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php - 获取数据并将其提交到数据库

转载 作者:行者123 更新时间:2023-11-28 03:11:39 25 4
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我的代码从数据库中获取条目并将它们打印成单独的表单,这样我就可以使用不同操作的按钮来接受或拒绝它们中的每一个。每当我对其中一个表单执行操作时,它不会对同一表单执行操作,而是从数据库中的第一个条目开始按顺序移动。

enter image description here

在上图中,如果我接受了第二种形式,它将执行第一种形式的功能。

代码:

<?php
} else if ($usertype == 1) {
ini_set('display_errors',1);
ini_set('display_startup_errors',1);
error_reporting(-1);
$server = "";
$user = "";
$pass = "r=~";
$db = "";
$db3 = "aukwizcq_professors";
$user1 = $_SESSION['username'];
$mysqli = new Mysqli($server, $user, $pass, $db) or mysqli_error($mysqli);
$mysqli5 = new Mysqli($server, $user, $pass, $db3) or mysqli_error($mysqli);
$name= $mysqli5->query("SELECT name FROM professor WHERE username= '$user1'")->fetch_object()->name; //taking name from list of professors that are equal to the userid logged in (if hes a professor, it will show, if not 0)
$overrides = $mysqli->query("SELECT * FROM Overrides WHERE professor= '$name' AND status ='1'");
$num_rows = mysqli_num_rows($overrides);



echo "&nbsp;Pending Overrides: " . $num_rows;
?>
<?php
while($row = mysqli_fetch_array($overrides)) { ?>
<fieldset>
<form method="post" action="dbheads.php" name="HF" id="HF" autocomplete="off">

<?php
echo "First Name:&nbsp;&nbsp; " . $row['name'] . "<br />";
echo "<br />Mid. Name:&nbsp;&nbsp; " . $row['mname'] . "<br />";
echo "<br />Fam. Name:&nbsp;&nbsp; " . $row['fname'] . "<br />";
echo "<br />Student ID:&nbsp;&nbsp;&nbsp;&nbsp;" . $row['sid'] . "<br />";
echo "<br />Scolarship:&nbsp;&nbsp;&nbsp;&nbsp; " . $row['sc'] . "<br />";
echo "<br />Phone No:&nbsp;&nbsp;&nbsp;&nbsp; " . $row['phone'] . "<br />";
echo "<br />Email:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; " . $row['email'] . "<br />";
echo "<br />Subject:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; " . $row['subject'] . "<br />";
echo "<br />Section:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; " . $row['section'] . "<br />";
echo "<br />Semester:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; " . $row['semester'] . "<br />";
echo "<br />Professor:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; " . $row['professor'] . "<br />";

$id = $row['id'];
echo "<input type='hidden' name='id' value='$id'>";
$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'>";
//$professor = $row['professor'];
// echo "<input type='hidden' name='professor' value='$professor'>";



?>
<br />
<div>
<label for="comments" accesskey="c">Notes & Comments:</label><br />
<textarea name="comments" id="comments" cols="50" rows="5"></textarea>
<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
}

?>

数据库头文件

<?php 
include_once 'includes/db_connect.php';
include_once 'includes/functions.php';
sec_session_start();
?>
<html>

<?php
$mysql_host = "localhost";
$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 = 2;
$id = $_POST['id'];

$stmt = $mysqli->prepare("UPDATE Overrides SET status=? WHERE id='$id'");
$stmt->bind_param("s", $status);
$stmt->execute();
print 'Error : ('. $mysqli->errno .') '. $mysqli->error;
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>

请帮忙解决问题?

最佳答案

这是因为您通过 ID 发送表单,并且您对所有表单都有相同的 ID(无效)。

如果您要为这样的表单创建唯一 ID:

<form method="post" name="HF" id="HF_<?php echo $row['id'] ?>" autocomplete="off">

像这样更改 JavaScript 之后:

document.getElementById('HF_<?php echo $row['id'] ?>').action = action;
document.getElementById('HF_<?php echo $row['id'] ?>').submit();

它会起作用

如果我不能改进你的代码,我有以下建议给你:

如果您将使用 ,则不需要更多 JavaScript,您可以测试用 PHP 提交的按钮(最佳实践)。示例:

<form method="post" name="HF" id="HF_<?php echo $row['id'] ?>" autocomplete="off">
<input type="hidden" name="HF[id]" />
<input type="hidden" name="HF[name]" />
<input type="hidden" name="HF[mname]" />
<input type="hidden" name="HF[fname]" />
<input type="hidden" name="HF[sid]" />
<button type="submit" value="1" name="HF[accept]">Accept</button>
<button type="submit" value="1" name="HF[reset]">Reset</button>
</form>

和PHP测试

if(isset($_POST['HF'])){
if($_POST['HF']['accept']){
// do accept
}
elseif($_POST['HF']['reset']){
// do reset
}
}

如果你看到我不需要为每个 Action 设置单独的 Action ,并且我有表单前缀,这意味着我可以在此 url 上发送多个表单,但只会处理一个表单

关于php - 获取数据并将其提交到数据库,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30256297/

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