Question

我是这方面的新手，并编写了下面的代码来从MySQL数据库中获取用户数据并将其显示在表单中以进行编辑和保存。问题是，它不起作用。任何帮助将不胜感激。

<html>
    <head>
        <title> Delegate edit form</title>
    </head>
    <body>
        Delegate update form  <p>
            <?php
            $usernm   = "root";
            $passwd   = "";
            $host     = "localhost";
            $database = "swift";

            //$Name=$_POST['Name'];
            //$Username=$_POST['User_name'];
            //$Password=$_POST['Password'];

            mysql_connect($host,$usernm,$passwd);

            mysql_select_db($database);

            $sql    = "SELECT * FROM usermaster WHERE User_name='$Username'";
            $result = mysql_query($sql) or die(mysql_error());
            while ($row    = mysql_fetch_array($result))
            {

                $Name     = $row['Name'];
                $Username = $row['User_name'];
                $Password = $row['User_password'];
            }
            ?>
        <form action="Delegate_update.php" method="post">
            Name
            <input type="text" name= "Name" value= "<?php echo $row ['Name']; ?> "size=10>
            Username
            <input type="text" name= "Username" value= "<?php echo $row ['Username']; ?> "size=10>
            Password
            <input type="text" name= "Password" value= "<?php echo $row ['Password']; ?>" size=17>
            <input type="submit" name= "submit" value="Update">
        </form>
    </body>
</html>

Answer 1

玩这段代码。专注于概念，在必要时进行编辑，以便

<html>
<head>
    <title> Delegate edit form</title>
</head>
<body>
    Delegate update form  </p>

    <meta name="viewport" content="width=device-width; initial-scale=1.0">

    <link rel="shortcut icon" href="images/favicon.ico" type="image/x-icon" />


    <link href='http://fonts.googleapis.com/css?family=Droid+Serif|Ubuntu' rel='stylesheet' type='text/css'>

    <link rel="stylesheet" href="css/normalize.css">
    <link rel="stylesheet" href="js/flexslider/flexslider.css" />
    <link rel="stylesheet" href="css/basic-style.css">




    <script src="js/libs/modernizr-2.6.2.min.js"></script>

    </head>

    <body id="home">

        <header class="wrapper clearfix">



            <nav id="topnav" role="navigation">
                <div class="menu-toggle">Menu</div>
                <ul class="srt-menu" id="menu-main-navigation">
                    <li><a href="Swift_Landing.html">Home page</a></li>

        </header>
        </section>

        <style>
            form label {
                display: inline-block;
                width: 100px;
                font-weight: bold;
            }
        </style>
        </ul>

        <?php
        session_start();
        $usernm="root";
        $passwd="";
        $host="localhost";
        $database="swift";

        $Username=$_SESSION['myssession'];



        mysql_connect($host,$usernm,$passwd);

        mysql_select_db($database);

        $sql = "SELECT * FROM usermaster WHERE User_name='$Username'";
        $result = mysql_query ($sql) or die (mysql_error ());
        while ($row = mysql_fetch_array ($result)){

        ?>

        <form action="Delegate_update.php" method="post">
            Name
            <input type="text" name="Namex" value="<?php echo $row ['Name']; ?> " size=10>
            Username
            <input type="text" name="Username" value="<?php echo $row ['User_name']; ?> " size=10>
            Password
            <input type="text" name="Password" value="<?php echo $row ['User_password']; ?>" size=17>
            <input type="submit" name="submit" value="Update">
        </form>
        <?php
        }
        ?>
        </p>
    </body>
</html>

Answer 2

<form action="Delegate_update.php" method="post">
Name
<input type="text" name= "Name" value= "<?php echo $row['Name']; ?> "size=10>
Username
<input type="text" name= "Username" value= "<?php echo $row['Username']; ?> "size=10>
Password
<input type="text" name= "Password" value= "<?php echo $row['Password']; ?>" size=17>
<input type="submit" name= "submit" value="Update">
</form>

你没有首先关闭你的开场表格，而且你的代码非常混乱。我不会进入“使用pdo或mysqli语句，而不是mysql”，这样你就可以了解自己。你还有一个php标签在它下面打开和关闭，不知道那里需要什么。还有一件事是你的代码引用了一个外部页面，你没有发布，所以如果有什么东西不在那里工作，也可以方便地发布它。

