需要一些幫助解決PHP樹視圖中的錯誤（填充SQL）

Question

我試過在網站上實現樹視圖，但我有一些錯誤atm的時刻，我不能指出是什麼導致他們，也許有人可以爲我闡明一些情況。

我有如下因素表：

CREATE TABLE Rubriek(
rubrieknaam  char(50)  not null, /* char(24) */ 
rubrieknummer numeric(38)  not null, /* numeric(3) */ 
rubriek   numeric(38)  null, /* numeric(3) */ 
volgnr   numeric(38)  not null, /* nuemric(2) */ 
constraint pk_rubrieknummer primary key(rubrieknummer), 
constraint fk_rubriek foreign key(rubriek) references Rubriek (rubrieknummer) 
) 

rubrieknaam - >類別的名稱
rubrieknummer - >類的ID
rubriek - >告訴我們，如果一個類是子類，如果不那麼值爲null

我用下面的TESTDATA填充它：

Rubrieknaam   | rubrieknummer | rubriek 
    (= Name of Category) | (=Category ID)| (=Category is a subcategory of the following category): 
    ---------------------------------------------------------------------------------- 
    Cars    | 1    | Null 
    Audio    | 2    | Null 
    Ford    | 3    | 1 (subcategory of Cars) 
    Toyota    | 4    | 1 (subcategory of Cars) 
    Speakers   | 5    | 2 (subcategory of Audio) 
    Microphones   | 6    | 2 (subcategory of Audio) 
    Instruments   | 7    | Null 
    Guitar    | 8    | 7 (subcategory of Instruments) 

我實現下面的代碼的functions.php：然後

<?php 
function connection($sql){ 
ini_set('display_errors',true); 
$serverName = "WHS\sqlexpress"; 
$uid = "sa"; 
$pwd = "projectgroep37"; 
$databaseName = "EenmaalAndermaal"; 

//connection to SQL Database 
$connectionInfo = array("UID"=>$uid, 
"PWD"=>$pwd, 
"Database"=>$databaseName); 

//Connecting with SQL Authentication 
$conn = sqlsrv_connect($serverName, $connectionInfo); 


    $qry = sqlsrv_query($conn, $sql); 
    $rs = sqlsrv_fetch_array($qry, SQLSRV_FETCH_BOTH); 
    return $rs; 
} 
function hasChild($rubriek) 
    { 
    $sql = "SELECT COUNT(*) as count FROM Rubriek WHERE rubriek = ' " . $rubriek . " ' "; 
    $rs = connection($sql); 
    return $rs['count']; 
    } 

    function CategoryTree($list,$parent,$append) 
    { 
    $list = '<li>'.$parent['rubrieknaam'].'</li>'; 

    if (hasChild($parent['rubrieknummer'])) // check if the id has a child 
    { 
     $append++; // this is our basis on what level is the category e.g. (child1,child2,child3) 
     $list .= "<ul class='child child".$append." '>"; 
     $sql = "SELECT * FROM Rubriek WHERE rubriek = ' " . $parent['rubrieknummer'] . " ' "; 

     $child = connection($sql); 
     do{ 
     $list .= CategoryTree($list,$child,$append); 
     }while($child = connection($sql)); 
     $list .= "</ul>"; 
    } 
    return $list; 
    } 
    function CategoryList() 
    { 
    $list = ""; 
    $sql = "SELECT * FROM Rubriek WHERE (rubriek = 0 OR rubriek IS NULL)"; 
    $parent = connection($sql); 
    $mainlist = "<ul class='parent'>"; 
    do{ 
     $mainlist .= CategoryTree($list,$parent,$append = 0); 
    }while($parent = connection($sql)); 
    $list .= "</ul>"; 
    return $mainlist; 
    } 
?> 

的functions.php包含網頁上（頂部）。然後，我贊同在div下面的函數我希望它是：

<?php echo CategoryList(); ?> 

，我發現了以下錯誤：
頁面現在只是不斷加載，但沒有出現在屏幕上。

我可能做一些新手的錯誤，但一些幫助，將不勝感激

Answer 1

變量$ conn是功能之外定義。在函數內部它是不可見的，儘管它在前面已經定義。您可以執行以下任一選項：

• 傳遞$ conn作爲函數的參數。

  功能所屬分類（$康恩） { .... }

• 定義$康恩作爲函數內的全局變量，你需要它

  功能所屬分類（）{

  global $ conn; ...

  }