Question

所以，我有这段代码可以为我返回一个带有正确db值的简单表：

<?php
    echo "<table border='1'>
          <tr>
            <th>Código</th>
            <th>Nome</th>
            <th>Indicou</th>
          </tr>";

    while($coluna_bd_tabela = mysqli_fetch_array($sql_indicador_resul)){
        echo "<tr>";
        echo "<td>" . $coluna_bd_tabela['usu_codigo'] . "</td>";
        echo "<td>" . $coluna_bd_tabela['usu_nome'] . "</td>";
        echo "<td>" . $coluna_bd_tabela['usu_indicador_codigo'] . "</td>";
        echo "</tr>";
    }

    echo "</table>";
?>

这是没有查询功能的样式表：

<table class="table table-striped projects">
  <thead>
    <tr>
      <th style="width: 1%">#</th>
      <th style="width: 20%">Nome</th>
      <th>Membros Recentes</th>
      <th>Project Progress</th>
      <th>Status</th>
      <th style="width: 20%">#Edit</th>
    </tr>
  </thead>

  <tbody>
    <tr>
      <td>echo $coluna_bd_tabela['usu_codigo']</td>
      <td>
        <a> echo $coluna_bd_tabela['usu_nome']</a>
        <br />
        <small>echo $coluna_bd_tabela['usu_indicou']</small>
      </td>
      <td>
        <ul class="list-inline">
          <li>
            <img src="images/user.png" class="avatar" alt="Avatar">
          </li>
          <li>
            <img src="images/user.png" class="avatar" alt="Avatar">
          </li>
          <li>
            <img src="images/user.png" class="avatar" alt="Avatar">
          </li>
          <li>
            <img src="images/user.png" class="avatar" alt="Avatar">
          </li>
        </ul>
      </td>
      <td class="project_progress">
        <div class="progress progress_sm">
          <div class="progress-bar bg-green" role="progressbar" data-transitiongoal="57"></div>
        </div>
        <small>57% Complete</small>
      </td>
      <td>
        <button type="button" class="btn btn-success btn-xs">Success</button>
      </td>
      <td>
        <a href="#" class="btn btn-primary btn-xs"><i class="fa fa-folder"></i> View </a>
        <a href="#" class="btn btn-info btn-xs"><i class="fa fa-pencil"></i> Edit </a>
        <a href="#" class="btn btn-danger btn-xs"><i class="fa fa-trash-o"></i> Delete </a>
      </td>
    </tr>
  </tbody>

我想显示带有查询值的风格化表，但这毁了我。数据库连接很好，这些是包含连接上的查询，它们也可以正常工作，如简单表所示：

$sql_indicador = "SELECT * FROM esc_usuarios WHERE usu_indicador_codigo = '" . $_SESSION['codigo'] . "'";
$sql_indicador_resul = mysqli_query($conexao, $sql_indicador);

Answer 1

要提供完整的解决方案：

<table class="table table-striped projects">
    <thead>
        <tr>
            <th style="width: 1%">#</th>
            <th style="width: 20%">Nome</th>
            <th>Membros Recentes</th>
            <th>Project Progress</th>
            <th>Status</th>
            <th style="width: 20%">#Edit</th>
        </tr>
    </thead>

    <?php while($coluna_bd_tabela = mysqli_fetch_array($sql_indicador_resul)){ ?>
            <tbody>
                <tr>
                    <td><?php echo $coluna_bd_tabela['usu_codigo']; ?></td>
                    <td>
                        <a><?php echo $coluna_bd_tabela['usu_nome']; ?></a>
                        <br />
                        <small><?php echo $coluna_bd_tabela['usu_indicou']; ?></small>
                    </td>
                    <td>
                        <ul class="list-inline">
                            <li>
                                <img src="images/user.png" class="avatar" alt="Avatar">
                            </li>
                            <li>
                                <img src="images/user.png" class="avatar" alt="Avatar">
                            </li>
                            <li>
                                <img src="images/user.png" class="avatar" alt="Avatar">
                            </li>
                            <li>
                                <img src="images/user.png" class="avatar" alt="Avatar">
                            </li>
                        </ul>
                    </td>
                    <td class="project_progress">
                        <div class="progress progress_sm">
                            <div class="progress-bar bg-green" role="progressbar" data-transitiongoal="57"></div>
                        </div>
                        <small>57% Complete</small>
                    </td>
                    <td>
                        <button type="button" class="btn btn-success btn-xs">Success</button>
                    </td>
                    <td>
                        <a href="#" class="btn btn-primary btn-xs"><i class="fa fa-folder"></i> View </a>
                        <a href="#" class="btn btn-info btn-xs"><i class="fa fa-pencil"></i> Edit </a>
                        <a href="#" class="btn btn-danger btn-xs"><i class="fa fa-trash-o"></i> Delete </a>
                    </td>
                </tr>
            </tbody>
    <?php } ?>
</table>

您忘记在echo语句周围使用<?php和?>开头和结尾标签。此外，您在每个语句末尾都错过了;。我也将表的开头和结尾从PHP的echo中移出了，因为我认为这样看起来更清楚了。

如何用php和sql连接html样式表

1 个答案: