如何使用mysql数据填充select标签，然后在html表中显示它？

Question

我在使用HTML / PHP表时遇到问题我正在尝试填充html选择标记 使用数据库中的数据并将其显示在表中。但是具有数据库数据的select标记仅显示在表的第一行中。 这是我的代码：

$result2  = mysqli_query($con,"SELECT vozac FROM vozaci");

$result = mysqli_query($con,"SELECT registracija,tip,status,vozac,napomena2,IDV FROM status_vozila order by status desc")

or die("Error: ".mysqli_error($con));

while($row = mysqli_fetch_array($result))
  {
  echo '<form action="" method="post">';
  echo '<tr>';
  echo '<td>'.$row['registracija'].'</td>';
  echo '<td>'.$row['tip'].'</td>';
  echo '<td>'.$row['status'].'</td>';
  echo '<td>
  <select name="statusi">
  <option value="Na parkingu" name="Na parkingu">Na parkingu</option>
  <option value="Na turi" name="Na turi">Na turi</option>
  <option value="Potreban radionici" name="Potreban radionici">Potreban radionici</option>
  <option value="Potreban radionici-na parkingu" name="Potreban radionici-na parkingu">Potreban radionici-na parkingu</option>
  <option value="Potreban radionici-na turi" name="Potreban radionici-na putu">Potreban radionici-na turi</option>
  <option value="Razvoz" name="Razvoz">Razvoz</option>
  </select>
  <input type="submit" name="update" value="Promeni" /></td>';
  echo '<td>'.$row['vozac'].'</td>';
  echo '<td>';
  echo '<select name="list">';

   while ($row = mysqli_fetch_array($result2)) {
  echo '<option value="' . $row['vozac'] .'">' . $row['vozac'] .'</option>';
  }
  echo '</select>';
  echo '</td>';
  echo '<td>'.$row['napomena2'].'</td>';
  echo '<td><input type="text" name="napomena2" value=""><input type="submit" name="update3" value="Dodaj" /></td>';
  echo'<td><input type="hidden" name="IDV" value="'.$row['IDV'].'"></td>';

  echo '</tr>';
  echo '</form>';
  }

mysqli_close($con);
?>

Answer 1

您正在覆盖row变量。在其他循环更改变量名称

 while ($row1 = mysqli_fetch_array($result2)) {
  echo '<option value="' . $row1['vozac'] .'">' . $row1['vozac'] .'</option>';
  }

此外，您需要先将所有值存储在一个数组中，然后遍历它。而不是多次调用相同的查询。获取数据后。在您再次调用（mysqli_query()）查询之前，它不会再次执行。

修改

我已经提到首先将所有值存储在一个数组中，然后遍历它。试试这段代码：

$result2  = mysqli_query($con,"SELECT vozac FROM vozaci");
$list_vozaci = array();
while ($row = mysqli_fetch_array($result2)) {
  $list_vozaci[] = $row['vozac'];
}

$result = mysqli_query($con,"SELECT registracija,tip,status,vozac,napomena2,IDV FROM status_vozila order by status desc")

or die("Error: ".mysqli_error($con));

while($row = mysqli_fetch_array($result))
  {
  echo '<form action="" method="post">';
  echo '<tr>';
  echo '<td>'.$row['registracija'].'</td>';
  echo '<td>'.$row['tip'].'</td>';
  echo '<td>'.$row['status'].'</td>';
  echo '<td>
  <select name="statusi">
  <option value="Na parkingu" name="Na parkingu">Na parkingu</option>
  <option value="Na turi" name="Na turi">Na turi</option>
  <option value="Potreban radionici" name="Potreban radionici">Potreban radionici</option>
  <option value="Potreban radionici-na parkingu" name="Potreban radionici-na parkingu">Potreban radionici-na parkingu</option>
  <option value="Potreban radionici-na turi" name="Potreban radionici-na putu">Potreban radionici-na turi</option>
  <option value="Razvoz" name="Razvoz">Razvoz</option>
  </select>
  <input type="submit" name="update" value="Promeni" /></td>';
  echo '<td>'.$row['vozac'].'</td>';
  echo '<td>';
  echo '<select name="list">';

  foreach ($list_vozaci as $key => $value) {
    echo '<option value="' . $value .'">' . $value .'</option>';
  }
  echo '</select>';
  echo '</td>';
  echo '<td>'.$row['napomena2'].'</td>';
  echo '<td><input type="text" name="napomena2" value=""><input type="submit" name="update3" value="Dodaj" /></td>';
  echo'<td><input type="hidden" name="IDV" value="'.$row['IDV'].'"></td>';

  echo '</tr>';
  echo '</form>';
  }

mysqli_close($con);
?>