查询和显示mysql搜索的结果[关闭]

Question

我正在尝试在我的网站上创建一个搜索栏，它将查询 mysql 数据库并以时尚的方式返回结果。搜索将用于查找名称。例如，用户可以输入“John Doe”，任何关于 John Doe 的信息都会被返回。不幸的是，每当我单击提交按钮并且我的站点转换为“search.php”时，页面都是完全空白的。我已经测试了我与数据库的连接并且知道它正在工作。我的代码有什么问题，以至于完全没有出现吗？它甚至不会将用户恢复到 index.html。

我的 index.html 上有一个搜索栏，如下所示：

       <form id="navsearch" class="form-group" role="search" action="search.php" method="post">
            <div class="input-group" display="inline">
                <input name="search" type="text" value="Search" class="form-control" placeholder="Search">
                <span class="input-group-btn">
                    <button class="btn btn-default btn" name="submit" class="btn btn-lg" type="submit">
                        <span class="glyphicon glyphicon-search" name="submit"></span>
                    </button>    
            </div>  
        </form>

这是“search.php”上的 php：

 <?php

 //make connection to database

  $connection = mysqli_connect('xxx.xxxx', 'root', 'hello', 'awstutorial', 3306);

 $search_word = '';
 if (isset($_POST['submit'])) {
   if (!empty($_POST['search'])) {
    $search_word = $_POST['search'];
    $search = mysqli_real_escape_string($connection,      $_POST['search']);

    $query = "SELECT * FROM data WHERE name LIKE '%$search%' ";
    $search_query = mysqli_query($connection, $query);

    // START RESULTS

    if ($result = mysqli_query($connection, $query)) {
        $to_display = "<div>";


        while ($row = mysqli_fetch_array($result)) {
            if ((substr($row['file'], -3) == 'jpg') || (substr($row['file'], -3) == 'JPG') || (substr($row['file'], -3) == 'PNG') || (substr($row['file'], -3) == 'png') || (substr($row['file'], -3) == 'GIF') || (substr($row['file'], -3) == 'gif'))
                $img_url = "uploads/" . $row['file'];
            else {
                if ((strtolower(substr($row['file'], -3) == 'png')) || (strtolower(substr($row['file'], -3) == 'doc')) || (strtolower(substr($row['file'], -3) == 'docx') ))
                    $img_url = "uploads/pdf_logo.png";
                else
                    $img_url = "uploads/no_doc.png";
            }
            $to_display .= "<a href='story.php?id=".$row['id']."'  class='col-md-3 asdivs portfolio-box fancybox fancybox.iframe'><img alt='' class='img-responsive' src='" . $img_url . "'><div class='portfolio-box-caption'>
                            <div class='portfolio-box-caption-content'>
                                <div class='project-category text-faded'>
                                    Read
                                </div>
                                <div class='project-name'>"
                        . $row['name'] . "'s Story
                                </div>
                            </div>
                        </div>" . $row['name'] . " - " . $row['email'] . "</a>";

        }
    }

    $to_display .= "</div><div style='clear:both;'></div>";
} else {
    header("Location: index.html");
   }
 }

?>

Answer 1

在提交表单时，我不确定fashionable way 的含义是什么——我能想到的最好的方法是它指的是一个 ajax 请求，它在不重新加载页面的情况下提交搜索，结果是然后以某种方式在同一页面上向用户显示？由于没有 ajax 的证据，所以很难说。

正如我在评论中提到的，您的查询指的是$search_query，但通过记录集的迭代指的是$result——这些变量必须是同一个变量，所以这可能就是你面对空白屏幕的原因——那和事实上变量$to_display 从未回显到浏览器。

如有疑问，请使用error_reporting 突出显示可能出错的地方。

<?php
    /* Sessions? */
    session_start();

    error_reporting( E_ALL ); /* Change this for live site! */

    if( $_SERVER['REQUEST_METHOD']=='POST' ){

        $conn = mysqli_connect('xxx.xxxx', 'root', 'hello', 'awstutorial', 3306);
        $exts_img=array('jpg','png','gif','jpeg');
        $exts_doc=array('pdf','doc','docx');
        $html=array();  /* use this to construct output. More efficient that string concatenation. */

        if ( isset( $_POST['submit'] , $_POST['search'] ) && !empty( $_POST['submit'] ) && !empty( $_POST['search'] ) ) {

            /* It is better to use prepared statements or pdo */
            $search = mysqli_real_escape_string( $conn,$_POST['search'] );  
            $sql = "SELECT * FROM `data` WHERE `name` LIKE '%".$search."%';";
            $results = mysqli_query( $conn, $sql );


            if( $results && mysqli_num_rows( $results ) > 0 ){
                $html[]="<div>";

                while( $row = mysqli_fetch_object( $results ) ) {

                    $id     =   $row->id;
                    $file   =   $row->file;
                    $email  =   $row->email;
                    $name   =   $row->name;
                    $ext    =   strtolower( pathinfo( $file, PATHINFO_EXTENSION ) );



                    if( in_array( $ext, $exts_img ) ) {
                        $img_url = "uploads/" . $file;
                    } else {
                        if( in_array( $ext, $exts_doc ) ) $img_url='uploads/pdf_logo.png';
                        else $img_url='uploads/no_doc.png';
                    }


                    $html[]="
                        <a href='story.php?id=".$id."' class='col-md-3 asdivs portfolio-box fancybox fancybox.iframe'>
                            <img alt='' class='img-responsive' src='" . $img_url . "' />
                            <div class='portfolio-box-caption'>
                                <div class='portfolio-box-caption-content'>
                                    <div class='project-category text-faded'>Read</div>
                                    <div class='project-name'>{$name}'s story</div>
                                </div>
                            </div>
                            {$name} - {$email}
                        </a>";
                }
                $html[]="</div><div style='clear:both'></div>";
            }
            /* No results */
            $html[]='<h1>Sorry, no results...</h1>';

        } else {
            header("location: index.html");
        }

        /* free resources */
        mysqli_free_result( $results );
        mysqli_close( $conn );  

        /* output stuff to the browser */
        echo implode( PHP_EOL, $html );
    }
?>