在 PHP 中选择多个数据库，但单独打印 ID

Question

我想执行多数据库选择，但要单独打印ID。

$sql = "SELECT * FROM peron, peronmedia WHERE peronmedia.PID=5" and peron.ID=5"; $xc = mysqli_query($baglanti, $sql); $rs=mysqli_fetch_array($xc);
echo $rs["peron.PID"];
echo $rs["peronmedia.ID"];

但是此代码不起作用。我该怎么做？

Answer 1

在表后定义别名是更好的理解方式，并且必须指定数据库名称，否则它将上升field list is ambiguous

$sql = "SELECT pr.PID AS PERON_PID,pm.ID AS PERONMEDIA_ID FROM peron pr, peronmedia pm WHERE databasename1.peronmedia.PID=5 AND databasename2.peron.ID=5"; 
$xc = mysqli_query($baglanti, $sql);
$rs=mysqli_fetch_array($xc);
echo $rs["PERON_PID"];
echo $rs["PERONMEDIA_ID"];