Question

我有一个用于在SQL表中搜索的PHP搜索表单。所有这一切都很有效，但有一件事我想改变。

在搜索之前，整个表格都会显示在屏幕上。我想在搜索后仅提及记录。

有人知道用PHP隐藏表吗？非常感谢提前！

HTML

<form action="" method="post">
<input type="text" name="search" placeholder="Search">
<input type="submit" value="Submit" />
</form>

PHP

<?php 
$host = "******";
$user = "******";
$password = "******";
$database_name = "vangsten";
$pdo = new PDO("mysql:host=$host;dbname=$database_name", $user, $password, array(
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION
));
$search=$_POST['search'];
$query = $pdo->prepare("select * FROM meldingen WHERE soort LIKE '%$search%' OR zone LIKE '%$search%'  LIMIT 0 , 10");
$query->bindValue(1, "%$search%", PDO::PARAM_STR);
$query->execute();
         if (!$query->rowCount() == 0) {

                echo "<table style=\"margin:50px auto;\">"; 
                echo "<tr><td>VISSOORT</td><td>LENGTE</td><td>AANTAL</td><td>ZONE</td></tr>";
            while ($results = $query->fetch()) {
                echo "<tr><td>";            
                echo $results['soort'];
                echo "</td><td>";
                echo $results['lengte'];
                echo "</td><td>";
                echo $results['aantal'];
                echo "</td><td>";
                echo $results['zone'];
                echo "</td></tr>";              
            }
                echo "</table>";        
        } else {
            echo 'Nothing found';
        }
?>

Answer 1

您可以检查用户是否点击了搜索按钮：

if (isset($_POST['search'])) {
    // do your table generation here
}

Answer 2

这是因为即使没有搜索过，您也会最终运行查询：WHERE sort LIKE '%%'

您应该检查是否先通过了搜索

if(array_key_exists('search',$_POST) && !empty($_POST['search'])){
    $search=$_POST['search'];
    $query = $pdo->prepare("select * FROM meldingen WHERE soort LIKE '%$search%' OR zone LIKE '%$search%'  LIMIT 0 , 10");
    $query->bindValue(1, "%$search%", PDO::PARAM_STR);
    $query->execute();
    if (!$query->rowCount() == 0) {
        echo "<table style=\"margin:50px auto;\">"; 
        echo "<tr><td>VISSOORT</td><td>LENGTE</td><td>AANTAL</td><td>ZONE</td></tr>";
        while ($results = $query->fetch()) {
            echo "<tr><td>";            
            echo $results['soort'];
            echo "</td><td>";
            echo $results['lengte'];
            echo "</td><td>";
            echo $results['aantal'];
            echo "</td><td>";
            echo $results['zone'];
            echo "</td></tr>";              
        }
        echo "</table>";        
    } else {
        echo 'Nothing found';
    }
}

array_key_exists('search',$_POST)检查是否有值关键'搜索;'
!empty($_POST['search'])检查它不仅仅是一个空字符串。（你可能想允许这个）

您可以使用isset($_POST['search'])代替array_key_exists('search',$_POST)，但array_key_exists是更好的做法，因为isset如果值为NULL

仍会返回false

在PHP搜索之前隐藏完整的表

2 个答案: