请看我的代码:
function getShopConfig()
{
$sql = "SELECT sc_name, sc_address, sc_phone, sc_email, sc_shipping_cost, sc_order_email, cy_symbol, sc_currency
FROM kol_shop_config , kol_currency
WHERE sc_currency = cy_id";
$result = dbQuery($sql);
$row = dbFetchAssoc($result);
if ($row) {
//extract($row);
$shopConfig = array('name' => $row['sc_name'],
'address' => $row['sc_address'],
'phone' => $row['sc_phone'],
'email' => $row['sc_email'],
'sendOrderEmail' => $row['sc_order_email'],
'shippingCost' => $row['sc_shipping_cost'],
'currency' => $row['sc_currency']);
}
return $shopConfig;
}
Run Code Online (Sandbox Code Playgroud)
然后我称之为,
<td colspan="4" align="right"><?php getShopConfig(); echo $shopConfig['name'];?></td>
Run Code Online (Sandbox Code Playgroud)
但没有显示任何内容..错误在哪里?请帮忙.
注意:两者都在同一页面中.dbQuery()和dbFetchAssoc()函数是预定义的,之前已正常工作.如果我在函数内部回显它然后只是调用它然后它正常工作.
<?php
$shopConfig = getShopConfig();
echo $shopConfig['name'];
?>
Run Code Online (Sandbox Code Playgroud)
您必须$shopConfig先将返回的数组分配给some()变量.