replace(body=response.body.replace( <br /> ,  
 ))

    hxs = HtmlXPathSelector(response)

    date = hxs.select("//div[@id= content ]").extract()[0]
    dateStrip = re.search(r"([A-Z]*|[A-z][a-z]+)sd*d,s[0-9]+", date) 
    newDate = dateStrip.group()


    content = hxs.select("//div[@id= content ]") 
    content = content.select( string() ).extract()[0]

    filename = ("/path/to/a/folder/ %s.txt") % (newDate) 


    with codecs.open(filename,  w , encoding= utf-8 ) as output:
        output.write(content)

Answer 1

另一项答复指出,我方向正确,在平线检查了变形工具,但我认为,我发现的方法可能更直截了当。参看。我如何检查是否有多份使用沙尔吗? 。

以下是我提出的法典:

    existence = os.path.isfile(filename)

    if existence == False:
        with codecs.open(filename,  w , encoding= utf-8 ) as output:
            output.write(content)
    else:
        newFilename = ("/path/.../.../- " +  %s  ".1.txt") % (newDate)
        with codecs.open(newFilename,  w , encoding= utf-8 ) as output:
            output.write(content)

Edited to Add:

我不喜欢这一解决办法,认为其他解决办法可能更好,但却不奏效。我对我的解决方案持类似看法的主要部分是,它只处理2份同名档案;如果3或4份档案的名称与最初的问题相同的话。以下是我介绍的:

filename = ("/Users/path/" + " " + "title " +  %s  + " " + "-1.txt") % (date) 
filename = str(filename)

    while True:
        os.path.isfile(filename)
        newName = filename.replace(".txt", "", filename)
        newName = str.split(newName)
        newName[-1] = str(int(newName[-1]) + 1)
        filename = " ".join(newName) + ".txt"
        if os.path.isfile(filename) == False:
            with codecs.open(filename,  w , encoding= utf-8 ) as output:
                output.write(texts)
            break

这可能是最棘手的问题,可能是一种黑客主义做法,但迄今为止,这种做法已经奏效,似乎已经解决了我的问题。

Answer 2

你们可以使用os。收到现有档案清单并分配不会引起冲突的档案名称。

import os
def get_file_store_name(path, fname):
    count = 0
    for f in os.listdir(path):
        if fname in f:
            count += 1
    return os.path.join(path, fname+str(count))

# This is example to use 
print get_file_store_name(".", "README")+".txt"

Answer 3

检查C图书馆档案的通常方式是:>>。灰尘以<代码>os.stat(<>/code>的形式,围绕这一功能提供微薄的包装。我建议你这样做。

rel=“nofollow>http://docs.python.org/library/stat.html

def file_exists(fname):
    try:
        stat_info = os.stat(fname)
        if os.S_ISREG(stat_info): # true for regular file
            return True
    except Exception:
        pass
    return False

Answer 4

one other solution is you can append time with date, for naming file like

from datetime import datetime

filename = ("/path/to/a/folder/ %s_%s.txt") % (newDate,datetime.now().strftime("%H%M%S"))

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