Question

我目前正在尝试在PHP中输出MySQL表的结果。我对基于Web的编程有一般的了解，但还不足以调试我的代码。我知道SQL很好，并且数据库链接到我的网站，这只是将它发布到表格的问题。我将发布代码，并希望得到一些帮助：

<?php 

        $sql = "SELECT player_name AS 'Name',
        position AS 'Position',
        team AS 'Team',
        opp AS 'Opponent'
        FROM `dbname`
        WHERE position = 'QB'";

        $stmt = $db->query($sql);

        if($stmt-> num_rows > 0) {
            echo "<table class='table'>";
            echo "<thead class='thead-inverse'>";
            echo "<tr><th>Name</th><th>Position</th><th>Team</th><th>Opponent</th>";
            echo "</thead>";
            echo "<tbody>";
            while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
                echo "<tr><td>";
                echo $row['Name'];
                echo "</td><td>";
                echo $row['Position'];
                echo "</td><td>";
                echo $row['Team'];
                echo "</td><td>";
                echo $row['Opponent'];
                echo "</td></tr>";

         }
         echo "</tbody>";
         echo "</table>";
      }
      else {
          echo "No Results";
      }

我从中获得的所有结果都是无结果输出。

Answer 1

您忘记使用双引号关闭查询。

<?php 

    $sql = "SELECT player_name AS 'Name',
    position AS 'Position',
    team AS 'Team',
    opp AS 'Opponent'
    FROM `dbname`
    WHERE position = 'QB'";

    $stmt = $db->query($sql);
    $stmt -> execute();

    if($stmt-> num_rows > 0) {
        echo "<table class='table'>";
        echo "<thead class='thead-inverse'>";
        echo "<tr><th>Name</th><th>Position</th><th>Team</th><th>Opponent</th>";
        echo "</thead>";
        echo "<tbody>";
        while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
            echo "<tr><td>";
            echo $row['Name'];
            echo "</td><td>";
            echo $row['Position'];
            echo "</td><td>";
            echo $row['Team'];
            echo "</td><td>";
            echo $row['Opponent'];
            echo "</td></tr>";

     }
     echo "</tbody>";
     echo "</table>";
  }
  else {
      echo "No Results";
  }

添加$stmt -> execute();

您也可以在$stmt -> store_result();部分

之后使用execute

使用PHP从链接的MySQL表输出表

1 个答案: