我有一些代码可以从数据库中提取所有结果,并显示与用户搜索相关的结果。我有一些代码可以计算项目数量,并根据与用户搜索相关的项目数生成一定数量的页面。问题如下。如果我全部搜索,我的代码会在11页上显示数据库中的所有内容。如果我搜索汽车,它仍将显示11页,但只有2个结果在标题中有单词car。问题是这些结果显示在第八页上,所有其他页面都是空白的。在搜索期间,所有两个结果与标题中的汽车也显示在第八页上。搜索全部基于项目在数据库中的顺序。这是我目前的代码:
$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>";
}
}
$totalPages = $pageNum;
$currentPage = $page;
$numPagesToShow = 10;
if($currentPage > $totalPages) {
$currentPage = $totalPages;
}
if($numPagesToShow >= $totalPages) {
$numMaxPageLeft = 1;
$numMaxPageRight = $totalPages;
} else {
$pagesToShow = ceil($numPagesToShow/2);
$numMaxPageLeft = $currentPage - $pagesToShow;
$numMaxPageRight = $currentPage + $pagesToShow;
if($numMaxPageLeft <= 0) {
$numMaxPageRight = $numMaxPageRight - $numMaxPageLeft +1;
$numMaxPageLeft = 1;
} elseif($numMaxPageRight >= $totalPages) {
$numMaxPageLeft -= ($numMaxPageRight - $totalPages);
$numMaxPageRight = $totalPages;
}
}
for ($i=$numMaxPageLeft; $i<=$numMaxPageRight; $i++) {
echo "<a id =\"pagenationlink\" href=\"searchresults.php?search=".$terms."&page=".$i."\">".$i."</a>";
}
如何只显示一页上有两个结果的页面,而不是第八页上显示两个相关结果的11页?感谢
答案 0 :(得分:0)
请更新您的代码,如下所示 但是尝试使用mysqli_ ()作为mysql ()被删除
$cond = "";
if(!empty($_POST["search"]))
{
$cond = " write your search condition " ;
}
$start = (($page-1)*5);
$query = mysql_query("SELECT SQL_CALC_FOUND_ROWS * FROM posts where $cond LIMIT $start, 5");
$TotalDataQuery = mysql_query("SELECT FOUND_ROWS() tot;");
$rsVal = mysql_fetch_array($pagesQuery);
$pagesQuery = $rsVal['tot'];
$pageNum = ceil($pagesQuery/5);
while ($row = mysql_fetch_array($query)) {
//continue your code
}