選擇onchange更新其他輸入字段

Question

我有一個選擇選項和3個輸入文本框的表單。我從數據庫中獲取select選項的值。我想根據我選擇的內容更改3個輸入字段的數據。我嘗試了許多方法，但表現很好。

<Select onchange="update(this)"> 
<?php 
    $query = "SELECT * FROM wheel1 ORDER BY chair_no ASC"; 
    $run = mysqli_query($connect, $query); 
    $i = 0; 
    while ($chair = mysqli_fetch_assoc($start)) { 
    $i++; 
    ?> 
<option value="<?php echo $chair['chair_no'];?>"><?php echo $i ?></option> 
<?php 
    $name = $chair['name']; 
    $phone = $chair['phone']; 
    $detail = $chair['detail']; 
    } 
?> 

它爲Select菜單提供完美的數據，但只存儲姓名電話和詳細變量的最後一個值。這裏是JavaScript代碼。

function update(elem) { 
      var name = "<?php echo $name; ?>"; 
      var phone= "<?php echo $phone; ?>"; 
      var detail = "<?php echo $detail; ?>"; 
      document.getElementById("name").innerHTML = name; 
      document.getElementById("phone").innerHTML = phone; 
      document.getElementById("detail").innerHTML = detail; 
} 

任何幫助是appricated。提前致謝。在姓名，電話和細節選項

Answer 1

集向度屬性

演示：https://jsfiddle.net/sjhhv4pw/

<select onchange="update(this)"> 
<?php 
    $query = "SELECT * FROM wheel1 ORDER BY chair_no ASC"; 
    $run = mysqli_query($connect, $query); 
    $i = 0; 
    while ($chair = mysqli_fetch_assoc($start)) { 
    $i++; 
    ?> 
<option data-name="<?php echo $chair['name']; ?>" data-phone="<?php echo $chair['phone']; ?>" data-detail="<?php echo $chair['detail']; ?>" value="<?php echo $chair['chair_no'];?>"><?php echo $i ?></option> 
<?php 
    } 
?> 
</select> 

的jQuery：

function update(elem) { 
    var name = $(elem).find("option:selected").attr("data-name"); 
    var phone= $(elem).find("option:selected").attr("data-phone"); 
    var detail = $(elem).find("option:selected").attr("data-detail"); 
} 

Answer 2

試試下面的代碼

<Select onchange="update(this)"> 

「數據電話=」」數據椅的細節= 「」 值= 「」>

JavaScript代碼，考慮到您正在使用jQuery

function update(elem) { 
     var name = ele.data('chair-name'); 
     var phone= ele.data('phone'); 
     var detail = ele.data('chair-detail'); 

}