mysql_fetch_array() expects parameter 1 to be resource, string given

Question

I have two classes DBConn and DBQueries. DBQueries inherits after DBConn. When I'm going to display all users in users database, I see message:

mysql_fetch_array() expects parameter 1 to be resource, string given.

Thanks for any suggestions.

class DBConn /*extends Config*/ {
  public function dbConnection(){
    $db_host = 'localhost';
    $db_login = 'root';
    $db_password = '';
    $db_name = "database";
    $conn = mysql_connect($db_host, $db_login, $db_password);
    $db = mysql_select_db($db_name);
  }  
}

class DBQueries extends DBConn {
  function displayUsers(){
    $this->dbConnection();
    $query = "SELECT * FROM users";
    $result = mysql_query($query);
    while ($row = mysql_fetch_array($query)) {
      echo $row['password'];
    }
  }
}

jondavidjohn · Accepted Answer · 2011-02-04 18:45:43Z

16

You are passing in $query which is your sql string, when you need to pass in $result which is your db result object.

class DBQueries extends DBConn {
    function displayUsers(){
        $this->dbConnection();
        $query = "SELECT * FROM users";
        $result = mysql_query($query);
        while ($row = mysql_fetch_array($result)) {
            echo $row['password'];
        }
    }
}

answered Feb 4, 2011 at 18:45

jondavidjohn

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Comments

Matt K · Accepted Answer · 2011-02-04 18:45:43Z

4

You need to pass in the result handle $result returned by mysql_query().

answered Feb 4, 2011 at 18:45

Matt K

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Comments

feihtthief · Accepted Answer · 2011-02-04 18:46:16Z

0

You need to pass $result into the mysql_fetch_array function call as the first parameter, not $query.

class DBQueries extends DBConn {
  function displayUsers(){
    $this->dbConnection();
    $query = "SELECT * FROM users";
    $result = mysql_query($query);
    while ($row = mysql_fetch_array($result)) {
      echo $row['password'];
    }
  }
}

answered Feb 4, 2011 at 18:46

feihtthief

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