通过将PHP数组与MySQL数据进行比较来填充它 - Fill PHP Array by comparing it with MySQL data

Fill PHP Array by comparing it with MySQL data

Guys这是mysql表的输出。它表明month(Value)得到特定疾病的count(Total)。现在我有一个12个月的php array()，我想与该数组进行比较，发现如果特定月份没有从表中返回的数据，则在其他12个月新的array()中添加该月份的0，或者从表中找到的值。

Value | dname      | Total            
5     | Root Canal | 1         
8     | Root Canal | 1

我在php中的代码

foreach($ResultArray2[1] as $key => $val){       
if(empty($row_d['Value'])){
$ResultArray2[$i][$key] = '0';
else
$ResultArray2[$i][$row_d['Value']] = $row_d['Total'];

这就是我得到的

     Array (
[1] => Array
    (
        [1] => 0
        [2] => 0
        [3] => 0
        [4] => 0
        [5] => 0
        [6] => 0
        [7] => 0
        [8] => 0
        [9] => 0
        [10] => 0
        [11] => 0
        [12] => 0
    )
[2] => Array
    (
        [5] => 1
        [8] => 1
    )

)

这就是我需要的阵列（

[1] => Array
    (
        [1] => 0
        [2] => 0
        [3] => 0
        [4] => 0
        [5] => 0
        [6] => 0
        [7] => 0
        [8] => 0
        [9] => 0
        [10] => 0
        [11] => 0
        [12] => 0
    )
[2] => Array
    (
      [1] => 0
        [2] => 0
        [3] => 0
        [4] => 0
        [5] => 1
        [6] => 0
        [7] => 0
        [8] => 1
        [9] => 0
        [10] => 0
        [11] => 0
        [12] => 0
    )

)

如果我理解你的问题是正确的，那么

$finalResult = array();
while($row = mysql_fetch_object($querResult)) {
   if(!is_array($finalResult[$row->dname]) {
      $finalResult[$row->dname] = array_pad(array(), 13, 0);
   } 
   $finalResult[$row->dname][$row->Value] = $row->Total;
}

基本上，代码检查是否已经存在针对特定疾病的数组。如果不是，它会创建一个大小为13的数组[忽略0]，填充值为0，并在移动中更新值。这里疾病成了外阵的关键。如果你愿意，你可以使用从疾病名称到整数ID的映射。在这种情况下，

$finalResult[$map[$row->dname]][$row->value] = $row->Total;

$counts = array_fill(1,12,0);
while ($row = mysql_fetch_assoc($res)){
  $counts[$row['Value']] = $row['Total'];
}

编辑：因为你似乎想得到所有的疾病，所以把整个事情作为一个查询可能更容易，比如：

SELECT name,
SUM(IF(Value=1,1,0) as month1,
SUM(IF(Value=2,1,0) as month2,
SUM(IF(Value=3,1,0) as month3,
...
FROM tablename
GROUP BY name