塊1PHP的錯誤
<?php
// Filter our input.
$dID = filter_input(INPUT_GET, 'dID', FILTER_SANITIZE_NUMBER_INT);
if(!$dID) {
echo "<h2 style='color:red;'>Invalid Department</h2>";
exit;
}
$username = "###";
$password = "####";
$pdo = new PDO('mysql:host=localhost;dbname=####', $username, $password);
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sth = $pdo->prepare('
$某物= $ pdo->準備(」 SELECT 名,FNAME,LNAME,picpath,電子郵件 FROM系教授 WHERE Department.dID =? '); '); $ sth-> execute(array( $ dID ));
?>
塊2
<?php
echo "<span>{$row['name']}";
// Did we get any professors in this dept?
while($row2 = $sth->fetch(PDO::FETCH_ASSOC)) {
echo "<span>{$row['fname']} | {$row['lname']} | </span>"
; // echo
}
unset($sth);
?>
輸出給定:
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DEPT表:
表教授:
它似乎不喜歡我的查詢,不知道爲什麼,但它不從數據庫中拉東西。它可能是一個SQL錯誤?我很爲難
當我在phpMyAdmin測試它給了我:
更新的查詢: SELECT名字,FNAME,LNAME,picpath,電子郵件 FROM系教授 WHERE Department.dID = Professor.dID AND Department.dID =?
雖然它仍然沒有工作。基本上這個頁面例如在url中:'/dept.php?dID=30'它抓取了dID,並且應該顯示部門中的所有教授。任何人???
嘗試使用'WHERE Department.dID =?'而不是 – yoavmatchulsky 2011-04-02 14:11:34
什麼的錯誤代碼說? – BugFinder 2011-04-02 14:13:19
@yoavmatchulsky - 是的,我嘗試過。我的查詢現在已更新。 – Jshee 2011-04-02 14:16:03