CODEWITHSUNDEEP
phpmysql
Avdhut K
Avdhut K

Reputation: 74

Mysql select query results in null row values using php

** Following query executes properly and gives exact result for login. But when I trying to access fields returned from the query, it shows me null. In below code, I've checked for correct email id and my column name is 'user_email', it validates the user login correctly, but when I was trying to access the same email id from returned result($checkLogin->user_email), it gives me null ** 

 $checkLogin=$db->query("SELECT * from `users` where
    `user_email`='$u_email' and `user_password`='$u_password'");
      if($checkLogin)
      {
        $data[]=array(
         'status'=>'success',
         'email'=>$u_email,
         'pass'=>$u_password,
         'email1'=>$checkLogin->user_email
        );
      }

Please give me solution over this

Upvotes: 0

Views: 168

Answers (2)

pgross
pgross

Reputation: 509

The query() function returns a result object. You need to fetch your result. There are a lot of functions for this: fetch_row(), fetch_assoc(), fetch_array(), fetch_all(). You can find examples of how to use those at php.net.

You could for example write the following:

$checkLogin = $db->query("SELECT * from `users` where `user_email`='$u_email' and `user_password`='$u_password'");

if($row = $checkLogin->fetch_object())
{
    echo $row->user_email;
}

Upvotes: 2

Avdhut K
Avdhut K

Reputation: 74

I found the Solution. I've to use $db->get_row instead of $db->query as follows

  $checkLogin=$db->get_row("SELECT * from `users` where `user_email`='$u_email' and `user_password`='$u_password'");
  if($checkLogin)
  {
    $data[]=array(
     'status'=>'success',
     'email'=>$u_email,
     'pass'=>$u_password,
     'email1'=>$checkLogin->user_email
    );
  }

Upvotes: 0

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