Question

我有很多这样的记录：

2011-01-01, a, 1
2011-01-01, c, 5
2011-01-01, d, 3
2011-01-02, a, ...

第一个值是日期，第二个值是字符串（可以是a或b或c或d），第三个是数字。

如何在PHP中将所有记录添加到数组中，并且数组的结构如下：

array('2011-01-01' => array('a' => '1','b' => '0','c' => '5','d' => '3'), '2011-01-02' => ... ,)

这样日期变量是一个索引，而键是一个四元素数组，每个元素都是相应的记录号（第三个值）？

Answer 1

假设这是一个大字符串，将它分成行，然后在逗号上。对于每一行，测试日期是否已作为键存在，如果存在，则添加到它，否则，创建它：

$string = "2011-01-01, a, 1
2011-01-01, c, 5
2011-01-01, d, 3
2011-01-02, a, 1";

// Array to hold it all...
$array = array();
$lines = explode("\n", $string);
foreach ($lines as $line) {
  // Explode the line on the comma into 3 variables
  list($date, $key, $num) = explode(",", $line);
  $date = trim($date);
  $key = trim($key);
  $num = intval($num);

  // Create an array key for the current $date if it doesn't exist
  // as a new array
  if (!isset($array[$date])) $array[$date] = array();

  // And assign the number with the corresponding letter key to the array
  // for this date.
  $array[$date][$key] = $num;
}

Answer 2

$records = array(
    0 => array('2011-01-01', 'a', '1'),
    1 => array('2011-01-01', 'c', '5'),
    2 => array('2011-01-01', 'd', '3'),
    3 => array('2012-01-01', 'a', '1')
);

$tmp_arr   = array();
$tmp_dates = array();
$tmp_str   = '';

foreach ($records as $record) {
    $tmp_arr[$record[0]][$record[1]] = $record[2];
}

foreach ($tmp_arr as $date => $record) {
    $tmp_params = array();
    foreach ($record as $key => $val) {
        $tmp_params[] = $key . ':' . $val;
    }
    $tmp_dates[] = $date . ':' . '{' . implode(',', $tmp_params) . '}';
}

$tmp_str = '{' . implode(',', $tmp_dates) . '}';

echo $tmp_str; // {2011-01-01:{a:1,c:5,d:3},2012-01-01:{a:1}}

Answer 3

这个怎么样（假设输入数据的格式）：

$input = array(
    array('2011-01-01', 'a', '1'),
    array('2011-01-01', 'c', '5'),
    array('2011-01-01', 'd', '3'),
    array('2012-01-01', 'a', '1')
);

// used to populate the default strings (a b c d)
$strings = range('a', 'd');
$stringDefaults = array_combine($strings, array_fill(0, count($strings), '0'));

$output = array();

foreach ($input as $row) {
    list ($date, $string, $number) = $row;

    if (!isset($output[$date])) {
        $output[$date] = $stringDefaults;
    }

    $output[$date][$string] = $number;
}

echo json_encode($output);

Answer 4

如果您的数据位于文本文件（.txt）中：

$records = file_get_contents("filename.txt");

// $records will be equal to the following: (on windows, it would be \r\n instead of just \n)
// $records = "2011-01-01, a, 1\n2011-01-01, c, 5\n2011-01-01, d, 3\n 2011-01-02, d, 3";

$records = explode("\n", $records);

$output = array();

for ($i = 0; $i < count($records); $i++) {

    $rec = explode(",", $records[$i]);
    $key = $rec[0];

    if ($output[$key] == null)
        $output[$key] = array();

    if (!array_key_exists("a", $output[$key]))
        $output[$key]["a"] = 0;

    if (!array_key_exists("b", $output[$key]))
        $output[$key]["b"] = 0;

    if (!array_key_exists("c", $output[$key]))
        $output[$key]["c"] = 0;

    if (!array_key_exists("d", $output[$key]))
        $output[$key]["d"] = 0;

    $elem = trim($rec[1]);
    $value = trim($rec[2]);
    $output[$key][$elem] = intval($value);

}

var_dump($output);

Answer 5

好的，这就是你做的。

$records = array(
    0 => array('2011-01-01', 'a', '1'),
    1 => array('2011-01-01', 'c', '5'),
    2 => array('2011-01-01', 'd', '3'),
    3 => array('2012-01-01', 'a', '1')
);

$dateArray = array();
$baseArray = array('a' => '0','b' => '0','c' => '0','d' = '0');

foreach($records as $record)
{
     $dateArray[$record[0]][$record[1]] = $record[2];  //You could actually do this while pulling in the info from the DB
}

foreach($dateArray as $date => $array)
{
     $missingArray = array_diff_key($array,$baseArray);  //Determine what keys are missing using the base array, get keys and default values
     $dateArray[$date] = array_merge($array,$missingArray);  //Merge the missing parts with the current array
}

请注意，部件会从其他帖子中复制一段时间。

PHP数组作为元素，变量作为索引

5 个答案: