php切换到mysqli：num_rows问题

我最近开始更新一些代码到 MySQL改进的扩展,并且直到这一点已经成功：

// old code - works
$result = MysqL_query($sql);
    if(MysqL_num_rows($result) == 1){
    $row = MysqL_fetch_array($result);
    echo $row['data'];
    }


// new code - doesn't work
$result = $MysqLi->query($sql) or trigger_error($MysqLi->error." [$sql]"); 
    if($result->num_rows == 1) {
    $row = $result->fetch_array();
    echo $row['data'];
    }

如图所示,我试图使用面向对象的风格.
我没有得到MysqLi错误,vardump说没有数据……但db表中肯定有数据.

解决方法

试试这个：

<?PHP

// procedural style

$host = "host";
$user = "user";
$password = "password";
$database = "db";

$link = MysqLi_connect($host,$user,$password,$database);

IF(!$link){
    echo ('unable to connect to database');
}

ELSE {
$sql = "SELECT * FROM data_table LIMIT 1";
$result = MysqLi_query($link,$sql);
    if(MysqLi_num_rows($result) == 1){
    $row = MysqLi_fetch_array($result,MysqLI_BOTH);
    echo $row['data'];
    }
}
MysqLi_close($link);


// OOP style 

$MysqLi = new MysqLi($host,$database);
$sql = "SELECT * FROM data_table LIMIT 1";
$result = $MysqLi->query($sql) or trigger_error($MysqLi->error." [$sql]"); /* I have added the suggestion from Your Common Sence */
    if($result->num_rows == 1) {
    $row = $result->fetch_array();
    echo $row['data'];
    }

    $MysqLi->close() ;

// In the OOP style if you want more than one row. Or if you query contains more rows.    

$MysqLi = new MysqLi($host,$database);
$sql = "SELECT * FROM data_table";
$result = $MysqLi->query($sql) or trigger_error($MysqLi->error." [$sql]"); /* I have added the suggestion from Your Common Sence */
    while($row = $result->fetch_array()) {
     echo $row['data']."<br>";
    }

    $MysqLi->close() ;    

?>