從 MySQL 表中選擇所有列

Question

我很難從某些表中回顯某些列的內容取決於用戶選擇 - 現在我只有一個表和下面的代碼工作完美但當然其他表沒有相同的列名和值所以我嘗試修改它 - 到目前為止我被卡住了。我怎樣才能讓它動態？

謙虛地尋求一些幫助:)

<?php
function showTables($table, $order) {
    global $conn;
    $query = ("SELECT * FROM $table ORDER BY $order ASC LIMIT 0, 30 ");
    $returnTables = mysqli_query($conn, $query);

    while ($row = mysqli_fetch_assoc($returnTables)) {
        $id = $row['id'];
        $name = $row['name'];
        $surrname = $row['surrname'];
        $email = $row['email'];
        $firstTime = $row['firstTime'];
        $comment = $row['comment'];

        echo "<tr>";    
            echo "<td>$name</td>";
            echo "<td>$surrname</td>";
            echo "<td>$email</td>";
            echo "<td>$firstTime</td>";
            echo "<td>$comment</td>";         
            echo "<td class='no-print'><a href='admin.php?delete={$id}'>Usuń</a></td>"; 
        echo "</tr>";
    }
}
?>

Answer 1

簡單地從您的 TABLE 中獲取列名，

例如，

$sql = "SHOW COLUMNS FROM your-table";
$result = mysqli_query($conn,$sql);

while($row = mysqli_fetch_array($result)){
  echo $row['Field']."<br>";
}

編輯 2：

完整的動態示例（需要格式化 tr 和 td ）

function showTables($table, $order) {

    global $conn;

    $query = "SELECT * FROM $table ORDER BY $order ASC LIMIT 0, 30 ";
    $returnTables = mysqli_query($conn, $query);

    $sql_columns_name = "SHOW COLUMNS FROM $table";

    $sql_columns_result = mysqli_query($conn,$sql_columns_name);

    $column_arr = array();

    while($row = mysqli_fetch_array($sql_columns_result)){
      $column_arr[] = $row['Field'];
    }

    echo '<table>';

    while ($row = mysqli_fetch_array($returnTables)) {

        for ($i=0; $i < count($column_arr); $i++) {
            echo "<tr>";  
              echo "<td>".$row[$column_arr[$i]]."</td><br>";
            echo "</tr>";
        }

    }

    echo '</table>';
}