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问题描述
我有一个表单,我想在从数据库中选择一个选项时提交详细信息。 mysql数据库表[USERS]包含[email],[age],[name]等字段。我希望能够在菜单中选择一个字段时填充其他字段。
< form> User < select name =userid =user> < option> - 选择用户 - < / option> < option value =标记>标记< / option> < option value =Paul> Paul< / option> < option value =Hannah> Hannah< / option> < / select> < p> 年龄< input type =textname =ageid =age> < / p> < p> 电子邮件< input type =textname =emailid =email> < / p> < / form>如何使用jquery或javascript实现此功能。
解决方案请在您的HTML页面上试试这个,
:
将此内容写入您的html页面,
< script> $(变量),函数(){ var user = $(this).val(); $(document).ready(function(){ $('#user')。 (); $ .ajax({ url:getUser.php, dataType:'json',类型:'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: getUser .php
<?php $ link = mysqli_connect(localhost,user ,pass,mydb); $ user = $ _REQUEST ['user']; $ sql = mysqli_query($ link,SELECT FROM,email FROM userstable WHERE name ='。$ user。'); $ row = mysqli_fetch_array($ sql); json_encode($ row); die;
I have a form that i would like to submit details when an option is selected from the database. The mysql database table [USERS]contains fields like [email],[age],[name].I want to be able to populate the other fields 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>How do i acheive this using jquery or javascript.
解决方案Please Try this,
on your HTML page:
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>in getUser.php:
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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当选择选项时使用值填充输入字段
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