在一个有缺口的数组中查找总的年份跨度

Question

我需要为这个数组绘制一个堆叠条形图，它应该有2个条形，用2年的空白间隔（从1998年到2000年）

问题：第一条应该是

*1998-1997* - *2 years gap*  - *2000-2008* 

条形图应合并较短的年份跨度，就像从array0中提取2000和从array 1 中提取2008一样

Array
(
    [comp_name] => C++
    [parent_cat_name] => Information Technology
    [sub_cat_name] => Programming
    [total_years] => 6
    [last_year] => 2006
    [start_year] => 2000
)
Array
(
    [comp_name] => .NET
    [parent_cat_name] => Information Technology
    [sub_cat_name] => Programming
    [total_years] => 7
    [last_year] => 2008
    [start_year] => 2001
)
Array
(
    [comp_name] => API
    [parent_cat_name] => Information Technology
    [sub_cat_name] => Programming
    [total_years] => 1
    [last_year] => 1998
    [start_year] => 1997
)

Answer 1

如果两个数组项根据某些条件相邻，并且在其他字段上有一些相等条件，则需要合并它们。

你可以这样做：

for(;;)
{
    $merge = array();
    $n     = count($data);
    for ($i = 0; empty($merge) && $i < $n-1; $i++)
    {
        for ($j = $i+1; empty($merge) &&  $j < $n; $j++)
        {
            // Are $i and $j congruent?
            if ($data[$i]['parent_cat_name'] != $data[$j]['parent_cat_name'])
               continue;
            if ($data[$i]['sub_cat_name'] != $data[$j]['sub_cat_name'])
               continue;
            // Are $i and $j adjacent?
            if ($data[$i]['last_year']+1 == $data[$j]['start_year'])
            {
                $merge = array($i, $j);
                break;
            }
            if ($data[$j]['last_year']+1 == $data[$i]['start_year'])
                $merge = array($j, $i);
                break;
            }
            // They are congruent but not adjacent, try the next
        }
    }
    // If we get to the end and find nothing mergeable, exit.
    if (empty($merge))
        break;
    list($i, $j) = $merge;
    // We add $j to $i
    $data[$i]['last_year'] = $data[$j]['last_year']
    $data[$i]['total_years'] += $data[$j]['total_years']
    // We destroy $j
    unset($data[$j]);
    // Dirty renumber
    $data = array_values($data);
    // Now data has been modified, let's do this again.
}