在Codeigniter中使用Mysqli,并将Mysql作为Mysqli的新手而来。
以下Mysql查询引发Mysqli错误。我认为这可能与“ SELECT x AS y”有关。
$sql = "SELECT uk_postcode_lat AS lat, uk_postcode_long AS long FROM uk_postcodes where uk_postcode_code = '$zip' LIMIT 1";
如何使此查询与Mysqli一起使用?
最佳答案
long
是保留关键字。
尝试这个
$sql = "SELECT uk_postcode_lat AS lat, uk_postcode_long AS longitude FROM uk_postcodes where uk_postcode_code = '$zip' LIMIT 1";
关于mysql - mysqli选择x AS y,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15374307/