PHP Page: No database selected issue

Question

I get No database selected on line 13 error. And this code line is

$result = $mysqli->query($query) or die($mysqli->error." on line ".__LINE__);

Database connected successfully but when I try to open "question.php" file I get this error.

<?php
     // Set question number
     $number = (int) $_GET['n'];

     /*
     * Get Question
     */
     $query = "SELECT * FROM questions WHERE questions.question_number = $number";

    // Get Result

    $result = $mysqli->query($query) or die($mysqli->error." on line ".__LINE__);

    $question = $result->fetch_assoc();


    /*
    * Get Choices
    */
    $query = "SELECT * FROM choices WHERE question_number = $number";

   // Get Results
   $choices = $mysqli->query($query) or die($mysqli->error.__LINE__);

?>

UPDATE!

My database connection:

<?php

    // Create connection credentials
    $db_host = 'localhost';
    $db_name = 'quizzer';
    $db_user= 'root';
    $db_pass='';

    // Create mysqli object

    $mysqli = new mysqli($db_host, $db_user,$db_pass,$db_pass);

    // Error handler
    if($mysqli->connect_error){
        printf("Connect failed: %s\n", $mysqli->connect_error);
        exit();
}

$mysqli = new mysqli($db_host, $db_user,$db_pass,$db_pass); 2 instances of $db_pass?? really?? $mysqli = new mysqli($db_host, $db_user,$db_pass,$db_name); perhaps — Professor Abronsius
– Professor Abronsius, Commented Aug 19, 2016 at 14:45
WARNING: When using mysqli you should be using parameterized queries and bind_param to add user data to your query. DO NOT use string interpolation or concatenation to accomplish this because you have created a severe SQL injection bug. NEVER put $_POST or $_GET data directly into a query, it can be very harmful if someone seeks to exploit your mistake. Using intval to scrub values is not a general purpose solution. — tadman
– tadman, Commented Aug 19, 2016 at 15:29

vincenth · Accepted Answer · 2016-08-19 14:45:56Z

2

You have a typo, write that :

$mysqli = new mysqli($db_host, $db_user,$db_pass,$db_name);

instead of :

$mysqli = new mysqli($db_host, $db_user,$db_pass,$db_pass);

answered Aug 19, 2016 at 14:45

vincenth

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