html表

Question

Var转储结果

我将MySQL中的数据提取到HTML表格中：

<div id="toggleDiv" class="">
  <div class="box-body" id="toggleDiv_2">
    <div class="row">
      <div class="col-md-6">
        <?php
        $select_patient_info_cash =
            "SELECT * FROM patient_info WHERE id_logged = :id_logged".
            " AND patient_id = :patient_id AND payment_type = :pt";
        $select_patient_info_cash_stmt = 
            $conn->prepare($select_patient_info_cash);
        $select_patient_info_cash_stmt->bindValue(":id_logged", $id_logged);
        $select_patient_info_cash_stmt->bindValue(":patient_id", $patient_id);
        $select_patient_info_cash_stmt->bindValue(":pt", "cash");
        $select_patient_info_cash_stmt->execute();
        $select_patient_info_cash_stmt->fetch(PDO::FETCH_ASSOC);
        $select_patient_info_cash_stmt_count =
            $select_patient_info_cash_stmt->rowCount();
        if($select_patient_info_cash_stmt_count > 0){ ?>
        <table style="text-align:center"
               class="table table-bordered table-striped table-hover">
          <thead>
            <tr>
              <th>Project Description</th>
              <th>Project Cost</th>
              <th>Date Of Pay</th>
            </tr>
          </thead>
            <?php foreach ($select_patient_info_cash_stmt as $cash) { ?>
            <tr>
              <td><?php echo $cash['project'] ?></td>
              <td><?php echo $cash['project_cost'] ?></td>
              <td><?php echo $cash['date_now'] ?></td>
            </tr>
            <?php } ?>
        </table>
    <?php } else { ?>

    <?php } ?>
</div><!-- /.col -->

现在，我为患者信息中包含payment_type != cash，<thead>未显示的数据的用户进行测试。但是，当我在payment_type=cash <thead>显示<thead>时，但没有数据回显到行中。

它应该在dialog.setOnDissmissListener(){ void onDismiss(){ inputManager.hideSoftInputFromWindow(getCurrentFocus().getWindowToken(), 0); inputManager.hideSoftInputFromInputMethod(getCurrentFocus().getWindowToken(), 0); } } dialog.dismiss(); 之后显示2个新行，我无法弄清楚为什么数据不会显示在页面上

Answer 1

我想你在准备好的陈述中错过了->fetch()。根据PHP文档：

<?php
$stmt = $dbh->prepare("SELECT * FROM REGISTRY where name = ?");
if ($stmt->execute(array($_GET['name']))) {
  while ($row = $stmt->fetch()) {
    print_r($row);
  }
}
?>

因此，您需要将代码修改为：

<?php while ($cash = $select_patient_info_cash_stmt->fetch()) {  ?>
  <tr>
    <td><?php echo $cash['project'] ?></td>
    <td><?php echo $cash['project_cost'] ?></td>
    <td><?php echo $cash['date_now'] ?></td>
  </tr>
<?php } ?>

我还建议您使用MVC框架或某种模板引擎。混合PHP和HTML是一种非常糟糕的做法。

参考文献：http://php.net/manual/en/pdo.prepared-statements.php