我正在尝试从MySQL数据库中获取值。我需要获取特定项目的价格,这是我的表格的样子:
我不确定如何解决这个问题,我唯一能想到从表中获取价格的方法是在查询中添加一个条件,它只会抓取与domain_address
匹配的结果。
我通过调用返回<?php echo getPageURL();?>
的函数http://vauxhallpartswarehouse.co.uk/
来进行测试(它应该与数据库条目匹配。
//table which calls function
<td bgcolor="#999999" align="center">Price: <?php echo getPrice();?></td>
<?php
function getPrice() {
$con=mysqli_connect("host","user","pass","db");
$price = mysqli_query($con, "SELECT price FROM domains WHERE domain_address='<?php echo getPageURL();?>'");
mysqli_close($con);
return $price; //should only be 1 result returned since domain_address is unique
}
?>
这似乎不想工作,我真的不知道如何抓住价格。
我知道连接很好,我已经在脚本中进一步编写了这个代码,它创建了一个包含数据库所有结果的表,所以也许我可以用下面的代码做些什么?
<?php
$con=mysqli_connect("host","user","pass","db"));
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$results = mysqli_query($con, "SELECT domain_name, domain_address, price FROM domains");
echo "<table border='1'>
<tr><th>Domain</th>....
</tr>";
while($row = mysqli_fetch_array($results))
{
echo "<tr>";
echo "<td>" . $row['domain_name'] . "</td>";
echo "<td><a target='_blank' href=" . $row['domain_address'] . ">"....
//etc
mysqli_close($con);
?>
答案 0 :(得分:4)
那是因为在这一行:
$price = mysqli_query($con, "SELECT price FROM domains WHERE domain_address='<?php echo getPageURL();?>'");
getPageURL()
未执行,它只是整个字符串的静态部分。试试这个:
$price = mysqli_query($con, "SELECT price FROM domains WHERE domain_address='" . getPageURL() . "'");