我正在尝试运行以下内容.
id
name
parent_id
parent_name
level
email ";
while ($row = mysqli_fetch_assoc($result))
{
$aid = $row["id"];
$sql2 = "SELECT * FROM members WHERE MEMNO = '$aid'";
$result2 = mysqli_query($db,$sql2) or exit(mysqli_error($db));
while ($newArray = mysqli_fetch_array($result2)) {
$fname = $newArray['FNAME'];
$lname = $newArray['LNAME'];
$mi = $newArray['MI'];
$address = $newArray['ADDRESS'];
$city = $newArray['CITY'];
$state = $newArray['STATE'];
$zip = $newArray['ZIP'];
$kdate = $newArray['KDATE'];
$date = abs(strtotime(date('m/d/Y')) - strtotime(date($kdate))) / (60 * 60 * 24);
}
echo sprintf("%s %s %s %s %s %s ",
$row["id"],$row["name"],
$row["parent_id"],$row["parent_name"],
$row["level"],$row["email"]);
}
echo "";
}
mysqli_free_result($result);
mysqli_close($db);
?>
如果我删除以下行:
$aid = $row["agent_id"];
至….
$date = abs(strtotime(date('m/d/Y')) - strtotime(date($kdate))) / (60 * 60 * 24);
}
一切都会好起来的.如果没有,我收到以下错误:
Commands out of sync; you can’t run this command now
在研究中,我认为这可能是由于多个MySQLi查询同时运行,其中使用mysqli_multi_query但是对于the guide中的所有样本和一般数据似乎都不适用.
有任何想法吗?