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我正在构建一个页面，管理员用户可以在其中更新客户的信息。我首先查询客户信息并将其显示在文本字段中。然后我可以为该客户输入我想要的任何新信息。当我在 sql 中执行更新命令时，它会将我带到显示更新成功的页面，但是当我回顾数据库时，客户信息并没有改变。

Edit-client.php（显示客户信息的页面）

<?php


//Function to sanitize values received from the form. Prevents SQL injection
function clean($str) {
    $str = @trim($str);
    if(get_magic_quotes_gpc()) {
        $str = stripslashes($str);
    }
    return mysql_real_escape_string($str);
}

//define username variable and sanitize
$username = clean($_POST['username']);

//Run query for selected user and store in an array
$result = mysql_query("select * from members where username='".$username."'");
$row = mysql_fetch_array($result);

//display all clients information in a form to edit
echo '<h1>'.$username.'</h1>';
echo '<form name="update-client" action="update-client.php" />';
echo '<table>';
echo '<tr><td>';
echo '<input type="hidden" name="member_id" value="'.$row['member_id'].'"';
echo '</td></tr>';
echo '<tr><td>';
echo 'Username: <input name="username" type="text" value="'.$username.'" />';
echo '</td></tr>';
echo '<tr><td>';
echo 'Password: <input name="password" type="text" value="'.$row['password'].'" />';
echo '</td></tr>';
echo '<tr><td>';
echo 'Business Name: <input name="bizname" type="text" value="'.$row['bizname'].'" />';
echo '</td></tr>';
echo '<tr><td>';
echo 'Phone: <input name="phone" type="text" value="'.$row['phone'].'" />';
echo '</td></tr>';
echo '<tr><td>';
echo 'Email: <input name="email" type="text" value="'.$row['email'].'" />';
echo '</td></tr>';
echo '<tr><td>';
echo 'Website Address: <input name="url" type="text" value="'.$row['url'].'" />';
echo '</td></tr>';
echo '<tr><td>';
echo 'Contact: <input name="contact" type="text" value="'.$row['contact'].'" />';
echo '</td></tr>';
echo '<tr><td>';
echo 'Notes: <input name="notes" type="text" value="'.$row['notes'].'" />';
echo '</td></tr>';
echo '<tr><td>';
echo 'Sales Representative: <input name="sales_rep" type="text" value="'.$row['sales_rep'].'" />';
echo '</td></tr>';
echo '<tr><td>';
echo '<input name="submit" type="submit" value="Edit" />';
echo '</td></tr>';
echo '</table>';
echo '</form>';


?>

更新-client.php

<?php


//Function to sanitize values received from the form. Prevents SQL injection
function clean($str) {
    $str = @trim($str);
    if(get_magic_quotes_gpc()) {
        $str = stripslashes($str);
    }
    return mysql_real_escape_string($str);
}

//define variables
$member_id = $_POST['member_id'];
$username = $_POST['username'];
$password = $_POST['password'];
$bizname = $_POST['bizname'];
$phone = $_POST['phone'];
$email = $_POST['email'];
$url = $_POST['url'];
$contact = $_POST['contact'];
$notes = $_POST['notes'];
$sales_rep = $_POST['sales_rep'];
$member_type = $_POST['member_type'];

//encrypt the password
$password = md5($password);

//Check for duplicate username
if($username != '') {
    $qry_uname = "SELECT * FROM members WHERE username='".$username."'";
    $result = mysql_query($qry_uname);
    if($result) {
        if(mysql_num_rows($result) > 0) {
            $errmsg_arr[] = 'Username already in use';
            $errflag = true;
        }
        @mysql_free_result($result);
    }
    else {
        die("Query failed1");
    }
}

//update customers information
$qry = "update members set username='".$username."',password='".$password."',bizname='".$bizname."',phone='".$phone."',email='".$email."',url='".$url."',contact='".$contact."',notes='".$notes."',sales_rep='".$sales_rep."',member_type='".$member_type."' where member_id='".$member_id."'";

//Check whether the query was successful or not
/*if(mysql_query($qry)) {
    header("location: update-success.php");
exit();
    }
else {
    die("Query failed2");
    }*/

echo $qry;

?>

我的代码有问题吗？我正在使用 apache 服务器

4

1 回答 1

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//检查重复的用户名

检查用户名和 ID，因为如果没有更改，用户名已经存在。

于 2012-06-19T20:34:10.317 回答