我有一个返回值的递归函数。
public function getparent ($user)
{
$referrer=''; $pos=0;$this->setFieldNames();
$result=mysql_query("select * FROM ". $this->dbtablename ." WHERE ".$this->id_field . " = '$user'", $this->dbConnectionID);
while($row=mysql_fetch_array($result, MYSQL_ASSOC))
{
$referrer=$row['referrer'];
$export=array();
foreach($this->tablefields as $val){
$export[$val] = $row[$val];
}
}
if($referrer != "0" && $referrer != "")
{
$result2=mysql_query("select * from " . $this->dbtablename . " where " . $this->id_field . " ='$referrer'", $this->dbConnectionID);
while($row2=mysql_fetch_array($result2, MYSQL_ASSOC))
{
$export=array();
foreach($this->tablefields as $val){
$export[$val] = $row2[$val];
}
}
}
$result2=mysql_query("select * from " . $this->dbtablename . " where " . $this->parent_id_field . " ='$referrer' order by id asc", $this->dbConnectionID);
while($row2=mysql_fetch_array($result2, MYSQL_ASSOC))
{
$usernames[]=$row2['username'];
}
$pos=array_search($user, $usernames);
if( $referrer == '0' || $pos >=1 )
{echo 'xxxxxxx';
return $export;
}
else
{
$this->getparent($referrer);
}
}
似乎当$ referrer为零时,它不返回任何值,但它回显'xxxxxxx'。问题是什么!?
答案 0 :(得分:2)
有一个问题:
while($row=mysql_fetch_array($result, MYSQL_ASSOC))
{
$referrer=$row['referrer'];
$export=array();
foreach($this->tablefields as $val){
$export[$val] = $row[$val];
}
}
你在每次迭代时都要替换你的导出数组 - 这是你的意图吗?除最后一行外,所有数据都被丢弃了。然后在这个块中再次执行:
if($referrer != "0" && $referrer != "")
{
$result2=mysql_query("select * from " . $this->dbtablename . " where " . $this->id_field . " ='$referrer'", $this->dbConnectionID);
while($row2=mysql_fetch_array($result2, MYSQL_ASSOC))
{
$export=array();
foreach($this->tablefields as $val){
$export[$val] = $row2[$val];
}
}
}
你在(a)只获得最后一行数据和(b)覆盖第一个数据块的数据(可能需要后者)。
最后,如果你碰到过这个区块(第一次通过你的代码,推荐人不是零):
else
{
$this->getparent($referrer);
}
你永远不会从递归调用中重新计算一个值,它应该是:
else
{
return $this->getparent($referrer);
}
我怀疑是实际问题。但我必须质疑这种递归方法是否最好,而不是迭代方法。
答案 1 :(得分:0)
尝试在while循环之前声明$ export