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HELP needed on this simple code no output no error


newbaba
Go to solution Solved by Jacques1,

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<?php
print_r ("
<form name=\"form1\" method=\"post\" action=\"quizt1a.php\">
<div align=\"center\">
<table width=\"699\" border=\"1\" cellspacing=\"1\" cellpadding=\"0\" bordercolor=\"#000000\">");

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "quizdb";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
//Get question Rows
$sql = mysqli_query ("SELECT * FROM quizt1");
while ($row = mysqli_fetch_array($sql)) {
$qno = $row["questionno"];
$question = $row["question"];
$qanswer1 = $row["answer1"];
$qanswer2 = $row["answer2"];
$qanswer3 = $row["answer3"];
$qanswer4 = $row["answer4"];
$qanswer5 = $row["answer5"];

print_r ("<tr>
<td width=\"26\">$qno</td>
<td width=\"360\">$question</td>
<td width=\"281\">
<select name=\"question$qno\" class=\"formbox1\">
<option value=\"0\" selected>-- Answers --</option>
<option value=\"1\">$qanswer1</option>
<option value=\"2\">$qanswer2</option>");
if ($qanswer3 <> ""){
print_r "<option value=\"3\">$qanswer3</option>";};
if ($qanswer4 <> ""){
print_r "<option value=\"4\">$qanswer4</option>";};
if ($qanswer5 <> ''){
print_r "<option value=\"5\">$qanswer5</option>";};
{
print_r "</select></td></tr>";
}
print_r '$quizid';
}
}
?>

 

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  • Solution

Several things:

  • Don't show your internal database errors to your visitors. It's very irritating for legitimate users, and it helps attackers gain information about your system. Instead, enable exceptions, so that error messages are automatically sent to the right device according to your PHP configuration (an error log in production, the developer's screen during development).
  • Your database structure is broken. Whenever you find yourself numbering table names or columns, there's something wrong. SQL is not Excel. It's a database system with specific rules how to store data.
  • Don't use SELECT *. Always select specific columns.
  • Don't copy every item of the $row array into an extra variable. Just access the array items whenever you need them: $row['index_of_the_item']. You'll want a more meaningful name than $row, though.
  • Stylistic HTML attributes like align, width, color etc. are dead since 1997. We have CSS now.
  • If this is the entire script, you're also failing to output a complete and valid HTML document.

Create a function for your database connection so that you can reuse this code in all your scripts:

 

database.php

<?php

/**
 * Establishes a MySQL database connection using mysqli
 *
 * @param $host     string the hostname or IP address of the database server (e. g. "localhost")
 * @param $user     string the name of the database user
 * @param $password string the password of the database user
 * @param $database string the database name
 * @param $charset  string the character encoding of the connection (e. g. "utf8)
 *
 * @return mysqli the connection
 *
 * @throws mysqli_exception if the connection failed
 */
function connect_to_database($host, $user, $password, $database, $charset)
{
    // make mysqli throw an exception whenever it encounters a problem
    $mysqli_driver = new mysqli_driver();
    $mysqli_driver->report_mode = MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT;

    $database_connection = mysqli_connect($host, $user, $password, $database);

    // specify the character encoding of the database connection
    $database_connection->set_charset($charset);

    return $database_connection;
}

Then create a configuration script to hold the connection parameters:

 

configuration.php

<?php

const DATABASE_HOST = 'localhost';
const DATABASE_USER = '...';
const DATABASE_PASSWORD = '...';
const DATABASE_NAME = '...';
const DATABASE_ENCODING = 'utf8';

Fix your database structure. For example:

TABLES

quizzes
- quiz_id

questions
- question_id
- question

answers
- question_id
- answer
- is_correct

quiz_questions
- quiz_id
- question_id
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