Check SQL database if value exists and then return value if it does

家住魔仙堡 提交于 2020-02-20 09:22:08

问题


I am pretty new to both php and SQL. I have a login page where I ask for the users username and password. I would like to search a database I have created for both the username and password to see if they exist. The database table has two columns, Username and Password. I don't care to much about security so a simple script will work. But I do want to be able to expand on it someday, so therefor I am using a database, because currently I just use an array in php to store usernames and passwords.

I am currently trying to get this code to work but am not getting the results I need.

$user_name = "db_username";
$password = "db_password";
$database = "db_name";
$server = "db_server";



$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);

if ($db_found) {
$result =mysql_query("SELECT 1 FROM my_table WHERE Username = $username");
if ($result>0)

    {
        echo 'Username and Password Found'; 
    }
else
    {
    echo 'Username and Password NOT Found';
    }
}
else {
print "Database NOT Found.";
mysql_close($db_handle);
}

This always returns Username and Password Found no matter is the username is in there or not. When printing $result I get Resource id #2. Thank you


回答1:


$result I think will evaluate to true even if the result set contains zero rows. It only returns boolean false on error according to the manual. Use mysql_num_rows to determine if you actually found anything with the query.

if ($db_found) {
$result =mysql_query("SELECT 1 FROM my_table WHERE `Username` = '$username'");
if ($result && mysql_num_rows($result) > 0)

    {
        echo 'Username and Password Found'; 
    }
else
    {
    echo 'Username and Password NOT Found';
    }
}
else {
print "Database NOT Found.";
mysql_close($db_handle);
}

EDIT: Of course, as of now (November 2013, and since long ago), the mysql_* functions have indeed been deprecated. Apparently you can now use identical mysqli_* functions (maybe just use find/replace), but most people are using PDO.




回答2:


Try this for your SQL:

$result = mysql_query("SELECT 1 FROM my_table WHERE Username = " . mysql_real_escape_string($username));

Note: You should consider using PDO.




回答3:


<?php
$user=$_POST["user"];
$pass=$_POST["pass"];
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "interior_design";
$db_handle = mysql_connect($servername,$username,$password);
$db_found = mysql_select_db($dbname,$db_handle);
if($db_found)
{
$result = mysql_query("SELECT ssn,fname FROM admin WHERE ssn='$pass' and fname='$user'");
if ($result && mysql_num_rows($result) > 0)
    {
        echo 'Username and Password Found'; 
    //header( 'Location: index.html' ) ;

    }
else
    {
    echo 'Username and Password NOT Found';
    echo $pass;
    echo $user;``
    }
}

else{
echo 'db nt found';   

mysql_close($db_hanle);
}

?>
<html>
<body>
<a href="login.html"><h2>LOGIN</h2></a>
</body>
</html>


来源:https://stackoverflow.com/questions/8088516/check-sql-database-if-value-exists-and-then-return-value-if-it-does

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