请注意，表单中的$ row数组变量之间有空格。您必须通过删除空格将这些链接在一起（请参阅我编辑的部分）。当涉及到这些错误时，PHP并不宽容。

然后你的HTML。我也冒昧地纠正了这个问题

<html>
    <head>
        <title> Delegate edit form</title>
    </head>

    <body>
          <p>Delegate update form</p>
<?php
$usernm="root";
$passwd="";
$host="localhost";
$database="swift";

mysql_connect($host,$usernm,$passwd); 
mysql_select_db($database);

$sql = "SELECT * FROM usermaster WHERE User_name='".$Username."'"; // Please look at this too.
$result = mysql_query($sql) or die (mysql_error()); // dont put spaces in between it, else your code wont recognize it the query that needs to be executed
while ($row = mysql_fetch_array($result)){     // here too, you put a space between it
    $Name=$row['Name'];
    $Username=$row['User_name'];
    $Password=$row['User_password'];
    }
?>

另外，尽量具体。 “它不起作用”对我们没什么帮助，一个特定的错误类型通常是有用的，加上任何指示代码应该做什么（好吧，这里有点明显，因为它的登录/注册编辑在这里，但对于更大的代码块应该总是解释一下）

无论如何，欢迎来到Stack Overflow

Answer 3

<form action="Delegate_update.php" method="post">
  Name
  <input type="text" name= "Name" value= "<?php echo $row['Name']; ?> "size=10>
  Username
  <input type="text" name= "Username" value= "<?php echo $row['Username']; ?> "size=10>
  Password
  <input type="text" name= "Password" value= "<?php echo $row['Password']; ?>" size=17>
  <input type="submit" name= "submit" value="Update">
</form>

看看这个

Answer 4

请尝试这些

<form action="Delegate_update.php" method="post">
Name
<input type="text" name= "Name" value= "<?php echo $row['Name']; ?> "size=10>
Username
<input type="text" name= "User_name" value= "<?php echo $row['User_name']; ?> "size=10>
Password
<input type="text" name= "User_password" value= "<?php echo $row['User_password']; ?>" size=17>
<input type="submit" name= "submit" value="Update">
</form>

Answer 5

<?php
 include 'cdb.php';
$show=mysqli_query( $conn,"SELECT *FROM 'reg'");


while($row1= mysqli_fetch_array($show)) 

{

                 $id=$row1['id'];
                $Name= $row1['name'];
                $email = $row1['email'];
                $username = $row1['username'];
                $password= $row1['password'];
                $birthm = $row1['bmonth'];
                $birthd= $row1['bday'];
                $birthy= $row1['byear'];
                $gernder = $row1['gender'];
                $phone= $row1['phone'];
                $image=$row1['image'];
}


?>


<html>
<head><title>hey</head></title></head>

<body>

<form>
<table border="-2" bgcolor="pink" style="width: 12px; height: 100px;" >

    <th>
    id<input type="text" name="" style="width: 30px;" value= "<?php echo $row1['id']; ?>"  >
</th>

<br>
<br>

    <th>

 name <input type="text" name=""  style="width: 60px;" value= "<?php echo $row1['Name']; ?>" >
</th>

<th>
 email<input type="text" name="" style="width: 60px;" value= "<?php echo $row1['email']; ?>"  >
</th>
<th>
    username<input type="hidden" name="" style="width: 60px;"  value= "<?php echo $username['email']; ?>" >
</th>
<th>
    password<input type="hidden" name="" style="width: 60px;"  value= "<?php echo $row1['password']; ?>">
</ths>

<th>
    birthday month<input type="text" name="" style="width: 60px;"  value= "<?php echo $row1['birthm']; ?>">
</th>
<th>
   birthday day<input type="text" name="" style="width: 60px;"  value= "<?php echo $row1['birthd']; ?>">
</th>
<th>
       birthday year<input type="text" name="" style="width: 60px;" value= "<?php echo $row1['birthy']; ?>" >
</th>
<th>
    gender<input type="text" name="" style="width: 60px;" value= "<?php echo $row1['gender']; ?>">
</th>
<th>
    phone number<input type="text" name="" style="width: 60px;" value= "<?php echo $row1['phone']; ?>">
</th>
<th>
    <th>
     image<input type="text" name="" style="width: 60px;"  value= "<?php echo $row1['image']; ?>">
</th>

<th>

<font color="pink"> <a href="update.php">update</a></font>

</th>
</table>

</body>
</form>
</body>
</html>

使用PHP从MySQL数据库中获取数据，以表格形式显示以进行编辑

5 个答案: