I have a single page php application to generate an 8 digit number, user inputs a value and page generates another value, both values are inserted into mysql db and displaying the generated value in an input box at a single button click. Insertion is happening, but generated value is not displaying in the inputbox at the first time, I also tried session its not helping too.
here is my php code,
<?php
require_once __DIR__ . '/db_connect.php';
$db = new DB_CONNECT();
?>
<!doctype html>
<html>
<head>
<meta charset="UTF-8">
<title>Untitled Document</title>
</head>
<body>
<div>
<?php
if(isset($_POST['luckyButton']) && $_POST['myLuckyNumber'] != ""){
//sleep(20);
$myLuckyNum = $_POST['myLuckyNumber'];
$resultmyLuckyNum = mysql_query("SELECT * FROM lucky where luckyNum='$myLuckyNum'") or die(mysql_error());
if (mysql_num_rows($resultmyLuckyNum) > 0) {
while ($rowmyLuckyNum = mysql_fetch_array($resultmyLuckyNum)) {
$generatedNumber = $rowmyLuckyNum["generatedNum"];
}
}else{
$gen = getGen();
$resultGenNum = mysql_query("INSERT INTO lucky (slno, luckyNum, generatedNum) VALUES ( NULL, '$myLuckyNum', '$gen')");
if (mysql_num_rows($resultGenNum) > 0) {
$generatedNumber = $gen;
}
}
}
?>
<form action="" method="post">
<input type="text"class="inputs" name="myLuckyNumber" onClick="this.select();" placeholder="Enter You Lucky Number" value="<?php echo $myLuckyNum; ?>" />
<input type="submit" class="css_button" name="luckyButton" value="Check"/>
</form>
</div>
<br><br><br>
<?php
if($generatedNumber != ""){
echo '<input type="text" name="myLuckyNumber" onClick="this.select();" id="generatedNumber" readonly="true" value="' . $generatedNumber . '" />';
}
echo '<p id="errText"> ' . $err . ' </p>';
function random_string($length) {
$key = '';
$keys = array_merge(range(0, 9));
for ($i = 0; $i < $length; $i++) {
$key .= $keys[array_rand($keys)];
}
return $key;
}
function getGen(){
$gen1 = random_string(8);
$result = mysql_query("SELECT * FROM lucky where generatedNum='$gen1'") or die(mysql_error());
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result)) {
if($row["generatedNum"] == $gen1){
getGen();
}
}
}else{
//echo $gen1;
return($gen1);
}
}
?>
</body>
</html>
my table,.
CREATE TABLE `lucky` (
`slno` int(11) NOT NULL,
`luckyNum` text NOT NULL,
`generatedNum` text NOT NULL
) ENGINE=InnoDB AUTO_INCREMENT=1 DEFAULT CHARSET=latin1;
