由于警告，带有分页（使用Php）的Php搜索代码未生成导航链接。如何更正错误？

我是php编程新手，我已经编写了php代码从数据库中搜索数据，然后用分页链接显示数据，以便在搜索结果过多的情况下轻松导航。但是由于上述警告，分页链接将不会显示。查询没有生成计数结果以使my$limit变量为true，从而仅显示第一页。这是下面的代码，请您帮我更正错误，以便显示分页链接

<?php
include_once("dbconnect.php");

if(isset($_GET["search"]))  
                 {  
                    $condition = '';  
                     $search_query = explode(" ", $_GET["search"]);  
                      foreach($search_query as $text)  
                      {  
                           $condition .= "search LIKE '%".mysqli_real_escape_string($dbconnect, $text)."%' OR ";  
                      }  


$sql = "SELECT COUNT(stockID) FROM stock WHERE search LIKE ". $condition ."GROUP BY stockID";
$query = mysqli_query($dbconnect, $sql);
$row = mysqli_fetch_row($query);

$rows = $row[0];

$page_rows = 12;

$last = ceil($rows/$page_rows);

if($last < 1){
$last = 1;
}

$pagenum = 1;

if(isset($_GET['pn'])){
$pagenum = preg_replace('#[^0-9]#', '', $_GET['pn']);
}

if ($pagenum < 1) { 
$pagenum = 1; 
} else if ($pagenum > $last) { 
$pagenum = $last; 
}

$limit = 'LIMIT ' .($pagenum - 1) * $page_rows .',' .$page_rows;
 $condition = ''; 
                      $search_query = explode(" ", $_GET["search"]);  
                      foreach($search_query as $text)  
                      {  
                           $condition .= "search LIKE 
'%".mysqli_real_escape_string($dbconnect, $text)."%' OR ";  
                      }  
                      $condition = substr($condition, 0, -4);  
                      $sql= "SELECT * FROM stock WHERE " . $condition." 
ORDER BY stockID ASC $limit";  

$query = mysqli_query($dbconnect, $sql);

$textline1 = "(<b>$rows</b>)";
$textline2 = "Page <b>$pagenum</b> of <b>$last</b>";
$paginationCtrls = '';
$search_query = $_GET['search'];
if($last != 1){

if ($pagenum > 1) {
    $previous = $pagenum - 1;
    $paginationCtrls .= '<a href="'.$_SERVER['PHP_SELF'].'?
pn='.$previous.'&search='.$search_query.'">Previous</a> &nbsp; &nbsp; ';

    for($i = $pagenum-4; $i < $pagenum; $i++){
        if($i > 0){
            $paginationCtrls .= '<a href="'.$_SERVER['PHP_SELF'].'?
pn='.$i.'">'.$i.'</a> &nbsp; ';
        }
    }
}

$paginationCtrls .= ''.$pagenum.' &nbsp; ';

for($i = $pagenum+1; $i <= $last; $i++){
    $paginationCtrls .= '<a href="'.$_SERVER['PHP_SELF'].'?
pn='.$i.'&search='.$search_query.'">'.$i.'</a> &nbsp; ';
    if($i >= $pagenum+4){
        break;
    }
}

if ($pagenum != $last) {
    $next = $pagenum + 1;
    $paginationCtrls .= ' &nbsp; &nbsp; <a href="'.$_SERVER['PHP_SELF'].'?
pn='.$next.'&search='.$search_query.'">Next</a> ';
}
}
$list = '';
$lists = '';

while($row = mysqli_fetch_array($query)){
$id = $row["stockID"];
$name = $row["stockName"];
$image = $row["thumbnail"];
$description = $row["description"];
$topline = $row['topline'];

$lists .= '<div class="col-md-3" style="margin-bottom:10px;">    
        <div class="panel panel-info"><a href="index.php?
page=item&stockID='.$id.'&sub_category="> 

    <div class="panel-body">

    <img class="img-responsive" style="height:100px;" 
src="resources/stocks_images/'.$image.'"  />
     </div>
    <div class="panel-footer" style="height:100px;">'.$name.'

    </div></a>
    </div>
    </div>';


                 }
                 }

mysqli_close($dbconnect);
?>

---

$dbconnect
在脚本中未定义。还要重写您的查询：

$sql=“从类似搜索的股票中选择COUNT（stockID）”$条件：“按stockID分组”我在“分组依据”之前添加了一个空格。希望这会有所帮助。
它产生了这个错误-->>警告：mysqli_fetch_row（）希望参数1是mysqli_result，在第17行的C:\xampp\htdocs\HTML\searchresult.php中给出的布尔值，$dbconnect变量是定义的，因为当我搜索结果时，而不是分页链接时，你仍然会得到警告吗？（添加空格后？）