php - mysqli 根据行值显示行 - (如果行值=1 显示，否则隐藏)

Question

我想在一个表上显示数据库中的 30 行，但只显示已批准的条目。 此“批准的条目”是为具有值“1 表示已批准”和“0 表示未批准”的行定义的

这是我需要更改的代码:

<?php
$servername = "localhost";
$username = "username ";
$password = "password";
$dbname = "name";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT id, label, title_id, season, episode, approved FROM links order by id desc LIMIT 30 OFFSET 1";
$result = $last_id = $conn->query($sql);


if ($result->num_rows > 0)
 {
    echo "<table><tr><th>ID</th><th>Audio</th><th>URL</th><th>Temporada</th><th>Episodio</th><th>Aprobado</th></tr>";
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "<tr><td>".$row["id"]."</td><td>".$row["label"]."</td>";
        echo "<td><a href=";
        if (empty($row['episode'])) {
     echo '/peliculas-online/'.$row["title_id"];
    }
    else {
     echo '/series-online/'.$row['title_id'].'/seasons/'.$row['season'].'/episodes/'.$row["episode"];
        }
        echo ">".$row["title_id"]."</a></td>";
        echo "<td>".$row["season"]."</td><td>".$row["episode"]."</td><td>".$row["approved"]."</td></tr>";
    }
    echo "</table>";
} else {
    echo "0 results";
}
$conn->close();
?>

如您所见，此代码列出了所有(已批准或未批准)...如何获得类似的内容:

if row approved=1 显示行，否则隐藏(隐藏所有而不仅仅是已批准的行)

如果条目未获批准，则需要隐藏此表行中的所有条目 -> id-label-title_id-season-episode-approved

谢谢

Answer 1

你应该知道 MySQL 中 SELECT 的 WHERE 子句。

类似于:

"SELECT id, label, title_id, season, episode, approved 
 FROM links 
 WHERE approved = 1 
 ORDER BY id desc 
 LIMIT 30 OFFSET 1";

引用:

• https://dev.mysql.com/doc/refman/5.0/en/select.html