Question

目前，我可以显示数据库中的图像，但是如何使用搜索栏搜索特定图像并显示到网页中，这是最好的方法，有人可以通过此代码帮助我，因为我PHP的基本学习者..

<html>
<head>
    <meta charset="UTF-8">
    <title></title>
</head>
<body>
    <?php
    mysql_connect("localhost","root","");
    mysql_select_db("display_images");
    $res=mysql_query("select *from table1");
    echo "<table>";

    while($row=mysql_fetch_array($res))
    {
        echo "<tr>";
        echo "<td>";?> <img src="<?php echo $row["images1"];?>" height="100" width="100"> <?php echo "</td>";
        echo "<td>"; echo $row["name"]; echo"</td>";

        echo "</tr>";
    }
    echo "</table>";
    ?>


    </table>
</body>

Answer 1

正如我所说的mysql_*现在已被弃用并删除了库，我将为您提供mysqli_*的示例示例

<html>
<head>
    <meta charset="UTF-8">
    <title></title>
</head>
<body>
    <form method="POST">
        <input type="text" name="image_name" placeholder="type a name and hit submit button to see particular image">
        <input type="submit" name="submit" value="submit">
    </form>
    <table>
        <?php

        $connection = mysqli_connect( "localhost", "root", "", "display_images" ) or die( mysqli_connect_error() );

        $query = "SELECT * FROM table1";
        if( !empty( $_POST['image_name'] ) ){
            $image_name = $_POST['image_name'];
            $query = "SELECT * FROM table1 WHERE images1 LIKE %$image_name%"; 
            //you can do $image_name% or %$image_name based on your requirement
        }
        $result = mysqli_query( $connection, $query ) or die( mysqli_error( $connection ) );

        while( $row = mysqli_fetch_assoc( $res ) ){ 
        ?>
            <tr>
                <td>
                    <img src="<?php echo $row["images1"];?>" height="100" width="100"> 
                </td>
                <td><?php echo $row["name"];?></td>

            </tr>
    <?php }?>
    </table>
</body>

注意： ：使用准备好的语句来防止SQL注入

参考：-

mysqli::prepare

PDO::prepare

Answer 2

尝试使用此示例代码

this

或尝试

this.myFormGroup

通过搜索栏获取图像并显示在网页上

2 个答案: