How to check if name already exist in MySQL Database?

<?php

$name=$_POST["name"];

$age=$_POST["age"];

mysql_select_db("test", $connect);

$query_check_name = "select * from name where name ='" . $name . "'";

$res_check_name = mysql_query($query_check_name);

$arr_check_name = mysql_fetch_array($res_check_name);

if(empty($arr_check_name)){

$query=("INSERT INTO name (id, name, age) VALUES('', '$name', '$age')");

if(mysql_query($query,$connect)){

echo "<h1>:)</h1>";

}else{

echo mysql_error();

}

}

else{

// redirect to your page (index.php?msg=name_exist);

header( 'Location: index.php?msg=name_exist' ) ;

}

?>

you need to check on index page that $_GET['msg'] variable (msg) is exist or not. For this you can use isset() function. So put below code in index.php file on top or where you want to display message.

<?php

if(isset($_GET['msg']) && $_GET['msg'] == 'name_exist'){

echo "Name Already Exist. Please Enter Unique Name.";

}

?>

I did not check this code. May be possible some syntax error.

Hope this will help :)

When the database isn't finding the '$name' value in your field it is returning 0 rows. I bet your php code assumes that you are getting at least one row back. Hence the error when the resource doesn't have anything in it. There are two ways around it. Either you can modify your sql to $query = "SELECT count(*) FROM users WHERE name='$name';"; which will guarantee you a result where you can just check that it's greater than 0 or you can use the mysql_num_rows() function to check that you actually have a row before you fetch the first row e.g. // does user exist? $name = mysql_real_escape_string($name); $query = "SELECT * FROM users WHERE name='$name';"; $res = mysql_query($query); if (mysql_num_rows($res) > 0) { // yes, pull in the user details } else // no, user doesn't exist }

First of all... create a unique key for name column

create table tablename (

...

user_id bigint, -- or whatever

age date,

unique key(name)

...

);

when you execute mysql_query then if name already exist then mysql return false other wise reture true..

$result= mysql_query(INSERT INTO table_name(age,name) values('23','ketan');

if(mysql_num_rows($result)==0){

echo 'name available';

}else{

echo '<script>alert("name already exist")</script>';

}

SORRY FOR MY BAD ENGLISH!

you need to select users from your database, that have you entered in the form...

mysql_query("SELECT id FROM name WHERE name = $name");

...then check if id is defined give an error using javascript.

send.php******************************…

<?php //please, DO NOT close and re-open php!

include_once("connect.php");

$name=$_POST["name"];

$age=$_POST["age"];

mysql_select_db("test", $connect);

$query = "select * from `name` where `ìd`='" . $name . "' limit 1";

$res = mysql_query($query);

$row = mysql_fetch_array($res);

if ($row['id'] == $name)

echo ("Name already exists");

else

...

v bcv