MySQL 即使没有结果也返回一个值 - IFNULL

Question

当 MySQL SELECT 语句中没有结果时，我想要某个值(55) .

我试过了，但没用。

  <?php
include_once("JSON.php");
$json = new Services_JSON();


$con = mysql_connect("XXXXX.com", "XXXXX", "XXXXXX");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("XXXXX", $con);

$arr = array();

//以下语句运行，但当我尝试使用 IFNULL 时..不起作用...

//$rs = mysql_query("SELECT * FROM partie WHERE statut = 'en cours' ORDER BY date ASC LIMIT 1");

$rs = mysql_query("SELECT IFNULL((SELECT * FROM partie WHERE statut = 'en cours' ORDER BY date ASC LIMIT 1),55)");


while($obj = mysql_fetch_object($rs)) {
    $arr[] = $obj;
}

Echo $json->encode($arr);

?>

Answer 1

您收到错误，请尝试echo mysql_error();

您可能想要的是这样的:

$rs = mysql_query("SELECT * FROM partie WHERE statut = 'en cours' ORDER BY date ASC LIMIT 1");
if( mysql_num_rows($rs) == 0 ) {
    echo 55;
}