下面是我编写的用于创建随机字符串的代码。效果很好。它还检查以确保生成的字符串不存在,如果存在,则生成另一个。但是,我还没有找到一种方法来检查生成的代码是否已经存在,以查看该代码是否存在。我最好做一个 elseif 语句吗?
PHP
<?PHP
require_once('dbConfig.php');
$randomstring = '';
$characters = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ';
for ($i = 0; $i < 12; $i++) {
$randomString .= $characters[rand(0, strlen($characters) - 1)];
}
//$generatedId = "SPI-E7HN2SBIIF5W";
$generatedId = 'SPI-'.$randomString;
//Prepare select query
$statement = $db->prepare("SELECT client_unique_id FROM clients WHERE client_unique_id = ? LIMIT 1");
//Determine variable and then bind that variable to a parameter for the select query ?
$id = $generatedId;
$statement->bind_param('s', $id);
//Execute and store result so that num_rows returns a value and not a 0
$statement->execute();
$statement->store_result();
//Bind result to a variable for easy management afterwards
$statement->bind_result($clientId);
// Generate a random ID for the user if the previously generated one already exists
if($statement->num_rows > 0){
$randomstring = '';
$characters = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ';
for ($i = 0; $i < 0; $i++) {
$randomString .= $characters[rand(0, strlen($characters) - 1)];
}
$generatedId = 'SPI-'.$randomString;
echo $generatedId;
} else {
// If there's no issue with what's been created, echo it out :)
echo $generatedId;
}
?>
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