php - 如何使用 PHP 检查数据是否已存在于 MySQL 中

Question

这是我的 PHP 代码。如何用PHP检查MySQL中是否存在这3个参数(user、agwera、vis)？如果已经存在，请不要第二次添加。 + 想要 echo "Already in Exists<br/>";

    <?
sleep(10);
define ('DBHOST', 'yyy');
define ('DBNAME', 'xxxx');
define ('DBUSER', 'xxxx');
define ('DBPASS', 'xxxxx');

mysql_pconnect(DBHOST, DBUSER, DBPASS) or die('DB');
mysql_select_db(DBNAME) or die ('USER');
mysql_query('SET NAMES utf8');

header('Content-Type: text/html; charset=utf-8');
error_reporting (0);
$act = isset($_GET['act']) ? $_GET['act'] : '';
switch($act)
{
case 'send':
$destination = $_POST['adress']; 
$signature = $_POST['text']; 
$text = "$signature ";



if (empty($_POST["text"])  or strlen($_POST["text"])==0){

@header("location:error5.php");
break;

}

$str = $_POST['text']; 
if (strlen($str)>160){
@header("location:error4.php");
break;

}

if (!preg_match('/^[0-9]{9}$/',$destination))
{
@header("location:error1.php");
break;
}

$str2 = $_POST['sender'];
if (strlen($str2)>11){
@header("location:error2.php");
break;
}

if (empty($_POST["sender"])  or strlen($_POST["sender"])==0){

@header("location:error3.php");
break;

}

else
{

    $password = 'zzzzzz';
    $destination = $_POST['adress']; 
  $nomeri = "995$destination";
    $nomeri2 = $_POST['sender']; 
    $source = $nomeri2;
    $signature = $_POST['text']; 
    $text = "$signature "; 

    $content = 
                'key='.rawurlencode($password). 
                '&destination='.rawurlencode($nomeri). 
                '&sender='.rawurlencode($source). 
                '&content='.rawurlencode($text); 

    $a = file_get_contents('http://example.com/api/send.aspx?'.$content); 

@header("location:ok.php");

mysql_query("INSERT `gagzavnili` SET
                      `user`        =   '".$source."',
                      `agwera`      =   '".$text."',
                      `vis`     =   '".$destination."',
                      `time_upload` =   '".time()."',
                      `time`        =   '".time()."'") or die(mysql_error());
                       $file_id = mysql_insert_id();



}
}
?>

Answer 1

$data = array('user', 'agwera', 'vis');

foreach( $data as $val ) {
$res = mysql_query("SELECT * FROM `table_name` WHERE `column_name` = '$val'");
$count = mysql_num_rows($res);

if( $count > 0 ) {
    echo "<script>alert('Already Exists')</script>";
} else {
    //insert query for $val...
}
}