所以我有一个漂亮的小代码集,当我给它一个特定的KittenID但它根本不起作用时,我可以运行它并找到信息,我很难过。哦,太累了,有人能告诉我哪里出了问题吗?是的,我确实有:
<?php
date_default_timezone_set('America/New_York');
//If statements:
//find:
date_default_timezone_set('America/New_York');
if(isset($_POST['Find']))
{
$connection = mysql_connect("ocelot.aul.fiu.edu","userName","password");
// Check connection
if (!$connection)
{
echo "Connection failed: " . mysql_connect_error();
}
else
{
//select a database
$dbName="spr15_xgotz001";
$db_selected = mysql_select_db($dbName, $connection);
//confirm connection to database
if (!$db_selected)
{
die ('Can''t use $dbName : ' . mysql_error());
}
else
{
$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID =<?php$_POST[KittenID]?>;)
while($row = mysql_fetch_array($result))
{
$Name = $row['Name'];
$KittenID = $row['KittenID'];
$KittenAge = $row['KittenAge'];
$Email = $row['Email'];
$Comments = $row['Comments'];
$Gender = $row['Gender'];
$Personality = $row['Personality'];
$Activity = $row['Activity'];
echo $row['Comments'];
}
}
}
mysql_close($connection);
}
?>
使用
$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID = " .$_POST['KittenID']);
而不是
$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID =<?php$_POST[KittenID]?>;)
注意:请将mysqli_用于您未来的项目
您需要拥有更多上下文的特权。你是如何设置$_GET['id']。。它是否真的被存储为$_GET["小猫ID"](例如。https://yoursite.com?view&小猫ID=1)。如果是…
您可以设置一个变量并声明"KittenID"
$kittenid = $_POST['KittenID'];
$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID = $kittenid");
我建议提供更多的上下文。你犯了什么错误?你的参数是什么样子的?
使用
$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID = " .$_SERVER['KittenID']);
而不是
$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID =<?php$_POST[KittenID]?>;)