從平面PHP SQL表制作JSON樹

Question

從涉及多個JOINS的MySQL查詢中創建代表表的JSON樹的最干凈，最有效的方法是什么？

到目前為止，php數組是通過以下循環創建的：

$rs = mysqli_query($connect, $query);
$arr = array();
while ($row = mysqli_fetch_array($rs, MYSQL_ASSOC)) {
    $arr[] = $row;
}

然后，我可以執行echo $ arr [2] [“ sold”]並獲得“ 2蘇打水”

Answer 1

好吧，這是我對自己問題的解決方案！...為了所有人的利益！ 它正在工作，但是誰能提出更好的更快方法？

    $rs = mysqli_query($connect, $query);
$arr = array();
while ($row = mysqli_fetch_array($rs, MYSQL_ASSOC)) {

    $arr[] = $row;
}
$dateLevel = 0;
$employeeLevel = 0;
$soldLevel = 0;
$tree = array();
$count = count ($arr);
for ($i = 0; $i < $count; $i++){

    $A = $tree[$dateLevel-1];
    if ($A["text"] != $arr[$i]["date"]){
        $tree[$dateLevel]= array("text" => $arr[$i]["date"],  "expanded" => true, items => array());
        $dateLevel++;
        $employeeLevel = 0;
        $soldLevel = 0;
    }

    $A = $tree[$dateLevel-1];
    $B = $A["items"];
    $C = $B[$employeeLevel-1];
    if ($C["text"]  != $arr[$i]["employee"]){
        $tree[$dateLevel-1]["items"][$employeeLevel] = array ("text" => $arr[$i]["employee"],  "expanded" => true, items => array());
        $employeeLevel++;
        $soldLevel = 0;
    }

    $A = $tree[$dateLevel-1];
    $B = $A["items"];
    $C = $B[$employeeLevel-1];
    $D = $A["items"];
    if ($D["text"]  != $arr[$i]["sold"]){
        $tree[$dateLevel-1]["items"][$employeeLevel-1]["items"][$soldLevel] = array ("text" => $arr[$i]["sold"],  "expanded" => true, items => array());
        $soldLevel++;
    }
}

echo json_encode($tree);