select based on random value if not exist change value php - php

Question

I'm trying to select values from database where date is equal to random date and sometimes there is no records how can I run the select again with another random value without refreshing website?
$start = "2017-08-12";
$now = date("Y-m-d");
$now = date('Y-m-d', strtotime($now. ' - 7 days'));
$daty = array();
while($start<$now){
$start = date('Y-m-d', strtotime($start. ' + 7 days'));
$daty[] = $start;
}
$rand_keys = array_rand($daty, 2);
$random_data = $daty[$rand_keys[0]] ;
$sql2= 'SELECT count(*) as total from tab where
Date="'.$random_data.'"';
$result12 = mysqli_query( $conn, $sql2 );
$row12 = $result12->fetch_assoc();
if($row12>0){
$sql1 = 'select * from premier1718 where Date="'.$random_data.'"';
$result1 = mysqli_query( $conn, $sql1 );
while($row1 = $result1->fetch_assoc()){
echo $row1["HomeTeam"]." ".$row1["AwayTeam"]."</br>";
}

Answer 1

I think you can possibly do this in one query like this
$sql = 'select * from `premier1718` where `date`=(
select `date` from `tab` where `date` in ( "'.implode( '","', $daty ).'" ) order by rand() limit 1
)';
$result = mysqli_query( $conn, $sql );
if( $result ){
while( $rs=$result->fetch_object() ) printf('%s %s<br />',$rs->hometeam,$rs->awayteam);
}

Answer 2

$rand_keys = array_rand($daty, 2);
$random_data = $daty[$rand_keys[0]] ;
$sql2= 'SELECT count(*) as total from tab where
Date="'.$random_data.'"';
$result12 = mysqli_query( $conn, $sql2 );
$row12 = $result12->fetch_assoc();
if($row12>0){
$success = false;
while ($success == false) {
$sql1 = 'select * from premier1718 where Date="'.$random_data.'"';
$result1 = mysqli_query( $conn, $sql1 );
while($row1 = $result1->fetch_assoc()){
$success = true;
echo $row1["HomeTeam"]." ".$row1["AwayTeam"]."</br>";
}
}