Question

我尝试在网站上实施树视图，但我现在有一些错误，我无法指出导致它们的原因，也许有人可以解决一些问题关于我的情况。

我已经得到了以下表格：

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 - ＆gt;类别名称
rubrieknummer - ＆gt;类别的id
rubriek - ＆gt;告诉我们一个类别是否属于子类别，如果不是，那么该值将为null

我已经填写了以下测试数据：

    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)

我已经实现了以下代码functions.php：

   <?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;
  }
?>

然后将

functions.php包含在网页上（顶部）。然后我在div中回显以下函数：

 <?php echo CategoryList(); ?>

我收到以下错误：
该页面现在只是继续加载，但屏幕上没有任何内容。

我可能会犯一些新手的错误，但一些帮助会非常感激

Answer 1

变量$ conn在函数外部定义。在函数内部，它是不可见的，即使它是先前定义的。您可以执行以下任一选项：

将$ conn作为参数传递给函数。

function CategoryList（$ conn） { .... }
将$ conn定义为您需要的函数内的全局变量

函数CategoryList（） {
全球$ conn; ...

}

需要一些帮助解决PHP Treeview中的错误（填充SQL）

1 个答案: