php - 为什么 mysql_fetch_array() 存储空结果？

Question

我成功地从数据库表中选择了数据，每当我尝试借助 mysql_fetch_array() 将该数据提取到数组中时，它不会将任何内容存储到数组中。

@session.start();
$name=$_SESSION['umailid'];
$chkname1 = "select * from ".USREG." where email='.$name.'";

echo $chkname1;
// it is printing like this:
// " select * from users_temp where email='.subbu66g@gmail.com.' "
// which means query was successful, above email is there in database table

$res1 = mysql_query($chkname1, $con) or die(mysql_error());

$chkresult1 = mysql_fetch_array($res1);
echo $chresult1['name']; //its not printing anything
if ($chkresult1) //it is storing null and entering into else block
{
    echo "query successful";
}
else {
    echo "query was not successful";
}

结果是“查询不成功”。我认为我的选择查询一切正常。那为什么这个 mysql_fetch_array() 没有获取数据呢？

Answer 1

在 $chkname 中更改为 email = $name(删除点)或 email = '$name'

而你正在使用 mysql，而是使用 mysqli 或 pdo sql