PHP AJAX基于组合框更改值

时间:2017-09-12 12:02:05

标签: php jquery ajax

我想要的是每次我从组合框中更改值(组合框中的值来自数据库)输入类型也会改变

这里

<div class="col-md-12" >
    <label>Charge</label>
    <select id="name" name="name" class="form-control">
    <?php
        while ($reserve=mysqli_fetch_array($charge)) { ?>
           <option value=" <?php echo $reserve['name']?>">
               <?php echo $reserve['name']; ?>
           </option><?php } ?>
    </select>
</div>
<div class="col-md-12">
<br>
    <label>Price</label>
    <input type="number" class="form-control" id="Price" name="Price" disabled>
</div>

在数据库上名称具有相应的价格,例如椅子有200个价格 我想要的是每次我改变组合框时,inputtype的值也会改变

3 个答案:

答案 0 :(得分:1)

如果你想显示选项值,我不确定你为什么使用ajax。请检查此代码。

 $('#name').change(function () {
            var option_value = $(this).val();
            $('#Price').val(option_value);
        });
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form>
    <select id="name" name="name">
        <option value="1">1</option>
        <option value="2">2</option>
        <option value="3">3</option>
    </select>


    <div class="col-md-12">
                <br>
                <label>Price</label>
                <input type="number" class="form-control" id="Price" name="Price" disabled>
              </div>
</form>

所以使用此代码更新选择代码

<select id="name" name="name" class="form-control">
                  <?php
                  while ($reserve=mysqli_fetch_array($charge)) { ?>
                    <option value=" <?php echo $reserve['name']?>">
                      <?php echo $reserve['name']; ?>
                    </option><?php } ?>
                </select>

答案 1 :(得分:0)

您需要添加Jquery代码才能完成此任务。 你的Html:

<div class="col-md-12" >
  <label>Charge</label>
  <select id="name" name="name" id="name" class="form-control">
    <?php while ($reserve=mysqli_fetch_array($charge)) { ?>
    <option value=" <?php echo $reserve['name']?>">
    <?php echo $reserve['name']; ?>
    </option><?php } ?>
  </select>
</div>

<div class="col-md-12"><br>
  <label>Price</label>
  <input type="number" class="form-control" id="Price" name="Price" disabled>
</div>

Jquery代码:

$( document ).ready(function() {
   $("#name").change(function () {
        var value = this.value;
        $('#Price').val(value);
    });
});

答案 2 :(得分:0)

  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>

                    <select id="name" name="name" class="form-control" onchange="callAjax(this)">
                      <?php
                      while ($reserve=mysqli_fetch_array($charge)) { ?>
                        <option value=" <?php echo $reserve['name']?>">
                          <?php echo $reserve['name']; ?>
                        </option><?php } ?>
                    </select>

<script>
    function callAjax(val){
        var selectedVal = val.value;
        alert(selectedVal+"  "+val)
        $.ajax({
            url: "<some_Url>", // Use PHP url from where you can get Data from Database
            data: selectedVal,
            success: function(result){
                $("#Price").val(result);
        }});


    }
</script>