draxu26480 2018-04-17 08:36
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SQL echo输出错误通过文本输入字段

i have written some sql code to output values from a database based on the first dropdown selection, but i can see the values being returned through inspect element in the console but my text field is blank

<?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);
?>
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1条回答 默认 最新

  • douyiqi9640 2018-04-17 08:42
    关注
    <?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);
    ?>
    

    Try to this added name="sites[]" to array and MYSQLI_ASSOC in getting result have association array Also added < input '/'>

    for Select Drop Down :

        $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);  ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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