0

如何将选定的数据库放入第二页，然后插入另一个表。

这是我的第一页：

    <?php
    session_start();

    $result = mysqli_query($con,"SELECT * FROM qwerty
    WHERE ID LIKE '$id' ");

while($row = mysqli_fetch_array($result))
  {
  echo "Name:" . $row['Fname'] . " " . $row['Lname'];
  } 

$_SESSION['fname']=$row['Fname'];
$_SESSION['lname']=$row['Lname']; 
?>

这是我的第二页：

<?php
session_start();


   $sql="INSERT INTO login (Fname, Lname)
    VALUES
    ('{$_SESSION['fname']}','{$_SESSION['lname']}')";
4

2 回答 2

0 
$sql="INSERT INTO login set
Fname='".$_SESSION['fname']."',
Lname='".$_SESSION['lname']."'";
于 2013-08-12T07:27:04.730 回答
0

请查看并尝试以下操作：

session_start();

$firstName = $_SESSION['fname'];
$lastName = $_SESSION['lname'];

$sql="INSERT INTO `login` (Fname, Lname)  VALUES ('$firstName','$lastName')";

还要确保在填充会话时

while($row = mysqli_fetch_array($result))
  {
  echo "Name:" . $row['Fname'] . " " . $row['Lname'];
  } 

$_SESSION['fname']=$row['Fname']; //CHECK
$_SESSION['lname']=$row['Lname'];

有数据分配。也许将其添加到循环中。如果您正在寻找最后一条记录，无论如何它都会被分配：

while($row = mysqli_fetch_array($result))
  {
     echo "Name:" . $row['Fname'] . " " . $row['Lname'];

     $_SESSION['fname']=$row['Fname']; //CHECK
     $_SESSION['lname']=$row['Lname'];
  }
于 2013-08-12T07:28:00.520 回答