Prevent display Duplicate entry in php via JavaScript

Question

$fname=$_POST['fname'];
$lname = $_POST['lname'];
$Phone = $_POST['Phone'];
$date = date("Y/m/d");
mysql_query ( "INSERT INTO example (fname,lname,Phone,date) VALUES ('$fname' ,'$lname', '$Phone', '$date')")

or die(mysql_error());

when i write a duplicate entry in my codes these Error will display: my PRIMARY key is Phone.

  Duplicate entry '' for key 'PRIMARY'

but i want to check and display by JavaScript to Prevent this Error before adding data. how is it possible? please help completely and wire source. thanks

there are plenty of examples of ajax on SO. No one wants to 'help completely and [write] source' for you, unless you are paying. Post what you have attempted. — Sean
– Sean, Commented Dec 7, 2014 at 20:40
Here are just 4 of many examples - stackoverflow.com/questions/9762478/… stackoverflow.com/questions/5679998/check-username-using-ajax stackoverflow.com/questions/3393148/… stackoverflow.com/questions/14840286/… Use one of these as an example, changing 'username' to 'Phone'. more can be found at stackoverflow.com/search?q=ajax+check+if+username+exists — Sean
– Sean, Commented Dec 8, 2014 at 4:06

Aria5h4h · Accepted Answer · 2014-12-08 16:45:26Z

0

$fname=$_POST['fname'];
$lname = $_POST['lname'];
$Phone = $_POST['Phone'];
$date = date("Y/m/d");

$query = "SELECT * FROM example WHERE Phone =    '".mysql_real_escape_string($_POST['Phone'])."'";
if (mysql_num_rows(mysql_query($query)) > 0)
{ 
print "inuse";
} 
 else
{
 mysql_query ( "INSERT INTO example (fname,lname,Phone,date) VALUES ('$fname'   ,'$lname', '$Phone', '$date')")

or die(mysql_error());  
}

answered Dec 8, 2014 at 16:45

Aria5h4h

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