<?php
if(isset($_POST["Add"]))
{
$name=$_POST["Emp_Username"];
$pass=$_POST["Emp_Password"];
$fname=$_POST["Emp_Fname"];
$lname=$_POST["Emp_Lname"];
$hph=$_POST["ContactNo_Home"];
$hp=$_POST["ContactNo_HP"];
$mail=$_POST["Emp_Email"];
$add=$_POST["Emp_Address"];
$age=$_POST["Emp_Age"];
$pos=$_POST["Position"];
$dept=$_POST["Dept_ID"];
mysql_query("
insert into employee(Dept_ID, Emp_Address, Emp_Age,
Position, >Emp_Username, Emp_Password, Emp_Fname, Emp_Lname,
ContactNo_Home, ContactNo_HP, Emp_Email)
>values('$dept','$add','$age','$pos','$name','$pass','$fname',
'$lname','$hph','$hp','$mail'>)
");
?>
<script type="text/javascript">
alert("Record saved.");
</script>
<?php } ?>
4 に答える
0
フォームを送信する前に、フィールド「user」をajaxで検証する必要があるかもしれません
$name = $_POST["Emp_Username"];
$results = mysqli_query($connecDB,"SELECT * FROM employee WHERE Emp_Username='$name'");
$name_exist = mysqli_num_rows($results);
if($username_exist) {
echo 'busy';
}else{
echo 'free';
}
于 2013-08-26T12:59:32.253 に答える