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我有一个简单的下拉菜单，可以将结果发布给自己，但是当我在下拉菜单中选择一个选项时，它不会按预期回显结果。

我确定我只是错过了一些简单但无法发现的东西。有任何想法吗？表单发布但不回显 $user_settings。

<?php
include "functions.php";
connect();

$sql="SELECT user_id, user_realname FROM users ORDER BY user_realname ASC";
    $result=mysql_query($sql);

    while ($row = mysql_fetch_array($result)) {
        $name=$row['user_realname'];
        $options.="<OPTION VALUE=>".$name.'</option>';
        }

if(isset($_POST['submit'])){
    $user_realname = $_POST['username_select'];
    $user_select = mysql_query("SELECT user_id, user_realname FROM users WHERE user_realname = '$user_realname'")
    or die ("Could not get user data");

    while($row = mysql_fetch_array($user_select)){

        $user_settings = $row['user_id'];
        echo $user_settings;
    }
}
?>

<html>
    <head>
        <body>
            <form action="<?php echo $PHP_SELF;?>" method="POST">
                <tr><label>Choose User to Edit</tr>
                <tr><SELECT NAME="username_select"><OPTION VALUE=""></option>User's Name<?php echo $options;?></SELECT></label></tr>
                <tr><input type="submit" value="submit" name="submit"></tr>
            </form>
            <?php echo $user_settings;?>
            <br/>
            <a href="admin.php">Go Back</a>
        </body>
    </head>
</html>
4

1 回答 1

2

用户$_SERVER['PHP_SELF']而不是 $PHP_SELF

于 2012-08-27T10:13:23.147 回答