这是代码。当我执行它时,它显示错误消息"可捕获的致命错误:类mysqli_result的对象无法转换为字符串" 请帮我解决这个问题。
$sql = "SELECT dname FROM bdonor WHERE DID = 2";
$dname = $con->query($sql);
echo $dname;
答案 0 :(得分:0)
您正在尝试将类 mysqli_result转换为字符串(显然它会返回错误),试试这个..
$sql = "SELECT dname FROM bdonor WHERE DID = 2";
$result= $con->query($sql);
while($row = $result->fetch_array(MYSQLI_ASSOC))
echo $row['dname '].'<br />'; // here is the output display line by line
答案 1 :(得分:0)
试试此代码
function mysqli_result($res,$row=0,$col=0){
$numrows = mysqli_num_rows($res);
if ($numrows && $row <= ($numrows-1) && $row >=0){
mysqli_data_seek($res,$row);
$resrow = (is_numeric($col)) ? mysqli_fetch_row($res) : mysqli_fetch_assoc($res);
if (isset($resrow[$col])){
return $resrow[$col];
}
}
return false;
}