从表中选择简单的SQL查询,其中email = email [英] Simple SQL query selecting from table where email = email
问题描述
我有两个名为$ email和$ pass的php值.
I've got two php values called $email and $pass.
email-MySQL数据库中的行名 密码-MYSQL数据库中的行名
email - Name of row in MySQL database password - Name of row in MYSQL database
我正在运行一个SQL查询,以从表成员中进行选择,其中email = $ email和password = $ pass.
I'm running a sql query to select from table member where email = $email and password =$pass.
然后我正在运行mysqli_query来查看是否存在行,但没有得到任何结果.回声肯定会回显出信息匹配的ID.
I'm then running mysqli_query to see if a row exists, I'm not getting any results. Surely the echo would echo out the ID of where the info matches.
//Get the connection info.
global $connect;
$sql = "SELECT FROM members WHERE email='$email' AND password='pass'";
//Fetch the row and store the ID of this row.
$row = mysqli_query($connect, $sql);
$id = $row['userID'];
echo $id;
推荐答案
除了该代码已大量暴露于SQL注入这一事实之外,您还在查询数据但未获取结果.
Besides the fact that this code is massively exposed to SQL injection.. you are querying the data but not fetching the results.
添加提取命令:
$data = mysqli_query($connect, $sql);
$row = mysqli_fetch_assoc($data);
$id = $row['userID'];
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