PHP: Mysql查询麻烦

Question

<?php
    session_start();
    include "dbconnect.php";

     echo "email=".$_SESSION['email'];
    $uid = mysql_query("SELECT uid FROM master WHERE emailid='{$_SESSION['email']}'");
     echo $uid;
?>

sql查询没有唤醒结果是email=asd@asda.com资源 id #5那就是它在回显电子邮件而不是 uid 。

Answer 1

mysql_query 方法返回一个资源，而不是查询结果中的值。

你需要这样做:

$result = mysql_query("SELECT uid FROM master WHERE emailid='{$_SESSION['email']}'");
$uid = mysql_result($result,0);
echo $uid;

不过，在调用 mysql_result 之前，您可能应该检查以确保结果有效:

$result = mysql_query("SELECT uid FROM master WHERE emailid='{$_SESSION['email']}'");
if (mysql_num_rows($result) > 0) {
    $uid = mysql_result($result,0);
    echo $uid;
} else {
    echo "Error";
}

旁注:您正在使用已弃用的 mysql_* 方法，应该切换到 mysqli_* 或 PDO。