[英]How to show PHP variable in hyperlink?
我试图回显一个变量,以便它显示在我的超链接中。 这是我当前的代码:
<?php
if (!defined("WHMCS"))
die("This file cannot be accessed directly");
$customerserviceid = mysql_query("SELECT id FROM `tblhosting` WHERE `userid` = '{$_SESSION['uid']}'");
function limitOrders($vars) {
if(mysql_num_rows(mysql_query("SELECT packageid FROM `tblhosting` WHERE `userid` = '{$_SESSION['uid']}'")) > 0) {
if($packageid = '1' || $packageid = '2' || $packageid = '3' || $packageid = '4' || $packageid = '5' || $packageid = '6' || $packageid = '7' || $packageid = '8' || $packageid = '9' || $packageid = '10') {
global $errormessage;
$errormessage = "<li>It looks like you already have an account! Please <a href='http://mywebsite.com/upgrade.php?type=package&id=$customerserviceid'>click here</a> to upgrade or downgrade your account.</li>";
}
}
}
add_hook("ShoppingCartValidateCheckout", 1, "limitOrders");
?>
我尝试将$customerserviceid
添加到第14行的URL中,但是它仅显示为空白,因此我猜测我没有正确添加一些内容。 当我在phpMyAdmin中运行查询时,它确实显示了我想要的内容,因此查询本身应该是正确的...
您需要从查询结果中获取结果,然后使用global
访问global变量。
<?php
if (!defined("WHMCS"))
die("This file cannot be accessed directly");
$result = mysql_query("SELECT id FROM `tblhosting` WHERE `userid` = '{$_SESSION['uid']}'");
$row = mysql_fetch_assoc($result);
$customerserviceid = $row['id'];
function limitOrders($vars) {
global $customerserviceid;
if(mysql_num_rows(mysql_query("SELECT packageid FROM `tblhosting` WHERE `userid` = '{$_SESSION['uid']}'")) > 0) {
if($packageid = '1' || $packageid = '2' || $packageid = '3' || $packageid = '4' || $packageid = '5' || $packageid = '6' || $packageid = '7' || $packageid = '8' || $packageid = '9' || $packageid = '10') {
global $errormessage;
$errormessage = "<li>It looks like you already have an account! Please <a href='http://mywebsite.com/upgrade.php?type=package&id=$customerserviceid'>click here</a> to upgrade or downgrade your account.</li>";
}
}
}
add_hook("ShoppingCartValidateCheckout", 1, "limitOrders");
?>
我认为您需要获取刚刚执行查询以从数据库中选择数据的数据,但未打印结果,可以使用以下方法:
$query = mysql_query("SELECT id FROM `tblhosting` WHERE `userid` = '{$_SESSION['uid']}'");
$customerserviceid=mysql_fetch_array($query);
mysql_query返回资源。 您需要从资源中获取数据。 您可以这样做。
$result= mysql_query("SELECT id FROM `tblhosting` WHERE `userid` = '{$_SESSION['uid']}'");
$row = mysql_fetch_assoc($result);
$customerserviceid = $row['id'];
这应该工作。
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