看起来一切都与此代码工作正常，但实际上无法更新数据库，在获取数据时可以正确显示数据，但是当我按下update按钮时，数据消失了，但未执行任何更新。对我来说看起来不错，但似乎我错了。
这是我教授的一个项目，因此我不关心SQL注入和其他注入。
<html>
<head>
<link rel="stylesheet" type="text/css" href="btnstyle.css">
<title>Managament System</title>
</head>
<body>
<h1>TU Chemnitz Student managament system</h1>
<br>
<a href="insert_form.html" class="myButton">ADD Person</a>
<a href="insert_form.html" class="myButton">Edit Person</a>
<a href="insert_form.html" class="myButton">Manage Boards</a>
<a href="insert_form.html" class="myButton">Manage Departments</a>
<a href="search_personel.html" class="myButton">Search N&S</a>
<a href="three_search.html" class="myButton">Triple Search</a>
<a href="mtable.php" class="myButton">Membership</a>
<br>
<br>
<?php
// set database server access variables:
$host = "localhost";
$user = "";
$pass = "";
$db = "";
// open connection
$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!");
// select database
mysql_select_db($db) or die ("Unable to select database!");
// create query
$querys = "SELECT * FROM tblperson";
// execute query
$result = mysql_query($querys) or die ("Error in query: $query. ".mysql_error());
echo "<table border=1 align=center>
<tr>
<th>Personal ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>Deparment</th>
<th>Board</th>
<th>Marticulation Number</th>
<th>Reg Date</th>
<th>Action</th>
</tr>";
while($row = mysql_fetch_array($result)) {
?>
<?php
echo '<tr>';
echo '<td>'. $row['personid'].'</td>';
echo '<td>'. $row['personname'].'</td>';
echo '<td>'. $row['personsurname'].'</td>';
echo '<td>'. $row['persondepartment'].'</td>';
echo '<td>'. $row['personboard'].'</td>';
echo '<td>'. $row['martinumber'].'</td>';
echo '<td>'. $row['personregdate'].'</td>';
echo '<td>'.'<a href="edit20.php?edit='.$row['personid'] . '"> EDIT </a> '.'</td>';
}
?>
</body>
</html>
这是似乎有问题的编辑文件。
<?php
include_once('coneksioni.php');
if(isset($_GET['edit']))
{
$personid = $_GET['edit'];
$res = mysql_query("SELECT * FROM tblperson WHERE personid='$personid'");
$row = mysql_fetch_array($res);
}
if(isset($_POST['newpersonname']))
{
$newpersonname = $_POST['newpersonname'];
$personid = $_POST['personid'];
$sql = "UPDATE tblperson SET personname = '$newpersonname' WHERE personid = '$personid'";
$res = mysql_query($sql) or die ("Cant be updated");
echo "< meta http-equiv='refresh' content='0;url=home.php'>";
}
?>
<form action="edit20.php" method="POST">
<table border="0">
<tr>
<td>First Name</td>
<td><input type="text" name="newpersonname" value="<?php echo $row[1];?>" maxlength="30" size="13"></td>
</tr>
<tr>
<td>Last Name</td>
<td> <input type="text" name="personsurname" value="<?php echo $row[2];?>" maxlength="30" size="30"></td>
</tr>
<tr>
<td>Department</td>
<td>
<select name='persondepartment'>
<option>Production</option>
<option>Sales</option>
</select>
</td>
</tr>
<tr>
<td>Board</td>
<td>
<select name='personboard'>
<option>Evaluation</option>
<option>Executive</option>
<option>Research</option>
</select>
</td>
</tr>
<tr>
<td>Marticulation Number</td>
<td> <input type="text" name="martinumber" maxlength="60" size="30"></td>
</tr>
<tr>
<td>Date of Registration</td>
<td><input type="date" name="personregdate" maxlength="7" size="7"></td>
</tr>
<tr>
<td colspan="2"><input type="submit" value=" Update"></td>
</tr>
</table>
</form>
最佳答案
当您在edit20.php中的表单上按下“更新”按钮时，您正在寻找personid，但该值从未设置过，因此它将为空并且更新将失败。
后
<form action="edit20.php" method="POST">
加：
<input type="hidden" name="personid" value="<?php echo $personid; ?>">
关于php - 无法更新管理员输入的新数据，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/38566317/