PHP表单，用于更新MySQl数据库以进行Ajax聊天

Question

I am using a open source ajax chat program for some users at work for group chat. The program works fine but it was not built with a way for the users to change there passwords or for me to manage the users without doing it from the database directly. So I have put together this PHP script. It is able to display the data from the database but it will not update it. I am using Xampp for MySQl and Apache server. I plan on moving to IIS once I have it up and running. Here is the layout of my table.

  ID Username Password Role Channels EMail

To save sometime I will only post the update and update_ac scripts. When using windows 7 as my server I get undefined variable after submitting the update. it still says update successful but the database is not updated. In windows xp as server I do not get undefined variable error. If some could please give me some advise on what i am doing wrong or point me to another solution here I would appreciate it thanks.

update.php
// get value of id that sent from address bar
$id=$_GET['id'];

// Retrieve data from database
$sql="SELECT * FROM $tbl_name WHERE id='$id'";
$result=mysql_query($sql);

$rows=mysql_fetch_array($result);
?>

<table width="400" border="0" cellspacing="10" cellpadding="0">
<tr>
<form name="form1" method="post" action="update_ac.php">
<td>
<table width="100%" border="10" cellspacing="1" cellpadding="10">

<tr>
<td colspan="3"><strong>Update User</strong> </td>
</tr>
<center>
<tr>
<td align="center"><strong>Username</strong></td>
<td align="center"><strong>Password</strong></td>
<td align="center"><strong>Role</strong></td>
<td align="center"><strong>Channels</strong></td>
<td align="center"><strong>EMail</strong></td>
</tr>
</center>
<tr>
<td align="center">
<input name="username" type="text" id="Username" value="<?php echo 
$rows['Username'];   
?>" size="15">
</td>

<td align="center">
<input name="password" type="Password" id="Password" value="<?php echo      
$rows['Password']; ?>" size="15">
</td>

<td>
<input name="role" type="text" id="Role" value="<?php echo $rows['Role']; ?>" size="1">
</td>

<td>
<input name="channels" type="text" id="Channels" value="<?php echo $rows['Channels']; 
?>" size="10">
</td>


<td>
<input name="EMail" type="text" id="EMail" value="<?php echo $rows['EMail']; ?>"  
size="25">
</td>

<tr>
<td>
<input name="id" type="hidden" id="ID" value="<?php echo $rows['ID']; ?>">
</td>
<td align="center">
<input type="submit" name="Submit" value="Submit">
</td>

</tr>
 update_ac.php
// update data in mysql database
$sql = "UPDATE $tbl_name SET Username='$Username', Password='$Password', Role='$Role', 
Channels='$Channels', EMail='$EMail' WHERE id='$id'";
$result = mysql_query($sql);

// if successfully updated.
if($result)
{

echo "Successful";
echo "<BR>";
echo "<a href='index.php'>View result</a>";

}

else
{
echo "ERROR";
}

?>

Answer 1

Here, You are not getting the value posted from from, i supposed to use $_POST to get the posted values.

// update data in mysql database
$sql = "UPDATE $tbl_name SET Username='$Username', Password='$Password', Role='$Role', 
Channels='$Channels', EMail='$EMail' WHERE id='$id'";

So you have to get the posted value as folows

$Username  = $_POST['Username'];
$Password  = $_POST['Password'];
$Role      = $_POST['Role'];
$Channels  = $_POST['Channels'];
$EMail     = $_POST['EMail'];
$id        = $_POST['id'];

Answer 2

我发现将name =“ password更改为name =” Password“修复了我的脚本，我必须对所有字段都执行相同操作，但现在可以使用了。