[英]HTML and PHP generate pagination links
我有一些PHP和html代碼可從數據庫中加載結果。 每頁顯示五個結果。 假設我有1000頁。 所有這些頁面的鏈接都將在屏幕上消失。 Google遇到了此問題,但他們僅通過顯示當前鏈接以及向后5個鏈接和向前5個鏈接來解決此問題。 我想做這樣的事情。 我不想顯示指向各個頁面的100個鏈接。 假設用戶在第100頁上。我想顯示第100頁的鏈接以及從95到105的鏈接。我該怎么做? 到目前為止,這是我的代碼:
$page = $_GET["page"];
$pagesQuery = mysql_query("SELECT count(id) FROM(`posts`)");
$pageNum = ceil(mysql_result($pagesQuery, 0)/5);
$start = (($page-1)*5);
$currentname = mysql_query("SELECT * FROM posts LIMIT $start, 5");
while ($row = mysql_fetch_array($currentname)) {
//recieve relevant data.
$title = $row[0];
$desc = $row[13];
$ID = $row[6];
$views = $row[3];
$user = $row[7];
//fetch the last id from accounts table.
$fetchlast1 = mysql_query("SELECT * FROM allaccounts WHERE id=(SELECT MAX(id) FROM allaccounts)");
$lastrow1 = mysql_fetch_row($fetchlast1);
$lastid1 = $lastrow1[6];
//acquire the username of postee.
for ($i1=1; $i1 <= $lastid1; $i1++) {
$currentname1 = mysql_query("SELECT * FROM allaccounts WHERE id=$user");
while ($row1 = mysql_fetch_array($currentname1)) {
$username1 = $row1[0];
}
}
//Format Title, description and view count.
$title2 = rtrim($title);
$donetitle = str_replace(" ", "-", $title2);
$url = "articles/".$ID."/".$donetitle."";
$donetitle = strlen($title) > 40 ? substr($title,0,40)."..." : $title;
$donedesc = '';
if(strlen($desc) > 150) {
$donedesc = explode( "\n", wordwrap( $desc, 150));
$donedesc1 = $donedesc[0] . '...';
}else{
$donedesc1 = $desc;
}
$finviews = number_format($views, 0, '.', ',');
//Give relevant results
if(stripos($title, $terms) !== false || stripos($desc, $terms) !== false || stripos($username1, $terms) !== false){
if($row[10] == null){
$SRC = "img/tempsmall.jpg";
}else{
$SRC ="generateThumbnailSmall.php?id=$ID";
}
echo "<div id = \"feature\">
<img src=\"$SRC\" alt = \"article thumbnail\" />
</div>
<div id = \"feature2\">
<a href= \"$url\" id = \"titletext\" alt = \"article title\">$donetitle</a>
<p id=\"resultuser\" >$username1</p>
<p id=\"resultp\">$donedesc1</p>
<a href = \"sendflag.php?title=$title&url=$url&id=$ID&userid=$user\" id = \"flag\" alt = \"flag\"><img src=\"img/icons/flag.png\"/></a><b id=\"resultview\">$finviews views</b>
</div>
<div id = \"border\"></div>";
}
}
for ($j=1; $j < $pageNum; $j++) {
echo "<a id =\"\" href=\"searchresults.php?search=".$terms."&page=".$j."\">".$j."</a>";
}
您要更改此內容:
for ($j=1; $j < $pageNum; $j++) {
echo "<a id =\"\" href=\"searchresults.php?search=".$terms."&page=".$j."\">".$j."</a>";
}
在這里,您列出所有鏈接$ j <$ pageNum,並且只列出10:$ j <= 10起始$ j = $ currentPage或$ j = $ currentPage-如果$ currentPage> 5,則為5
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.