Question

我使用树状视图，我想对数据进行计数并显示在子项上，但是该父项可以显示子项的总和，依此类推

我有2张桌子：

table_estruc has (id - text - parent_id)
table_resp has (field_id, field_name, id, parent_id)

我可以算出table_resp上table_estuc.id = table_resp.id上的所有记录，甚至算作要显示在父ID中，但我无法实现的是计数侦查并将其显示在父ID上当前的parent_id等。

$sql = "SELECT * FROM `estructura_set`";
$res = mysqli_query($conn, $sql) or die("database error:". mysqli_error($conn));?>

<?php $a = $row["area"];
      $cct = $row["cct"];
      $mid = $row["id"];
      $mparenid = $row["parent_id"];

    $promovidos_tot_query2 = "SELECT COUNT tbl_promotor_parent_id, (promov.tbl_promovido_ID) as promo2 FROM tbl_promovidos promov INNER JOIN tbl_promotores pmt ON pmt.tbl_promotor_ID = promov.tbl_promovido_promotor_ID
    WHERE pmt.tbl_promotor_eid= '$mid'";
    $res2 = mysqli_query($conn, $promovidos_tot_query2) or die("database error:". mysqli_error($conn));
    $total_promov2 = $res2->fetch_assoc();
    $promovidos2 = $total_promov2['promo2'];
    $promovidosp = $total_promov2['tbl_promotor_parent_id'];


$promo = $promovidos2;
$sualternos = $promovidos3;
$grantotal = $promo + $sualternos;

$promo = $promovidos2;
$sualternos = $promovidos3;
$grantotal = $promo + $sualternos;

$row["text"] = $row["text"].' - ('.$row["cct"].') <span class="badge badge-primary"> Promovidos <span class="badge badge-light">'. $promo.'</span></span> 
        <span class="badge badge-primary"> Subalternos <span class="badge badge-light">'. $sualternos .'</span></span>
        <span class="badge badge-primary"> Total <span class="badge badge-light">'. $grantotal .'</span></span>';

$data[] = $row; 
     }

如何计算另一个表中的子数据并显示在父表中？

0 个答案: