在php mysql中选择选项时，用值填充输入字段 [英] Populate input fields with values when option is selected in php mysql

问题描述

我有一个表格，当从数据库中选择一个选项时，我想提交详细信息.

I have a form that i would like to submit details when an option is selected from the database.

mysql数据库表:-

The mysql database table :-

用户包含[email]，[age]，[name]之类的字段.

USERS contains fields like [email],[age],[name].

我希望能够从菜单中选择一个字段时填充其他输入字段的值.

I want to be able to populate the other input fields values when one field is selected from the menu.

        <form>
        User
        <select name="user" id="user">
          <option>-- Select User --</option>
          <option value="Mark">Mark</option>
          <option value="Paul">Paul</option>
          <option value="Hannah">Hannah</option>
        </select>

        <p>

          Age 
          <input type="text" name="age" id="age">

        </p>
        <p>
          Email 
          <input type="text" name="email" id="email">
        </p>
        </form>

我如何使用jquery或javascript实现此目的.

How do i acheive this using jquery or javascript.

推荐答案

请尝试此操作

在您的HTML页面上:

on your HTML page:

将此内容写入您的html页面，

write this to your html page,

<script>
$(document).ready(function(){
    $('#user').on('change',function(){
        var user = $(this).val();
        $.ajax({
            url : "getUser.php",
            dataType: 'json',
            type: 'POST',
            async : false,
            data : { user : user},
            success : function(data) {
                userData = json.parse(data);
                $('#age').val(userData.age);
                $('#email').val(userData.email);
            }
        }); 
    });
});
</script>

在getUser.php中:

in getUser.php:

<?php

$link = mysqli_connect("localhost", "user", "pass","mydb");

$user = $_REQUEST['user'];    
$sql = mysqli_query($link, "SELECT age,email FROM userstable WHERE name = '".$user."' ");
$row = mysqli_fetch_array($sql);

json_encode($row);die;

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