Fetch All rows of column in a table

[Bala Kumaran]

Hello friends,
I am developing my own website with php and sql. I've a mistake in my codings when try to fetch all rows of a column in a table, It shows the result of only last row of the particular column

There are my codings

<?php
include_once("com_check_login.php");
include("test_connect_to_mysql.php");
$client_idd = $fgmembersite->UserClientID();
$sqliw = mysql_query("SELECT * FROM domains WHERE client_id LIKE '" . $client_idd . "'");
$notify_a = mysql_num_rows($sqliw);
if($notify_a > 0) {
	while($rowsj = mysql_fetch_array($sqliw)) {
		$dom_a_ins = $rowsj["domains"];
} }
?>
<?php echo $dom_a_ins; ?>

Below image shows table name : 'domains'



I am getting the result of the above coding is:

/* result starts here */

example4.com

/* result end here */

I want to get all the matching result of the id "1000".

Because, Now I've logged with id name 1000.

Please help me

[Bala Kumaran]

hello friends!
I got the result for my question by changing the script

<?php
include_once("com_check_login.php");
include("test_connect_to_mysql.php");
$client_idd = $fgmembersite->UserClientID();
$domain_list = "";
$sqliw = mysql_query("SELECT * FROM domains WHERE client_id LIKE '" . $client_idd . "'");
$notify_a = mysql_num_rows($sqliw);
if($notify_a > 0) {
    while($rowsj = mysql_fetch_array($sqliw)) {
        $dom_a_ins = $rowsj["domains"];
        $domain_list = $dom_a_ins;
} } else {
$domain_list = "No any domains added here";
}
?>
<?php echo $dom_a_ins; ?>