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我查看了很多許多指南,其中介紹瞭如何在結果超出後製作編號導航鏈接,比方說25個結果,因此url欄將具有類似的URL ./index.php?page=3和找不到東西容易根據我的代碼來實現:記分牌系統,php和mysql
<body>
<div id = 'scoreboardcontainer'>
<?php
$ranking = mysql_query("SELECT username, points\n"
. "FROM sector0_players\n"
. "WHERE points = (SELECT MAX(points) FROM sector0_players WHERE points< (SELECT MAX(points) FROM sector0_players WHERE points <(SELECT MAX(points) FROM sector0_players)))");
?>
<ul>
<li><a href="#scoreboard" title = "Point scoreboards">Scoreboards</a></li>
<li><a href="#Tour_boards" title = "Tournament socreboards" class = "tbr">Tournament</a></li>
</ul>
<div id="scoreboard">
<div class = "scoreboardtitles">
<div class = "player_score_rankinfl2">Rank</div>
<div class = "player_score_usnamefl2">Username</div>
<div class = "player_score_charnamefl2">Character</div>
<div class = "player_score_pointfl2">Points</div>
</div>
<br />
<hr />
<?php
$scoreboardquery = mysql_query("SELECT * FROM sector0_players ORDER BY points DESC");
while($scoreboard_fetch = mysql_fetch_array($scoreboardquery)){
?>
<div class="global_container_score_class"> <!--Start generating from here-->
<div class = "scoreboard_1l">
<div class = "player_rank">1st</div>
<div class = "player_name"><a href = "profile.php?id=<?php echo $scoreboard_fetch['id'];?>"><?php echo $scbdname = $scoreboard_fetch['username'];?></a></div>
</div>
<div class = "scoreboard_2r">
<div class = "player_charname"><?php echo $scbddspname = $scoreboard_fetch['displayname'];?></div>
<div class = "player_points"><?php echo $scbdpoints = number_format($scoreboard_fetch['points']);?></div>
</div>
<div class = "clear"></div>
</div>
<?php
}
?>
</div>
我不是OOP經驗,我知道如何把它編程,但不能修改它來滿足我的需求。如果這種情況需要OOP,我會很感激,如果有人會爲我提供一個學習高級面向對象的良好開端。 如果沒有,只需輸入我需要的內容(除非必要,否則不需要代碼)。
我真的很抱歉我急於發佈問題。我通過搜索「分頁」關鍵字找到了一套很好的教程。非常感謝。 – 2012-04-11 21:22:30
@哈立德 - 沒問題,自己學習的願望是相當令人欽佩的:] – orourkek 2012-04-11 21:48:09