编辑表单 PHP MySQL

Question

您好，我有一个编辑表单，但我收到一个无法解决的错误，您能帮我解决吗？ 我得到的错误来自第一个代码，就是这个“ 注意：未定义索引：第 5 行 C:\xampp\htdocs\industrial\CVTool\cv\edit_personal_information.php 中的用户名" 表单的代码如下..

<?php
mysql_connect('localhost', 'root', 'password') or die(mysql_error());
mysql_select_db("cvtool") or die(mysql_error());

$username = $_GET['username'];
$query = mysql_query("SELECT * FROM personal_information WHERE username = '$username'") or die(mysql_error());

if(mysql_num_rows($query)>=1){
    while($row = mysql_fetch_array($query)) {
        $name = $row['name'];
        $email = $row['email'];
        $city = $row['city'];
        $phone = $row['phone'];
    }
?>
<form action="update.php" method="post">
<input type="hidden" name="username" value="<?=$username;?>">
Name: <input type="text" name="name" value="<?=$name?>"><br>
Email: <input type="text" name="email" value="<?=$email?>"><br>
City: <input type="text" name="city" value="<?=$city?>"><br>
Phone: <input type="text" name="city" value="<?=$phone?>"><br>
<input type="Submit" name="Update!">
</form>
<?php
}else{
    echo 'No entry found. <a href="javascript:history.back()">Go back</a>';
}
?>

更新代码如下...

 <?php
     mysql_connect('localhost', 'root', 'smogi') or die(mysql_error());
     mysql_select_db("cvtool") or die(mysql_error());

     $username = $_POST["username"];

     $name = mysql_real_escape_string($_POST["name"]);
     $email = mysql_real_escape_string($_POST["email"]);
     $city = mysql_real_escape_string($_POST["city"]);
     $phone = mysql_real_escape_string($_POST["phone"]);


     $query="UPDATE personal_information
             SET name = '$name', email = '$email', phone = '$phone' 
             WHERE username='$username'";


 mysql_query($query)or die(mysql_error());
 if(mysql_affected_rows()>=1){
     echo "<p>($username) Record Updated<p>"; }else{
     echo "<p>($username) Not Updated<p>"; } ?>

Answer 1

这是因为当您第一次加载页面时，没有$_GET["username"] 变量。你应该用一个条件来测试它：

if (!empty($_GET["username"])) {
    //Do code here
}

只有在你提交表单时它才会有价值。