Question

我已根据第一个下拉选项编写了一些sql代码来输出数据库中的值，但是我可以看到通过控制台中的inspect元素返回的值，但我的文本字段为空白

<?php
$connection = mysqli_connect("localhost", "root", "");
$sn = $_REQUEST['get_option'];
mysqli_select_db($connection, "dqa");
$result = mysqli_query($connection, "SELECT * FROM action_plan WHERE sn = '".$sn."' GROUP BY site");
while($row = mysqli_fetch_array($result))
  {
  ?>
     <input  type="text" name="site"  value="<?php echo $row['site']; ?>">
  <?php
  }
mysqli_free_result($result);
mysqli_close($connection);
?>

Answer 1

<?php
$connection = mysqli_connect("localhost", "root", "");
$sn = $_REQUEST['get_option'];
mysqli_select_db($connection, "dqa");
$result = mysqli_query($connection, "SELECT * FROM action_plan WHERE sn = '".$sn."' GROUP BY site");
while($row = mysqli_fetch_array($result,MYSQLI_ASSOC))
  {
  ?>
     <input  type="text" name="site[]"  value="<?php echo $row['site']; ?>" />
  <?php
  }
mysqli_free_result($result);
mysqli_close($connection);
?>

尝试将此添加的名称=“sites []”添加到数组中并获取结果中的MYSQLI_ASSOC具有关联数组还添加了< input '/'>

选择下拉：

    $connection = mysqli_connect("localhost", "root", "");
 $sn = $_REQUEST['get_option'];
 mysqli_select_db($connection, "dqa"); 
$result = mysqli_query($connection, "SELECT * FROM action_plan WHERE sn = '".$sn."' GROUP BY site"); ?> 
<select name="site" class="form-control" id="new_select">
 <?php
 while($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) {
 ?> 
<option value="<?php echo $row['site']; ?>" ><?php echo $row['site']; ?> </option>
 <?php } ?> 
</select> 
<?php 
mysqli_free_result($result);
 mysqli_close($connection);  ?>

SQL echo输出错误通过文本输入字段

1 个答案: