上一個/下一個記錄按鈕php/mysql woes

Question

我的問題涉及我的下一個/上一個按鈕。我可以讓我的更新/刪除按鈕起作用，但在處理下一個/上一個按鈕時，我準備好撕掉我的頭髮。任何幫助將是壯觀的！這是我的代碼。另外，我對PHP很新，所以如果這是不好的編碼，請讓我知道並指出我正確的方向，以便我可以修復我的錯誤。謝謝！！！

session_start(); 

include "connectionfile.php"; 

if (isset($_POST['fname']) && 
isset($_POST['lname']) && 
isset($_POST['email']) && 
isset($_POST['login']) && 
isset($_POST['password']) && 
isset($_POST['super']) && 
isset($_POST['foldername'])) 
{ 
$id = get_post('id'); 
$fname = get_post('fname'); 
$lname = get_post('lname'); 
$email = get_post('email'); 
$login = get_post('login'); 
$password = hash('sha256', get_post('password')); 
$super = get_post('super'); 
$foldername = get_post('foldername'); 

if ($_POST['submit']==0){ 

$query = mysql_query("SELECT * FROM `Logins` WHERE ID < '".$id."' ORDER BY ID DESC LIMIT 1;"); 
while($row = mysql_fetch_array($query)){ 
    $id = $row['ID']; 
    $fname = $row['fname']; 
    $lname = $row['lname']; 
    $email = $row['email']; 
    $login = $row['login']; 
    $password = $row['password']; 
    $super = $row['super']; 
    $foldername = $row['foldername']; 

} 
}else if ($_POST['submit']==1){ 
$query = "UPDATE Logins SET fname = '$fname', lname='$lname', email='$email".'@carouselclinical.com'."', login='$login', password='$password', super='$super', foldername='$foldername'"; 
$query .= "WHERE ID = '$id';"; 
if (!mysql_query($query, $connect)) 
echo "INSERT failed: $query<br />" . 
mysql_error() . "<br /><br />"; 

}else if($_POST['submit']==2){ 

$delete_query = "DELETE FROM Logins WHERE ID = '".$id."';"; 
mysql_query($delete_query); 
$rc = mysql_affected_rows(); 
echo "Rows Affected " . $rc; 

} 


if ($_POST['submit']==3){ 

$query = mysql_query("SELECT * FROM `Logins` WHERE ID= '". $id ."' ORDER BY ID ASC LIMIT 1;"); 
while($row = mysql_fetch_array($query)){ 
    $id = $row['ID']; 
    $fname = $row['fname']; 
    $lname = $row['lname']; 
    $email = $row['email']; 
    $login = $row['login']; 
    $password = $row['password']; 
    $super = $row['super']; 
    $foldername = $row['foldername']; 

} 

} 
} 

mysql_close($connect); 

function get_post($var) 
{ 
return mysql_real_escape_string($_POST[$var]); 
} 
?> 

<form action="" method="post"><pre> 
id <input type="text" readonly="readonly" name="id" value="<?php echo "$id"; ?>" /> 
First Name <input type="text" name="fname" value="<?php echo "$fname"; ?>" /> 
Last Name <input type="text" name="lname" value="<?php echo "$lname"; ?>" /> 
Email <input type="text" name="email" value="<?php echo "$email"; ?>" /> There's no need to put @carouselclinical.com. 
Login <input type="text" name="login" value="<?php echo "$login"; ?>"/> 
Password <input type="text" name="password" value="<?php echo "$password"; ?>"/> 
Super? <input type="text" name="super" value="<?php echo "$super"; ?>" /> 
foldername <input type="text" name="foldername" value="<?php echo "$foldername"; ?>" /> 
<button name="submit" value="0">Previous</button> 
<button name="submit" value="1">UPDATE</button> 
<button name="submit" value="2">Delete</button> 
<button name="submit" value="3">Next</button> 
</pre> 
<a href="super.php">Home</a> <br /> 
<a href="logout.php">Log out</a> 
</form> 

Answer 1

嘗試添加else正上方mysql_close($connect);。我的猜測是，在初始頁面加載時，您沒有發佈任何值，因此不採取任何操作。這將創建一個默認ID，如果沒有在您的頂部if中定義。

else{ 
$query = mysql_query("SELECT * FROM `Logins` ORDER BY ID ASC LIMIT 1;"); 
while($row = mysql_fetch_array($query)){ 
    $id = $row['ID']; 
    $fname = $row['fname']; 
    $lname = $row['lname']; 
    $email = $row['email']; 
    $login = $row['login']; 
    $password = $row['password']; 
    $super = $row['super']; 
    $foldername = $row['foldername']; 
} 

此外，您if ($_POST['submit']==3)，您需要更改=到>在$query這樣你就可以得到下一個記錄。目前你會選擇相同的ID，而不是更高的。

$query = mysql_query("SELECT * FROM `Logins` WHERE ID > '". $id ."' ORDER BY ID ASC LIMIT 1;"); 

最後，做上一頁/下一頁時，你還需要考慮你將如何處理Previous當你在第一ID和Next當你在最後id，你會返回一個空的結果集從MySQL。