如何在超链接中显示PHP变量？
如何在超链接中显示PHP变量？
提问于 2014-07-04 22:37:10
我试图回显一个变量，以便它显示在我的超链接中。以下是我的当前代码：
<?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，但它只是显示为空白，所以我猜我没有正确地添加一些内容。当我在phpMyAdmin中运行查询时，它确实显示了我想要的内容，所以查询本身应该是正确的。
回答 3
Stack Overflow用户
回答已采纳
发布于 2014-07-04 22:42:31
您需要从查询结果中获取结果，然后使用global访问全局变量。
<?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");
?>Stack Overflow用户
发布于 2014-07-04 22:41:31
我认为您需要获取数据--您刚刚执行了一个查询--从数据库中选择数据，但是没有打印结果，您可以使用以下方法：
$query = mysql_query("SELECT id FROM `tblhosting` WHERE `userid` = '{$_SESSION['uid']}'");
$customerserviceid=mysql_fetch_array($query);Stack Overflow用户
发布于 2014-07-04 22:43:15
mysql_query返回一个资源。您需要从资源中获取数据。你可以这样做。
$result= mysql_query("SELECT id FROM `tblhosting` WHERE `userid` = '{$_SESSION['uid']}'");
$row = mysql_fetch_assoc($result);
$customerserviceid = $row['id'];这应该能行。
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原文链接：
https://stackoverflow.com/questions/24581366
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