Question

我有一个表，其数据存储使用邻接列表，如下例所示

id     account parent
1      root    null
2      a1      1
3      b1      1
4      a2      2
5      a3      4
6      a4      2
7      b2      3

我保留这些帐户值的表格

id_account    value
2             10
2             10
4             20
5             30
6             10

我创建了一个函数，该函数返回给定父帐户的所有子帐户：

function getChildrenAccount($parent_id)
  {
    $query = "SELECT id, account FROM accounts WHERE parent='{$parent_id}' ORDER BY account";
    $result = mysql_query($query) or die(mysql_error());
    while($r[]=mysql_fetch_assoc($result));
    return $r;
  }

我正在尝试做的是一个函数，它不仅返回子帐户，还返回所有值的总和，包括每个结果的子项。例如

getChildrenAccount(4)

将返回具有以下语法的数组

array(1) { 
  [0]=> array(3) { 
     ["id"]=> 5
     ["account"]=> "a3"
     ["sum"]=> 50 //a2 + a3
}

并且

getChildrenAccount(2)

array(2) { 
  [0]=> array(3) { 
     ["id"]=> 4
     ["account"]=> "a2"
     ["sum"]=> 70 //a1 + a2 + a3
  [1]=> array(3) { 
     ["id"]=> 6
     ["account"]=> "a4"
     ["sum"]=> 30 //a1 + a4
}

我想我必须在我的while语句中使用某种递归但我有点困惑。你能帮帮我吗？

由于

Answer 1

function getChildrenAccount($accountID){

   $query = ' SELECT id,account,sum(value) as `SUM`
    FROM accounts,accountsValues
    WHERE accounts.id = accountsValues.id_accounts
    AND id = $accountID 
    OR id IN (SELECT id FROM accounts where parent = $accountID) ';
....
}

Answer 2

您需要迭代结果，然后为每个id调用getChildrenAccount作为其父ID。

Answer 3

我找到了一个在没有递归查询的情况下根据需要获得结果的函数。

$nodeList = array();
$tree     = array();
$query = mysql_query("SELECT A.guid, A.name, A.parent_guid, SUM( S.value_num ) /100 AS suma FROM accounts AS A
    LEFT JOIN splits AS S ON S.account_guid = A.guid GROUP BY A.guid ORDER BY A.name");
while($row = mysql_fetch_assoc($query))
  $nodeList[$row['guid']] = array_merge($row, array('children' => array()));
mysql_free_result($query);

foreach ($nodeList as $nodeId => &$node) {
  if (!$node['parent_guid'] || !array_key_exists($node['parent_guid'], $nodeList))
    $tree[] = &$node;
  else 
    $nodeList[$node['parent_guid']]['children'][] = &$node;
}
unset($node);
unset($nodeList);

查询中的递归和

3 个答案: