如果 MySQL value(55)
声明没有结果,我想要某个SELECT
。
我尝试了这个,但它不起作用。
<?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);
$arr = array();
//以下语句运行,但是当我尝试使用IFNULL时......不起作用......
//$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);
?>
答案 0 :(得分:0)
您收到错误,请尝试echo mysql_error();
你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;
}