php递归查询

Posted 2023-05-09

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他们的关系是主导航婚嫁->婚嫁下面1 ->商家,活动,妹纸.
如何通过递归给你id=1 找到下面的关联id
答案是1,2,3,4,7
大家给个思路或大概代码就好了,谢谢

function myfun($id)

    $sql = "select `path` from `table` ";
    $rs = mysql_query($sql);
    while($rs =mysql_fetch_array($rs))
        $arr[]=explode(\'-\',$rs);

    $sql = "select $id from `table` ";
    $rs = mysql_query($sql);
    foreach($arr as $val)
        if($val==mysql_fetch_array($rs))
            echo $val;


  // 以上不是递归做的以下递归写的
        function myfun($id)

            $sql = "select `pid` from `table` ";
            $rs = mysql_query($sql);
         while($rs =mysql_fetch_array($rs))
            $arr[]=$rs;

        function myfun_2($id)
        if($id<=count($arr))
         foreach($arr as $val)
            if($val!=$id)
                echo $id_1.=myfun_1($id);
            else
                    $id++;

            return $id_1;
         参考技术A function findnum($arr,$pid=0)
    $result=array();
    foreach($arr as $v)
        if($v.['id']==$pid)
            $v.['id']['child']=$v;
            findnum($arr,$v.['id']);
            $result[]=$v;



    return $result;//将处理结果返回

其实你的path很有规律，用‘-’拆分后就是当前id的路径深度以及他的各个父类id

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这能运行?

参考技术B 1 <?php
2 require('includes/application_top.php');
3
4 function search_test($id,&$array)
5
6 $test_query = tep_db_query("select * from new_test where pid='".$id."'");
7 if(tep_db_num_rows($test_query) > 0)
8
9 while($test_array = tep_db_fetch_array($test_query))
10 $array[] = $test_array['id'];
11 search_test($test_array['id'],$array);
12
13
14
15
16 $array = array(1);
17 search_test(1,$array);
18 print_r($array);
19 ?> 参考技术C like '%-1-%' or 其他条件

php与mysql表中如何求递归求和汇总?

如图所示，数据表是递归数据，每个节点是所有下一级节点之和，如何用php&mysql操作数据表，使更新最后一级数据，前一级一级数据自动更新！？

function sumShuzi(&$tree, &$updateData = array())
  $sum = 0;
  // foreach($tree as $key => $item)   //这句话有毒
  foreach($tree as $key => &$item)
    if(isset($item[\'children\']))
      $oldPshuzi = $tree[$key][\'Pshuzi\'];
      $tree[$key][\'Pshuzi\'] = sumShuzi($item[\'children\'], $updateData);
      if($oldPshuzi != $tree[$key][\'Pshuzi\'])
        $updateData[$item[\'id\']] = array($tree[$key][\'Pshuzi\'], $tree[$key][\'Pname_ch\']);


    $sum += $tree[$key][\'Pshuzi\'];


  return $sum;

$tree = json_decode(\'["id":"1","Pid":"0","Pname_ch":"\\u6e20\\u9053\\u90e8","Pshuzi":"1638000","children":["id":"4","Pid":"1","Pname_ch":"\\u9500\\u552e\\u4e8c\\u90e8","Pshuzi":"895000","children":["id":"13","Pid":"4","Pname_ch":"\\u5468\\u7ecf\\u7406","Pshuzi":"28","children":["id":"28","Pid":"13","Pname_ch":"\\u6e56\\u5357","Pshuzi":"158000","id":"35","Pid":"13","Pname_ch":"\\u65b0\\u7586","Pshuzi":"19000"],"id":"40","Pid":"4","Pname_ch":"\\u9648\\u7ecf\\u7406","Pshuzi":"5000"]]]\', true);

//$tree是具有父子关系的数据树
sumShuzi($tree, $updateData);

foreach ($updateData as $id => $item)
  $sql = "update your_table set Pshuzi=$item[0] where id=$id";
  mysqli_query($db, $sql);    //$db是你的数据库连接结果
参考技术A function get_str($id = 0)
global $str;
$sql = "select pname_ch,sum(pshuzi)as z from class where pid= $id group by pname_ch";
$result = mysql_query($sql);
if($result && mysql_affected_rows())
  $str .= '<ul>';
  while ($row = mysql_fetch_array($result))
   $str .= "<li>" . $row['panem_ch'] . "--" . $row['z'] . "</li>";
   get_str($row['id']);

  $str .= '</ul>';

return $str;

给你个函数试试

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