PHP如何在mySQL中编辑现有数据和更新

Question

我正在尝试创建一个PHP页面，该页面显示数据库中的信息，并且可以通过单击“批准”按钮进行更新。我收到了数据库中的信息（名字，姓氏，请求）。我的问题是，当我单击“批准”按钮时，我收到404页面未找到错误，并且数据库也未更新。任何帮助，将不胜感激。

第一个cgi文件名为ApproveDenyPrayerRequest

<?php

$username="XXXXX";
$password="XXXXX";
$database="XXXXX";

$link = mysqli_connect('XXXXX', $username, $password, $database);

if (!$link) {
die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());
}

$query = "SELECT * FROM Request";
$result = mysqli_query($link,$query); //<----- Added link
$row = mysqli_fetch_array($result);

?>

<form method="post" action="PrayerRequest.php" />

<table>

<tr>
<td>First Name:</td>
<td><input type="text" name="first" value="<? echo "$row[Reg_F_Name]" ?>"></td>
</tr>

<tr>
<td>Last Name:</td>
<td><input type="text" name="last" value="<? echo "$row[Reg_L_Name]" ?>"></td>
</tr>

<tr>
<td>Prayer Request</td>
<td><input type="text" name="phone" value="<? echo "$row[Reg_Request]" ?>"></td>
</tr>

</table>
<input name="add" type="submit" id="add" value="Approve Prayer Request">

</form>

这是名为PrayerRequest的第二个文件。

<?php

$username="XXX";
$password="XXXXX";
$database="XXXXX";

$link = mysqli_connect('XXXXX', $username, $password, $database);

if (!$link) {
die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());
}

echo 'Success... ' . mysqli_get_host_info($link) . "\n";

$query = "SELECT * FROM Request";
$query2="INSERT INTO Request (Reg_F_Name,Reg_L_Name,Reg_Request)
     VALUES ( '$row[Reg_F_Name]', '$row[Reg_L_Name]', '$row[Reg_Request]' )";
$result = mysqli_query($query);
$row = mysqli_fetch_array($result);
?>

<form method="post" action="ApproveDenyPrayerRequest.php" />

<table>

<tr>
<td>First Name:</td>
<td><input type="text" name="first" value="<? echo "$row[Reg_F_Name]" ?>"></td>
</tr>

<tr>
<td>Last Name:</td>
<td><input type="text" name="last" value="<? echo "$row[Reg_L_Name]" ?>"></td>
</tr>

<tr>
<td>Prayer Request</td>
<td><input type="text" name="phone" value="<? echo "$row[Reg_Request]" ?>"></td>
</tr>

</table>

</form>