Php 如何查询一个表但忽略某些列，如ID="；1“&&引用；2“；_Php_Html_Mysql_Css

Php 如何查询一个表但忽略某些列，如ID="；1“&&引用；2“；

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Php 如何查询一个表但忽略某些列，如ID="；1“&&引用；2“；,php,html,mysql,css,Php,Html,Mysql,Css,我试图查询Mysql表以返回某些行。我将它设置为显示表中的每一行，但是我希望它不显示ID=1 ID=2 ID=3的行这是我用来查询数据库的search.php <?php $responce = "Please enter a search term"; if(!isset($_POST['searchterm'])) { ?> <br> <br> <h3 class="t

我试图查询Mysql表以返回某些行。我将它设置为显示表中的每一行，但是我希望它不显示ID=1 ID=2 ID=3的行

这是我用来查询数据库的search.php

<?php
$responce = "Please enter a search term";
        if(!isset($_POST['searchterm'])) {
        ?>
        <br>
        <br>
        <h3 class="text-center" style="color:white"><u><?php echo $responce;?></u></h3><?php
    }


    $names = array();
    $link = array();


    if(isset($_POST['searchterm'])) {
        mysql_connect("localhost", "root", "Password");
        mysql_select_db("videos");
        $search = mysql_real_escape_string(trim($_POST['searchterm']));


        $find_videos = mysql_query("SELECT * FROM `videos` WHERE `keywords` LIKE'%$search%'");
        while ($row = mysql_fetch_assoc($find_videos)) {
            $names[] = $row['name'];
            $link[] = $row['link'];
        }

    }
    $deletethen = "DELETE  FROM videos WHERE name = ''";

    include('session.php');
?>
<!DOCTYPE html>
<html>
  <head>
  <link href="http://vjs.zencdn.net/4.12/video-js.css" rel="stylesheet">
  <link rel="icon" type="image/ico" href="images/favicon.ico">
<script src="http://vjs.zencdn.net/4.12/video.js"></script>
<link href="http://vjs.zencdn.net/4.12/video-js.css" rel="stylesheet">
<script src="http://vjs.zencdn.net/4.12/video.js"></script>
<style type="text/css">
  .vjs-default-skin .vjs-control-bar { font-size: 125% }
</style>
    <meta charset="utf-8">
    <title>Network TV | search</title>
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <link rel="stylesheet" href="css/bootstrap.css" />
    <link href="css/font-awesome.css" rel="stylesheet" />
    <link href="css/3.1.1/animate.css" rel="stylesheet" />
    <link rel="stylesheet" href="css/styles.css" />
  </head>
  <body style="background:center no-repeat fixed url('images/bg.jpg'); background-size: cover;overflow-x: hidden; background-color: black">
    <nav class="navbar navbar-default navbar-fixed-top">
      <div class="container">
        <div class="navbar-header">
          <button type="button" class="navbar-toggle collapsed" data-toggle="collapse" data-target="#navbar" aria-expanded="false" aria-controls="navbar">
            <span class="sr-only">Toggle navigation</span>
            <span class="icon-bar"></span>
            <span class="icon-bar"></span>
            <span class="icon-bar"></span>
          </button>
          <a class="navbar-brand" href="#">Network TV</a>
        </div>
        <div id="navbar" class="navbar-collapse collapse">
          <ul class="nav navbar-nav">
            <li class=""><a href="index.php">Home</a></li>
          </ul>
        </div><!--/.nav-collapse -->
      </div>
    </nav>
        <section style="padding-top:100px; width:100%"class="container-fluid" id="section2">
                    <h1 class="text-center" style="color:white">Network TV</h1>
                <h3 class="text-center" style="color:white"><u>Search results</u></h3>




                                        <table align="center" class="table" style="color: white; width:50%">
                                                <thead>
                                                </thead>
                                                <tbody>
                                                    <tr>

                                                        <?php
                                                        $i=0;
                                                        foreach($names as $name) {
                                                         echo '<tr>
                                                              <td><h4>' . $name . '</h4></td>
                                                              <td><a href="' .$link[$i]. '"><strong><h4><u>Watch now!</u></h4></strong></a></td></tr>';
                                                        $i++;
                                                                }
                                                                ?>
                                                    </tr>
                                                </tbody>
                                            </table>
        </section>
    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
    <script src="//maxcdn.bootstrapcdn.com/bootstrap/3.3.4/js/bootstrap.min.js"></script>

    <script src="js/scripts.js"></script>
</html>






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从id不在（1,2,3）中的表中选择*

这应该就够了。

这段代码太过时了。请参阅关于现代API（PDO、mysqli）和准备好的语句。更改你的密码哦，是的，我忘了你可以组合它们…facepalm…我需要将你的代码添加到我现有的查询中$find_videos=mysql_query（“从

videos

WHERE

keywords

LIKE“%$search%”中选择*”；我该怎么做？@OliverKetley

$find\u videos=mysql\u query（“从视频中选择*，其中关键字“%$search%”和id不在（1,2,3）”中）谢谢你，它工作得很好，这么简单的修复程序，真不敢相信我竟然弄不明白。