需要一些帮助来解决PHP树视图中的错误(用SQL填充)
我曾尝试在一个网站上实现一个treeview,但目前我遇到了一些错误,我无法指出是什么导致了这些错误,也许有人可以为我解释一下情况 我有以下表格:需要一些帮助来解决PHP树视图中的错误(用SQL填充),php,sql,treeview,Php,Sql,Treeview,我曾尝试在一个网站上实现一个treeview,但目前我遇到了一些错误,我无法指出是什么导致了这些错误,也许有人可以为我解释一下情况 我有以下表格: CREATE TABLE Rubriek( rubrieknaam char(50) not null, /* char(24) */ rubrieknummer numeric(38) not null, /* numeric(3) */ rubriek numeric(38) null,
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->类别名称rubrieknummer->类别的id
rubriek->告诉我们如果一个类别是一个子类别,如果不是,那么该值将为空 我已经用以下测试数据填充了它:
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 $mainlist;
}
?>
变量$conn在函数外部定义。在函数内部,它是不可见的,即使它是在前面定义的。您可以执行以下任一选项:
- 将$conn作为参数传递给函数
函数类别列表($conn)
{
....
}
- 将$conn定义为所需函数内的全局变量
函数类别列表()
{
全球$conn;
}
这些看起来都是变量范围问题-您需要将变量传递给函数,否则函数中的代码无法“看到”这些变量;但是它给了我一个致命的错误,因为加载时间超过了300秒。这是一个完全不同的问题。这意味着你提到的错误已经解决了。要解决新问题,请检查这是否对您有帮助,请接受答案并为任何不同的问题打开一个新的答案
<?php echo CategoryList(); ?>