選擇並插入無法將會話保存到數據庫庫

Question

如何將選定的數據庫獲取到第二頁，然後插入到另一個表中。

這是我的第一頁：

<?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']; 
?> 

這裏是我的第二頁：

<?php 
session_start(); 


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

Answer 1

$sql="INSERT INTO login set 
Fname='".$_SESSION['fname']."', 
Lname='".$_SESSION['lname']."'"; 

Answer 2

請期待和嘗試以下操作：

session_start(); 

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

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

還要確保當你填充會話

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

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

有數據分配。也許在循環中添加它。如果您正在尋找最後一張唱片，無論如何都會被分配：

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

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