Undefined variable: mysqli & call to a member function query() on null

Question

I'm kinda new to mysqli and I'm trying to display data on a table.

Here's the code:

<tbody>
   <?php
   $query = $mysqli->query("SELECT username,password,FName,LName,userAddress FROM tbl_users");
   $no = 1;
   while($row = $query->fetch_assoc()){
   ?>
   <tr>
  <td><?php echo $no++ ?></td>
  <td><?php echo $row['username'] ?></td>
  <td><?php echo $row['password'] ?></td>
  <td><?php echo $row['FName'] .' '. $row['LName'] ?></td>
  <td><?php echo $row['userAddress'] ?></td>
  <td>
      <a href="update.php?id=<?php echo $row['userid'] ?>" class="btn btn-warning btn-sm"><span class="glyphicon glyphicon-pencil" aria-hidden="true"></span></a>
     <a onclick="return confirm('Are you sure you want to delete data')" href="delete.php?id=<?php echo $row['userid'] ?>" class="btn btn-danger btn-sm"><span class="glyphicon glyphicon-trash" aria-hidden="true"></span></a>
  </td>
  </tr>
  <?php
  }
  ?>
</tbody>

These are the errors:

Undefined variable: mysqli in C:\xampp\htdocs\company\admin\user_table.php on line 205

Fatal error: Call to a member function query() on null in C:\xampp\htdocs\company\admin\user_table.php on line 205

pass db connection in query : $mysqli->query($conn, "SELECT username,password,FName,LName,userAddress FROM tbl_users"); — Krishna Gupta
– Krishna Gupta, Commented Dec 2, 2015 at 12:01
@Krishna, no, $mysqli would be the connection resource, it's either the oop-style where the object is the connection resource or the procedual style where you pass the resource as a parameter. — VolkerK
– VolkerK, Commented Dec 2, 2015 at 12:03
If you haven't created a connection to your database, you need to do so. If you have, are you sure you named it $mysqli ? — Epodax
– Epodax, Commented Dec 2, 2015 at 12:03
Off topic, but this jumped out at me from your code: You should never have any reason to echo a user's password. If you're doing things properly, the passwords will hashed and unreadable anyway so no point outputting them. — Simba
– Simba, Commented Dec 2, 2015 at 12:24

VolkerK · Accepted Answer · 2015-12-02 12:19:49Z

2

Undefined variable: mysqli means you didn't create object 'mysqli' or not able to get object mysqli; i.e. you will have to establish a connection to the MySQL server by creating an instance of mysqli before you can use this instance to send queries to the server.

create object

$mysqli = new Mysqli("host","db user","db password","db name");

//conect with database

For how to check whether the connection was established successfully, see http://docs.php.net/manual/en/mysqli.construct.php .

edited Dec 2, 2015 at 12:19

VolkerK

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answered Dec 2, 2015 at 12:03

Parth Chavda

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2 Comments

VolkerK Over a year ago

For how to check whether the connection was established successfully, see docs.php.net/mysqli.quickstart.connections . And an explaination why and how this is relevant to the question would be nice, too ;-)

Parth Chavda Over a year ago

@VolkerK you can vote up if my code is solve issue of Undefined variable: mysqli

user3744076 · Accepted Answer · 2015-12-02 12:18:06Z

0

Thank you @Epodax for reminding that. Just changed my config to $mysqli.

<?php 
$mysqli = new mysqli('localhost','root','sa','company');
if($mysqli->connect_errno){
echo "Connection Failed".$mysqli->connect_error;
}
?>

answered Dec 2, 2015 at 12:18

user3744076

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1 Comment

VolkerK Over a year ago

You might want to add a bit more error handling. The remaining script can't "live" without the working connection, so, only printing the error message won't suffice (and might be dangerous). You should print a "sorry, unavailable" page instead of the users table in case something went wrong with the mysql connection. And $mysqli->query may also fail; you should add error handling there, too. if ( !$query ) { ....error handler ... } else { while( $row = ...

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