繁体   English   中英

带换行符的Python字符串格式化

[英]Python formatting of strings with line breaks

我正在尝试很好地格式化数据库中的一些文本数据,以便在Django中发送电子邮件。 我遇到带换行符的字符串问题。

特别是,我想要的输出是:

1)  t1                     t2
    t1_continue            t2_end
    t1_end

2)  s1                     s3
    s1_continue
    s1_end

这里从数据库获取的字符串是“ t1 \\ nt1_continue \\ nt1_end”,“ t2 \\ nt2_end”,“ s1 \\ ns1_continue \\ ns1_end”和“ s3”。

这个期望的输出就像我们在Excel中有3列时看到的那样。

我想说的是,有些字符串包含换行符,所以格式很简单,例如:

print str(index) + '\t' + col1 + '\t' + col2 

将无法正常工作。

请与我分享您的方法。

非常感谢。

\\t的显示方式因操作系统,应用程序等而异。 对于真正的字符串格式,我建议您检查docs 例如,

print('{0})  {1:<25} {2}'.format(index,col1,col2))

假设col1不能大于24个字符,将执行您想要的操作。 如果是这样,您可以随时根据自己的喜好调整该值。

您可以使用<table>在电子邮件正文中显示表格数据。

print每个表行的循环之前,请print表头:

print('<table>')

对于每个数据行,将其格式化为:

print('<tr><td>{0}</td><td>{1}</td><td>{2}</td></tr>'.format(index, col1, col2))   

最后,在print每个表格行的循环之后, print表格页脚:

print('</table>')

我以为我可以在几分钟之内将某些东西混在一起,但是将文本格式化为列比我想象的要困难得多。 然而,这是我到目前为止所得到的,仍然是越野车...

class format_lines(object):
    """
    1)  some text goes      2nd column of text
        here. But not       its not really that
        all text goes       hard to add more text
        in this column      in a given format

    2)  another point
        for text
    """

    def __init__(self, *args, **kargs):
        self.max_char_width = 30
        self.gutter_width = 15
        self.gutter = ' '*self.gutter_width
        self.previous_line1 = None
        self.previous_line2 = None
        self.p1_index = 0
        self.p2_index = 0
        print args
        print "len args = ", len(args)
        if len(args) == 2:
            print "Starting the test"
            self.print_lines_(args[0], args[1])

    def print_lines_(self, p1, p2):
        while self.p1_index < len(p1.split()): 
            this = self.format_line_(p1, p2)
            p1=this['p1']
            p2=this['p2']
            print this['line']

    #problem with this format is it's going to messup words by
    #pushing them to the next line (this could be fixed with a -
    #or this function could be rewritten
    def format_line_(self, p1, p2):
        #must first figure out amount of words that can fit on a line
        p1_words = [""]
        p2_words = [""]
        if p1:
            p1_words = p1.split(' ')
            p1_words = p1_words[self.p1_index:]
        if p2:
            p2_words = p2.split(' ')
            p2_words = p2_words[self.p2_index:]
        #format left side's output
        loutput = p1_words[0]
        if len(loutput) < self.max_char_width:
            for i, word in enumerate(p1_words[1:]):
                if (len(loutput) + len(word)) <= self.max_char_width:
                    loutput += (' {0}'.format(word))
                    self.p1_index = i
                else:
                    break
            self.p1_index+=1 #for loop iteration starts at index 1 not 0 so
                        #a +1 is required
        elif (len(loutput) > self.max_char_width):
            long_word = []
            long_word.append(loutput[:len(loutput)/2])
            long_word.append(loutput[len(loutput)/2:])
            long_word[0]+='-'
            p1_words[0]=long_word[0]
            p1_words.insert(1, long_word[1])
            p1 = ' '.join(p1_words)
        else:
            #the left output is a single word
            #equal to the max_char_width
            pass
        #need to add 1 to the index, because at least 1 element in the list is
        #going to be printed
        self.p1_index+=1
        #format right side's output
        routput = p2_words[0]
        if len(routput) < self.max_char_width:
            for word in p2_words[1:]:
                if (len(routput) + len(word)) <= self.max_char_width:
                    routput += (' {0}'.format(word))
                else:
                    break
            self.p2_index+=1
        elif len(routput) > self.max_char_width:
            long_word = []
            long_word.append(routput[:len(routput)/2])
            long_word.append(routput[len(routput)/2:])
            long_word[0]+='-'
            p2_words[0]=long_word[0]
            p2_words.insert(1, long_word[1])
            p2 = ' '.join(p2_words)
        else:
            #the right output is a single word
            #equal to the max_char_width
            pass
        self.p2_index+=1

        #determin gutter size
        if len(loutput) < self.max_char_width:
            loutput+=(' '*(self.max_char_width-len(loutput)))
        #do not need to append spaces to the right hand size

        output = ''
        #if previous_line1 is not defined then there should be no way that previous_line2 is defined
        if not self.previous_line1:
            #must be the first line of a bullet point
            output = '1){0}{1}{2}{3}'.format(self.gutter,
                                             loutput,
                                             self.gutter,
                                             routput)
            self.previous_line1 = loutput
            self.previous_line2 = routput
        else:
            p1_width = len(self.previous_line1)
            p2_width = len(self.previous_line2)
            if loutput<p1_width:
                loutput+=(' '*(len(self.previous_line1)-p1_width))
            if routput<p2_width:
                routput+=(' '*(len(self.previous_line2)-p2_width))
            output = '  {0}{1}{2}{3}'.format(self.gutter,
                                             loutput,
                                             self.gutter,
                                             routput)
        ret_val = {'line':output, 'p1':(' '.join(p1_words)), 'p2':(' '.join(p2_words))}
        return ret_val

if __name__ == '__main__':
    t1 = "this is a very long line of text much larger than you can fit on one line"
    t2 = "this is another long line, only this line is a little different from the first line"
    test = format_lines(t2, t1)

我想我明白您的要求。 您想在列中自动换行,就像在单元格中有一段文本时,Excel一样。 我认为这个食谱可以帮助您入门; 它似乎可以满足您的所有需求。

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM