2012-11-12 38 views
1

這可能是一個愚蠢的問題,但我想就如何去隻字不提所遇到的好幾篇文章,讓設置文本的價值澄清框取決於使用jQuery和Ajax下拉菜單中做出的選擇。我的問題是當我試圖做同樣的取決於5下拉菜單中做出的選擇。我有存儲在數據庫中的Id值應該被用來填充文本框。任何人都可以通過多次下拉菜單來指導如何解決這個問題。這是我的代碼到目前爲止。自動填寫文本框取決於下拉值

<?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> 
+0

你能告訴我輸出是怎麼出現的嗎? –

回答

1

爲了實現自動完成,你可以在HTML文件中創建一個選擇字段,在keyup事件中調用javascript函數並使用jQuery調用你的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> 
0

嘗試設置每個HTML選擇選項的值是文本

的ID,因此回聲應該是

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

然後在HTML中選擇添加事件平變化並添加JS函數foo(),通過使用JQuery或Javascript從頁面中獲取所有選定的id來填充隱藏的輸入。

還必須添加的ID或類每個HTML選擇以獲得其選擇的價值容易

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

然後在一個隱藏的輸入字段中插入這些值,以便隱藏輸入現在包含所有選定IDS

+0

好的。我會嘗試這個並回到你身邊。 – user1597398