当选择选项时使用值填充输入字段 - php mysql [英] Populate input fields with values when option is selected - php mysql
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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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