Question

我正在尝试从MySQL数据库中获取记录，以便使用PHP以HTML格式进行编辑。首先是用户点击EDIT超链接以编辑数据的代码。

<?php
$con=mysqli_connect("localhost","root","")or die("Connection error");
mysqli_select_db($con,"lecture7")or die("dbase error");

$sql="select * from students";
$result=mysqli_query($con,$sql);

   echo "<table border=1>
    <th>RollNo</th>
    <th>Name</th>
    <th>class</th>
    <th>Update</th>";

  while($row=mysqli_fetch_array($result, MYSQLI_ASSOC))


{

    echo "<tr>";

    echo "<td>";
    echo $row['RollNo'];
    echo "</td>";

    echo "<td>";
    echo $row['name'];
    echo "</td>";

    echo "<td>";
    echo $row['class'];
    echo "</td>";

    echo "<td>"; 
    echo "<a href='update.php?roll=".$row['RollNo']."'> 

Edit</a>";
    echo "</td>";


}

?>

Aftter点击EDIT用户将继续update.php进行编辑。但不是在文本框中显示记录而是发送错误警告：mysqli_fetch_array（）期望参数1为mysqli_result，第11行的C：\ xampp \ htdocs \ update.php中给出布尔值这里是代码for update.php

<?php
$con=mysqli_connect("localhost","root","")or die("Connection error");
mysqli_select_db($con,"lecture7")or die("dbase error");

$r=$_GET['roll'];

$sql="select * from students where roll=$r";
$result=mysqli_query($con, $sql);


while($row=mysqli_fetch_array($result, MYSQLI_ASSOC))
{

?>


<form name="f1" action="" method="GET">
RollNo:<input type="text" name="txtr" value="<?php echo $row['RollNo'];?>"
Name:<input type="text" name="txtn" value="<?php echo $row['name'];?>"
Class:<input type="text" name="txtc" value="<?php echo $row['class'];?>"

</form>
<?php
}

?>

无法从数据库中检索记录以进行编辑

0 个答案: