我有两页 add_admin.php 和 ajax_admin.php 当我在下拉列表中选择名称时,它会在TEXTBOXES中显示FIRSTNAME,MIDDLENAME和LASTNAME。
这是我的代码:
add_admin.php
<!-- Department -->
<div class="col-xs-12 col-sm-12 col-md-12" >
<div class="form-group">
<select id="faculty_name" name="faculty_name" class="form-control" onchange='fetch_select(this.value)' required>
<option selected="selected" disabled="disabled">Please Select Faculty</option>
<?php
$query = mysql_query("select * from faculty_details");
while($row = mysql_fetch_array($query))
{
?>
<option value="<?php echo $row['FACULTY_ID'];?>"><?php echo $row['FIRSTNAME']." ".$row['MIDDLENAME']." ".$row['LASTNAME'];?></option>
<?php
}
?>
</select>
</div>
</div>
<!-- First Name -->
<div class="col-xs-12 col-sm-4 col-md-4">
<div class="form-group">
<div class="input-group">
<div class="input-group-addon"><i class="fa fa-fw fa-user"></i></div>
<input type="text" id="fname" name="fname" class="form-control input-md" placeholder="First Name" value="">
</div>
</div>
</div>
其他文本框中有中间名,姓氏
Ajax代码
function fetch_select(val)
{
$.ajax
({
type: 'post',
url: 'ajax_admin.php',
data:
{
get_option:val
},
success: function (response)
{
$('#fname').val(response);
}
});
ajax_admin.php
<?php
if(isset($_POST['get_option']))
{
$state = $_POST['get_option'];
$find=mysql_query("select * from faculty_details where FACULTY_ID=$state");
while($row=mysql_fetch_array($find))
{
echo "$row[FIRSTNAME]";
}
}
?>
答案 0 :(得分:1)
从服务器获取结果为json,以便您可以在jQuery中轻松使用它:
的 Ajax的admin.php的强>
if(isset($_POST['get_option']))
{
$state = $_POST['get_option'];
$row1=array();
$find=mysql_query("select firstname,middlename,lastname from faculty_details where FACULTY_ID=$state");
while($row=mysql_fetch_array($find))
{
$row1[]=$row;
}
die(json_encode($row1));
}
Ajax代码
function fetch_val(val) {
$.ajax({
url:"ajax-admin.php",
type:"POST",
data:{"get_option":val},
dataType:"JSON",
success:function(data){
$('#fname').val((data[0].firstname));
$('#mname').val((data[0].middlename));
$('#lname').val((data[0].lastname));
}
});
}