表單正在向數據庫提交多個條目，而不是單個
[英]Form is submitting multiple entries to the database instead of single
問題描述
我的代碼基本上會覆蓋學生向特定教授提交的請求（針對全容量課程）。 假設有兩個學生要求對同一個班級進行覆蓋，當一位教授登錄時，代碼會提取兩個覆蓋要求，並帶有接受/拒絕的選項，當我作為教授接受/拒絕一個覆蓋要求時，它會這兩個替代請求的操作，而不是我選擇的操作。
基本上，它不接受/拒絕選定的請求，它對所有替代都執行相同的操作。
碼：
<?php
} else if ($usertype == 1) {
$server = "";
$user = "";
$pass = "";
$db = "";
$db2 = "";
$db3 = "";
$user1 = $_SESSION['username'];
$mysqli = new Mysqli($server, $user, $pass, $db) or mysqli_error($mysqli);
$mysqli2 = new Mysqli($server, $user, $pass, $db2) or mysqli_error($mysqli);
$mysqli3 = new Mysqli($server, $user, $pass, $db3) or mysqli_error($mysqli);
$status= $mysqli->query("SELECT status FROM Overrides WHERE professor = '$user1'")->fetch_object()->status;
$overrides = $mysqli->query("SELECT * FROM Overrides WHERE professor = '$user1'");
$num_rows = mysqli_num_rows($overrides);
?>
<form method="post" action="dbheads.php" name="HF" id="HF" autocomplete="off">
<script type="text/javascript">
function submitForm(action)
{
document.getElementById('HF').action = action;
document.getElementById('HF').submit();
}
</script>
<?php if ($status == 1) {
echo " Overrides today: " . $num_rows;
?>
<?php
while($row = mysqli_fetch_array($overrides)) { ?>
<fieldset> <?php
echo "First Name: " . $row['name'] . "<br />";
echo "<br />Mid. Name: " . $row['mname'] . "<br />";
echo "<br />Fam. Name: " . $row['fname'] . "<br />";
echo "<br />Student ID: " . $row['sid'] . "<br />";
echo "<br />Scolarship: " . $row['sc'] . "<br />";
echo "<br />Phone No: " . $row['phone'] . "<br />";
echo "<br />Email: " . $row['email'] . "<br />";
echo "<br />Subject: " . $row['subject'] . "<br />";
echo "<br />Section: " . $row['section'] . "<br />";
echo "<br />Semester: " . $row['semester'] . "<br />";
$name = $row['name'];
echo "<input type='hidden' name='name' value='$name'>";
$mname = $row['mname'];
echo "<input type='hidden' name='mname' value='$mname'>";
$fname = $row['fname'];
echo "<input type='hidden' name='fname' value='$fname'>";
$sid = $row['sid'];
echo "<input type='hidden' name='sid' value='$sid'>";
$sc = $row['sc'];
echo "<input type='hidden' name='sc' value='$sc'>";
$phone = $row['phone'];
echo "<input type='hidden' name='phone' value='$phone'>";
$email = $row['email'];
echo "<input type='hidden' name='email' value='$email'>";
$subject = $row['subject'];
echo "<input type='hidden' name='subject' value='$subject'>";
$section = $row['section'];
echo "<input type='hidden' name='section' value='$section'>";
$semester = $row['semester'];
echo "<input type='hidden' name='semester' value='$semester'>";
?>
<br />
<div>
<label for="comments" accesskey="c">Notes & Comments:</label><br />
<input type="textarea" name="comments" id="comments" cols="35" rows="10">
<br>
</div>
<br>
<script type="text/javascript">
function submitForm(action)
{
document.getElementById('HF').action = action;
document.getElementById('HF').submit();
}
</script>
...
<input type="button" onclick="submitForm('dbheads.php')" value="Accept" />
<input type="button" onclick="submitForm('dbheads2.php')" value="Deny" /></form>
</fieldset>
<br>
<?php } }
?>
<br />
dbheads.php
<?php
include_once 'includes/db_connect.php';
include_once 'includes/functions.php';
sec_session_start();
?>
<html>
<?php
$mysql_host = "";
$mysql_username = "";
$mysql_password = "r!~";
$mysql_database = "";
$user = $_SESSION['username'];
if (login_check($mysqli) == true) : ?>
<p>Welcome <?php echo htmlentities($user); ?>!</p>
<?php
$mysqli = new Mysqli($mysql_host, $mysql_username, $mysql_password, $mysql_database) or die(mysqli_error());
$status = 2;
$stmt = $mysqli->prepare("UPDATE Overrides SET status=? WHERE username='$user'");
$stmt->bind_param("s", $status);
$stmt->execute();
echo htmlentities(accepted);
?>
<?php else : ?>
<p>
<span class="error">You are not authorized to access this page.</span> Please <a href="index.php">login</a>.
</p>
<?php endif; ?>
</html>
bheads2.php
<html>
<?php
$mysql_host = "";
$mysql_username = "";
$mysql_password = "";
$mysql_database = "";
$user = $_SESSION['username'];
if (login_check($mysqli) == true) : ?>
<p>Welcome <?php echo htmlentities($user); ?>!</p>
<?php
$mysqli = new Mysqli($mysql_host, $mysql_username, $mysql_password, $mysql_database) or die(mysqli_error());
$status = 5;
$stmt = $mysqli->prepare("UPDATE Overrides SET status=? WHERE username='$user'");
$stmt->bind_param("s", $status);
$stmt->execute();
echo htmlentities(denied);
?>
<?php else : ?>
<p>
<span class="error">You are not authorized to access this page.</span> Please <a href="index.php">login</a>.
</p>
<?php endif; ?>
</html>
任何有關如何解決此問題的幫助？ 我是一個初學者，所以請忽略凌亂的代碼。
1 個解決方案
解決方案1
0 2015-05-09 15:57:20
似乎您正在使用以下查詢更新數據庫
$stmt = $mysqli->prepare("UPDATE Overrides SET status=? WHERE username='$user'")
這只是在說用戶登錄或使用頁面的用戶名將更新為您選擇的狀態，您是否對每行替代都有唯一的標識符？ 可能是Override_ID。
如果是這樣，我將在您的第一頁上獲取該數據，並將其放入其他數據一樣的隱藏輸入中，然后使用以下查詢：
$ovid = $_POST['ovid'];
$stmt = $mysqli->prepare("UPDATE Overrides SET status=? WHERE override_id='$ovid'")
編輯：
您似乎還在更新頁面上更新WHERE username='$user' ，而不是WHERE professor='$user'
使用 PHP 將 HTML 表單提交到數據庫，它會帶回 php 頁面而不是提交
[英]Submitting HTML form to database using PHP and it brings back php page instead of submitting
聲明:本站的技術帖子網頁，遵循CC BY-SA 4.0協議，如果您需要轉載，請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.