Question

这可能是一个愚蠢的问题，但我想澄清一下如何解决这个问题。我遇到过很多文章，可以根据drop中的选择设置文本框的值下面菜单使用jQuery和Ajax.My问题是当我尝试做同样的事情时，取决于5个下拉菜单中的选择。我有Id值存储在数据库中，应该用来填充文本框。可以任何人都指导如何通过多次下拉来解决这个问题。到目前为止，这是我的代码。

    <?php
     $sql1="SELECT Schlungen  FROM schulung as s"; 
     $result=mysql_query($sql1); 
     echo "<p align='left'> <strong>Schulung 1</strong> <select name='Schlungen1'>           <option value default></option>";
     while ($row = mysql_fetch_array($result)) 
     {
     echo "<option value='" . $row['Schlungen'] . "'>" . $row['Schlungen'] . "   </option>";
      }
      echo "</select>";
      ?>

      <?php
      error_reporting(0);
      //Drop Down for Schulung 2
      $sql2="SELECT Schlungen  FROM schulung as s"; 
      $result=mysql_query($sql2); 
      echo "<p align='left'> <strong>Schulung 2</strong> <select name='Schlungen2'>  <option value default></option>";
      while ($row = mysql_fetch_array($result)) 
      {
      echo "<option value='" . $row['Schlungen'] . "'>" . $row['Schlungen'] . " </option>";
       }
      echo "</select>";
      ?>


      <?php
      error_reporting(0);
      //Drop Down for Schulung 3 
      $sql3="SELECT Schlungen  FROM schulung as s"; 
      $result=mysql_query($sql3); 
      echo "<p align='left'> <strong>Schulung 3</strong> <select name='Schlungen3'> <option value default></option>";
      while ($row = mysql_fetch_array($result)) 
      {
      echo "<option value='" . $row['Schlungen'] . "'>" . $row['Schlungen'] . "</option>";
       }
      echo "</select>";
      ?>

      <?php
      error_reporting(0);
      //Drop Down for Schulung 4 
      $sql4="SELECT Schlungen  FROM schulung as s"; 
      $result=mysql_query($sql4); 
      echo "<p align='left'> <strong>Schulung 4</strong> <select name='Schlungen4'><option value default></option>";
      while ($row = mysql_fetch_array($result)) 
      {
      echo "<option value='" . $row['Schlungen'] . "'>" . $row['Schlungen'] . " </option>";
       }
      echo "</select>";
      ?>

      <?php
      error_reporting(0);
      //Drop Down for Schulung 5  
      $sql5="SELECT Schlungen  FROM schulung as s"; 
      $result=mysql_query($sql5); 
      echo "<p align='left'> <strong>Schulung 5</strong> <select name='Schlungen5'><option value default></option>";
      while ($row = mysql_fetch_array($result)) 
      {
      echo "<option value='" . $row['Schlungen'] . "'>" . $row['Schlungen'] . " </option>";
      }
      echo "</select>";
      ?>   
      <p align="left"><strong>Access_level </strong> 
      <input type="text" name="a_level" disabled="disabled">
      </p>

Answer 1

为了实现自动完成，你可以在html文件中创建一个select字段，在keyup事件上调用javascript函数并使用jQuery来调用你的php文件

<html>
<head>
    <script>
        $('.autosuggest').keyup(function(){
         $.post("<your file.php>",{any data you need},function(data){
          //echo the data
        //echo "<option value='" . $row['Schlungen'] . "'>" . //$row['Schlungen'] ."   </option>";
        $('.result').html(data)

});
        });
        $('.result option').click(function(){
            var rValue = $(this).text();
            $('.autosuggest').attr('value',rValue);        
            $('.result').html('');
        });

    </script>
</head>
<body>
    <input type='text' class='autosuggest'/>
    <select class='result'>
    </select>
</body>
</html>

Answer 2

尝试将每个html选择选项的值设置为该文本的ID

所以回声应该是

echo "<option value='" . $row['id'] . "'>" . $row['Schlungen'] . " </option>";

然后在html选择中添加事件 onchange 并添加JS函数foo（）以使用所有选定的id填充隐藏的输入，方法是使用JQuery或Javascript从页面获取它们

您还必须为每个html选择添加id或类以轻松获取其选定值

$("#select1").val();
$("#select2").val(); ...

然后在隐藏的输入字段中插入这些值，以便隐藏的输入现在包含所有选定的ID

自动填充文本框取决于下拉值

2 个答案: