【问题标题】:I'm getting this error, Fatal error: Call to a member function execute() on a non-object我收到此错误,致命错误:调用非对象上的成员函数 execute()
【发布时间】:2015-10-22 09:00:15
【问题描述】:

当我通过单击分页上的按钮运行代码时,页面上出现以下错误?

致命错误:在第 29 行的 /home/u111693093/public_html/fetch_pages.php 中对非对象调用成员函数 execute()

这是我的代码;

<?php



if(isset($_POST) && isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest'){

    include("config.php");  

    if(isset($_POST["page"])){
        $page_number = filter_var($_POST["page"], FILTER_SANITIZE_NUMBER_INT, FILTER_FLAG_STRIP_HIGH); 
        if(!is_numeric($page_number)){die('Invalid page number!');} 
    }else{
        $page_number = 1; 
    }


    $results = $mysqli->query("SELECT COUNT(*) FROM stories");
    $get_total_rows = $results->fetch_row(); 

    $total_pages = ceil($get_total_rows[0]/$item_per_page);


    $page_position = (($page_number-1) * $item_per_page);



    $results = $mysqli->prepare("SELECT id, heading, article FROM stories ORDER BY id ASC LIMIT $page_position, $item_per_page");
    $results->execute(); 
    $results->bind_result($id, $heading, $article); 


    echo '<ul class="contents">';
    while($results->fetch()){ 
        echo '<li>';
        echo  $id. '. <strong>' .$heading.'</strong> &mdash; '.$article;
        echo '</li>';
    }
    echo '</ul>';

    echo '<div align="center">';

    echo paginate_function($item_per_page, $page_number, $get_total_rows[0], $total_pages);
    echo '</div>';

    exit;
}

function paginate_function($item_per_page, $current_page, $total_records, $total_pages)
{
    $pagination = '';
    if($total_pages > 0 && $total_pages != 1 && $current_page <= $total_pages){ 
        $pagination .= '<ul class="pagination">';

        $right_links    = $current_page + 3; 
        $previous       = $current_page - 3; 
        $next           = $current_page + 1; 
        $first_link     = true; 

        if($current_page > 1){
            $previous_link = ($previous==0)? 1: $previous;
            $pagination .= '<li class="first"><a href="#" data-page="1" title="First">&laquo;</a></li>'; 
            $pagination .= '<li><a href="#" data-page="'.$previous_link.'" title="Previous">&lt;</a></li>'; 
                for($i = ($current_page-2); $i < $current_page; $i++){ 
                    if($i > 0){
                        $pagination .= '<li><a href="#" data-page="'.$i.'" title="Page'.$i.'">'.$i.'</a></li>';
                    }
                }   
            $first_link = false; 
        }

        if($first_link){ 
            $pagination .= '<li class="first active">'.$current_page.'</li>';
        }elseif($current_page == $total_pages){ 
            $pagination .= '<li class="last active">'.$current_page.'</li>';
        }else{ 
            $pagination .= '<li class="active">'.$current_page.'</li>';
        }

        for($i = $current_page+1; $i < $right_links ; $i++){ 
            if($i<=$total_pages){
                $pagination .= '<li><a href="#" data-page="'.$i.'" title="Page '.$i.'">'.$i.'</a></li>';
            }
        }
        if($current_page < $total_pages){ 
                $next_link = ($i > $total_pages) ? $total_pages : $i;
                $pagination .= '<li><a href="#" data-page="'.$next_link.'" title="Next">&gt;</a></li>'; 
                $pagination .= '<li class="last"><a href="#" data-page="'.$total_pages.'" title="Last">&raquo;</a></li>'; 
        }

        $pagination .= '</ul>'; 
    }
    return $pagination; 
}

?>

【问题讨论】:

    标签: php ajax mysqli


    【解决方案1】:

    你的 "$mysqli->prepare" 因为错误返回 false 或 null,所以变量 "$results" 为 null 或 false 并且应该是对象。 您应该修复您的查询,“$mysqli->prepare”将返回对象。

    【讨论】:

    • 感谢您的回复。我是php新手,能告诉我怎么解决吗?
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