2014-11-04 68 views
2

我有一個頁面,我正在動態生成文本框和下拉列表。我試圖從下拉列表中選擇一個名稱,然後在其旁邊的文本框中顯示它。問題是我能夠做到這一點只爲第一行,我的意思是我可以從第一個下拉列表中選擇一個名稱,然後在它旁邊的文本框中顯示它。但我無法爲其他人做到這一點。你能幫忙嗎?無法從文本框中的下拉列表中獲取名稱

這裏是我的代碼

請注意:我使用JavaScript來顯示什麼用戶從下拉列表中選擇,並顯示在 文本

<?php 
    if(isset($_POST['team'])){ 
     $selectedTeam = $_POST['team']; 
     $selectedWeek = $_POST['week']; 

     //getting team member names from resolver table 
     $query = "SELECT * FROM `names` WHERE `teamName` = '$selectedTeam'"; 
     $query_run_teamName = mysql_query($query); 
     while($query_row = mysql_fetch_assoc($query_run_teamName)){ 
      $verNAME[] = $query_row['verNAME']; 
     }//end of while 
    }//end of if condition 
?> 
<!DOCTYPE html> 
<html> 
    <head> 
    <title>Option Page</title> 
    <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.4.4/jquery.min.js"></script> 
    <script type="text/javascript" src="roster.js"></script> 
    <link rel="stylesheet" type="text/css" href="style.css"> 
    <script type="text/javascript"> 

     //Function below gets ID and assigns value to the left text field 
     function getValue(RightTextfield, leftTextfield){ 
       var TextfieldValue  = _(RightTextfield).value; 
       _(leftTextfield).value = TextfieldValue; 
     }//end of getValue() function 

     function _(x){ 
      return document.getElementById(x); 
     }//end of _(x) function 
    </script> 
    </head> 
    <body> 
    <div id="wrapper"> 
     <form action="teamRoster.php" enctype="multipart/form-data" method="POST"> 
      <div id="response"></div><!--End of response Div--> 
      <?php 

       for ($i = 0; $i <= $count; $i++) { 
        echo '<div class="STeam">'; 

        //code below to generate Text Fields dynamically 
        echo '<input type="text" id="r1Value1" name="r1Value1">'; 

         //code below to generate dropdown list dynamically 
         echo '<select id="ChooseMember" name="ChooseMember" onchange="getValue(\'ChooseMember\', \'r1Value1\')">';  
         echo '<option value="">Select Team Member</option>'; 
           foreach ($verNAME as $key) { 
            echo '<option value="'[email protected]$key.'">'[email protected]$key.'</option>'; 
         } 
         echo '</select>'; 
        echo '</div>'; 
       }//End of for loop 
      ?> 
      <!--Passing week and team name as hidden values--> 
      <input type="hidden" id="HiddenDay" name="HiddenDay" value="<?php echo $selectedWeek;?>"> 
      <input type="hidden" id="HiddenTeam" name="HiddenTeam" value="<?php echo $selectedTeam;?>"> 
     </form> 
    </body> 
</html> 

謝謝

+1

你有'name =「HiddenTeam」value =「<?php echo $ selectedTeam'然而你正在使用'$ selectedTeam = $ _POST ['team']'你的POST不匹配,檢查其他。 – 2014-11-04 00:49:38

回答

0

好,我想你應該檢查你的選擇框ID和名稱。您在選擇框中輸入相同的ID和名稱。我的建議是與選擇框ID和名稱附加的$ I值(特別是與JavaScript的ID,但你應該追加到名稱由於從PHP搶值)

For select box HTML - 
<select id="ChooseMember_<?=$i?>" name="ChooseMember_<?=$i?>" onchange="getValue(\'ChooseMember_<?=$i?>\', \'r1Value1\')"> 

這個問題創造了相同的ID和姓名的選擇框。

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