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我正在開發一個簡單的社交網絡系統。這是我的代碼，它顯示登錄用戶的所有朋友列表，並在每個姓名的前面有一個Unfriend按鈕。我需要一個只能顯示5個朋友的頁面。使用頁面底部的下一個和上一個鏈接。第一頁加載沒有「上一頁」鏈接，最後一頁不應該有「下一頁」鏈接。任何幫助，將不勝感激。用於社交網絡系統的php分頁
<?php
require_once("settings.php");
$conn = @mysqli_connect($host,$user,$pswd) or die('Failed to connect to server');//connecting to the database
@mysqli_select_db($conn,$dbnm) or die('Database not available');//unless error
//if ((isset($_POST ["$log "]) && (!empty ($log)))){
//getting profile names and ids of the friends of the logged in user
$query = "SELECT friends.profile_name,friends.friend_id FROM friends INNER JOIN myfriends ON friends.friend_id=myfriends.friend_id1
WHERE myfriends.friend_id2='$friendID'";
//unless error
$results = @mysqli_query($conn, $query) or die("<p>Unable to execute the query 011.</p>". "<p>Error code " . mysqli_errno($conn). ": " . mysqli_error($conn)) . "</p>";;
$count=mysqli_num_rows($results);//row count
$row = mysqli_fetch_row($results);//fetching a row from db and soring it inside a variable
if(isset($_GET ["unfriend"]))//if unfriend variable is set
{
$unfriend=$_GET["unfriend"];
//echo $friend_id = $row[1];
//$query ="DELETE FROM myfriends WHERE (friend_id1=".$unfriend." and friend_id2=".$friendID.") OR (friend_id1=".$friendID." and friend_id2=".$unfriend.")";
//deleting the mutual friendship
$query ="DELETE FROM myfriends WHERE (friend_id1=$unfriend and friend_id2=$friendID) OR (friend_id1=$friendID and friend_id2=$unfriend)";
$results = @mysqli_query($conn, $query) or die("<p>Unable to execute the query 011.</p>". "<p>Error code " . mysqli_errno($conn). ": " . mysqli_error($conn)) . "</p>";;
//getting num of friends of the deleted friend
$query = "SELECT num_of_friends FROM friends WHERE friend_id='$unfriend'";
$results = @mysqli_query($conn, $query) or die("<p>Unable to execute the query 013.</p>". "<p>Error code " . mysqli_errno($conn). ": " . mysqli_error($conn)) . "</p>";;
$row = mysqli_fetch_row($results);
$friendcount=$row[0];
$friendcount--;
//updating the number of friends of the deleted friend
$query2 ="UPDATE friends
SET num_of_friends='$friendcount'
WHERE friend_id='$unfriend'";
$results2= @mysqli_query($conn, $query2) or die("<p>Unable to execute the query.</p>". "<p>Error code " . mysqli_errno($conn). ": " . mysqli_error($conn)) . "</p>";;
//getting profile names and ids of the friends of the logged in user
$query = "SELECT friends.profile_name,friends.friend_id,friends.num_of_friends FROM friends INNER JOIN myfriends ON friends.friend_id=myfriends.friend_id1
WHERE myfriends.friend_id2='$friendID'";
$results = @mysqli_query($conn, $query) or die("<p>Unable to execute the query 013.</p>". "<p>Error code " . mysqli_errno($conn). ": " . mysqli_error($conn)) . "</p>";;
$count=mysqli_num_rows($results);
//updating the number of friends of the logged in user.
$query1 ="UPDATE friends
SET num_of_friends='$count'
WHERE friend_id='$friendID'";
$results1= @mysqli_query($conn, $query1) or die("<p>Unable to execute the query 012.</p>". "<p>Error code " . mysqli_errno($conn). ": " . mysqli_error($conn)) . "</p>";;
//updating the session variable
$numoffriends=$count;
$_SESSION["numoffriends"]=$numoffriends;
$row = mysqli_fetch_row($results);
}
echo "<p>Total Number of friends is $count</p>";
echo "<table width='50%' border='1'>";
while ($row) {
echo "<tr><td>{$row[0]}</td>";
?>
<td><button onclick = "window.location.href='friendlist.php?unfriend=<?php echo $row[1];?>'">Unfriend</button></td></tr>
<?php
$row = mysqli_fetch_row($results);
}
echo "</table>";
echo"<p><a href =\"friendadd.php\">Add Friends</a><a href =\"logout.php\">Log Out</a></p>";
mysqli_free_result($results);
mysqli_close($conn);
?>
請格式化你的代碼 –