我想在'$ class'變量上使用!isset來查看它是否有值,然後將mysql_query函數作爲基礎。但這是不行的。看到有什麼不對?如果語句在mysql查詢
<?php session_start();
$heyyou = $_SESSION['usern'];
$points = $_SESSION['points'];
$school = $_SESSION['school'];
$class = $_POST['class'];
$prof = $_POST['prof'];
$date = $_POST['dater'];
$fname = $_FILES['fileToUpload']["name"];
?>
<div id='contenttext' class='contenttext'>
<?php
@mysql_select_db($database) or die("Unable to select database");
$query = "INSERT INTO uploadedfiles (usename, filename, date, teacher, class) VALUES ('$heyyou', '$fname', '$date', '$prof', '$class')";
if (!isset($class)){
echo 'You need to pick a class for the content'; }
else{
mysql_query($query); }
mysql_close();
?>
<?php
if (($_FILES["fileToUpload"]["type"] == "image/gif" || $_FILES["fileToUpload"]["type"] == "image/jpeg" || $_FILES["fileToUpload"]["type"] == "image/png") && $_FILES["fileToUpload"]["size"] < 10000000)
{
move_uploaded_file($_FILES["fileToUpload"]["tmp_name"],
"upload/" . $_FILES["fileToUpload"]["name"]);
echo "Your file has successfully been uploaded, and is awaiting moderator approval for points." . "<html><br><a href='uploadfile.php'>Upload more.</a>";
}
else
{
echo "Files must be either JPEG, GIF, or PNG and less than 10,000 kb";
}
?>
</div>
</body>
</html>
最新問題? probs最好不要發佈您的數據庫細節了;) – 2011-03-21 18:26:37