我有一个表格，当用户选择一个地方时，我需要在文本框中获取数据。如果用户选择一个地点，则以下名称和金额将出现在以下文本框中。我已经完成了以下操作，但它不起作用。有人可以帮我吗？

我尝试了以下内容：

html

<div>
    <label for="" class="">Place</label>
    <select  name="place" id="place"  onchange="myplace();">
        <option value="">--Option from db--</option>
    </select>       
</div>

<div>
    <label for="" class="">Name</label>
    <input type="text"  name="myname" id="myname" value="<?php //if there is data in databas then it is shown here...otherwise the value will change according to the above selection?>" >  
</div>

<div>
    <label for="" class="">Amount</label>
    <input type="text"  name="myamnt" id="myamnt" value="<?php //if there is data in databas then it is shown here...otherwise the value will change according to the above selection?>" >  
</div>

mysql：

if(!empty($_POST["myinitplace"]))
{
        
$sql = "SELECT * FROM all_data , my_place  WHERE all_data.place=my_place.place and all_data.place = ?";
$stmt = $conn->prepare($sql); 
$stmt->bind_param("s", $_POST["myinitplace"]);
$stmt->execute();
$result = $stmt->get_result(); 
while($rows = $result->fetch_assoc())   
{   
 $data['name']=$rows['name'] ;
 $data['myacc']=$rows['amount'] ;
                
}
echo json_encode($data);
                        
    }

ajax

function myplace() 
{
        
    var a = document.getElementById("place").value;
    //alert(a);
    $.ajax({
        type : "POST",
        data : {
            myinitplace : a 
        },
        dataType :  "JSON",
        success : function(data){
         document.getElementById("myname").value = data.name;
         document.getElementById("myamnt").value = data.myacc;
        }
    })  
}