Question

我有一个格式错误的文本文件，我想转换为csv。

以下是一个例子：

100910 NA/1-2013-99636 VIA DEI PESCATORI 2/A LODI APR 8 2013 4:24PM DANNEGGIATO -10% 200 2700 0 0 NO
148013 NA/1-2014-146194 CAVALLOTTI SNC LODI GEN 3 2014 3:37PM DANNEGGIATO -10% 0 0 2 0 NO
160032 NA/1-2014-158129 PAOLO GORINI SNC LODI MAG 6 2014 11:51AM DANNEGGIATO -10% 2 0 2 0 NO
54900 NA/1-2014-158070 STRADA VECCHIA CREMONESE SNC LODI MAG 6 2014 9:53AM DANNEGGIATO +10% 10 0 10 0 NO
100910 NA/1-2013-99636 VIA DEI PESCATORI 2/A LODI APR 8 2013 4:24PM DANNEGGIATO -10% 200 2700 0 0 NO
147959 NA/1-2014-146140 DOSSENA SNC LODI GEN 3 2014 10:45AM DANNEGGIATO -10% 200 0 200 0 NO

大致就是这种形式：

[number] [id] [awfully formatted street] ['LODI'] [timestamp] [damaged or not] [percentage] [squaremeters] [squaremeters] [squaremeters] [squaremeters] [asbest-crumbled or not]

我的问题是如何提取第三部分，[格式错误的街道]。基本上它是字符串['LODI']之前的[id]之后的字符串（但['LODI']必须在[timestamp]之前）

我应该用空格爆炸每个行，然后向后遍历数组，超过[timestamp]，超越['LODI']并加入array [id]之前的值，即array [1]？或者是否有一种更聪明（优雅）的方法来做到这一点，也许是使用preg_match（）？

感谢任何提示！

Answer 1

这应该可以从一行中提取地址。

<?php 
$row = "100910 NA/1-2013-99636 VIA DEI PESCATORI 2/A LODI APR 8 2013 4:24PM DANNEGGIATO -10% 200 2700 0 0 NO";
$row_array = preg_split('/\s+/', $row);


array_shift($row_array);
array_shift($row_array);

for($i=0; $i<12; $i++){
    array_pop($row_array);
}

$address = implode(" ", $row_array);

?>

Answer 2

我认为爆炸不会在这里做。我建议使用 regexp 。对于Instance，如果您将 .txt 文件作为一个字符串读取（其中数据字符串用\ n分隔）：

$f = fopen($fname="file.txt", "rt");
$str = fread($f, filesize($fname)));
fclose($f);

然后像这样使用preg_match_all()：

$re = "/^(\\d+)\\s*(.*)(LODI)\\s*(.+(?:AM|PM))\\s*(\\w+)\\s+(-?\\d{1,3}%)\\s+(\\d+)\\s+(\\d+)\\s+(\\d+)\\s+(\\d+)\\s+(\\w+)$/m"; 
preg_match_all($re, $str, $matches,PREG_SET_ORDER );
echo "<pre>\n";
print_r($matches);
echo "</pre>\n";

输出如下：

Array
(
    [0] => Array
        (
            [0] => 100910 NA/1-2013-99636 VIA DEI PESCATORI 2/A LODI APR 8 2013 4:24PM DANNEGGIATO -10% 200 2700 0 0 NO
            [1] => 100910
            [2] => NA/1-2013-99636 VIA DEI PESCATORI 2/A 
            [3] => LODI
            [4] => APR 8 2013 4:24PM
            [5] => DANNEGGIATO
            [6] => -10%
            [7] => 200
            [8] => 2700
            [9] => 0
            [10] => 0
            [11] => NO
        )

    [1] => Array
        (
            [0] => 148013 NA/1-2014-146194 CAVALLOTTI SNC LODI GEN 3 2014 3:37PM DANNEGGIATO -10% 0 0 2 0 NO
            [1] => 148013
            [2] => NA/1-2014-146194 CAVALLOTTI SNC 
            [3] => LODI
            [4] => GEN 3 2014 3:37PM
            [5] => DANNEGGIATO
            [6] => -10%
            [7] => 0
            [8] => 0
            [9] => 2
            [10] => 0
            [11] => NO
     )
..........// And so on

我使用了您在此示例中提供的文字。因此，在输出中，您可以将数据格式化为数组列表。所以你可以用它做任何你想做的事。 $ matches [$ i] [0] - 将存储整个匹配，所以只需跳过它并使用$ matches [$ i] [1] .... $匹配[$ i] [11]作为您的数据。

Answer 3

def ordinal(number)
  abs_number = number.to_i.abs

  if (11..13).include?(abs_number % 100)
    "th"
  else
    case abs_number % 10
      when 1; "st"
      when 2; "nd"
      when 3; "rd"
      else    "th"
    end
  end
end

def ordinalize(number)
  "#{number}#{ordinal(number)}"
end

结果是

<?php
    // read file line by line
    $line = '148013 NA/1-2014-146194 CAVALLOTTI SNC LODI GEN 3 2014 3:37PM DANNEGGIATO -10% 0 0 2 0 NO';

    //start by seperating the string on LODI
    $lodi_split = explode('LODI', $line);

    // Now split the first occ into an array on space
    $bits = explode(' ', $lodi_split[0]);

    $address = '';
    // start reading occurance from occ 2 to loose the first 2 fields
    for ($i=2; $i < count($bits); $i++ ) {
        $address .= $bits[$i] . ' ';
    }
    echo $address . PHP_EOL;

php - 将格式错误的txt转换为csv

3 个答案: