echo something out MySQL database - php

Question

I am trying to get a question with answers out of my database. I just want to get one thing out of the database and not with a row. I thought this would work but it puts out this: Resource id #4 can someone explains what I am missing.
Thanks :)
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db('lotto');
$sql = 'SELECT id, vraag, AntwA, AntwB, AntwC, AntwD FROM vraag1';
$test = mysql_query($sql);
echo $test;
?>

Answer 1

As said at least 10000 times everywhere in internet, never use MySQL_ ! (If your are trying to learn something new by using tutorials over internet, don't use old ones)
I recommend to use PDO which is modern API in PHP and a lot more secure when using it correctly with prepared statement ! But you can also use MYSQLI which is more similar to the MYSQL !
You have to export your data from return array :
Using PDO :
$db = new PDO ("mysql:host=".$hostname.";dbname=".$dbname, $username, $password);
$query = $db -> prepare ("SELECT * FROM vraag1");
$query -> execute (array ());
$rows = $query -> fetchAll (PDO::FETCH_ASSOC);
foreach ($rows as $row)
{
echo $id = $row["id"];
echo $vraag = $row["vraag "];
echo $AntwA = $row["AntwA "];
echo $AntwB = $row["AntwB "];
echo $AntwC = $row["AntwC "];
echo $AntwD = $row["AntwD "];
}
Using MYSQLI :
$db = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
$query = "SELECT * FROM vraag1";
$rows = mysqli_query($db, $query);
while($row = mysqli_fetch_assoc($rows))
{
echo $row["id"];
echo $row["vraag"];
echo $row["AntwA"];
echo $row["AntwB"];
echo $row["AntwC"];
echo $row["AntwD"];
}

Answer 2

First of all the mysql function you are using is depreciated and no longer supported. you should use mysqli or pdo instead with prepared statements.
<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "lotto";
// Create connection
$conn = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT id, vraag, AntwA, AntwB, AntwC, AntwD FROM vraag1";
$test = mysqli_query($conn, $sql);
if (mysqli_num_rows($test) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($test)) {
echo "ID : ".$row['id']."<br>";
echo "vraag :".$row['vraag']."<br>";
echo "AntwA :".$row['AntwA']."<br>";
echo "AntwB :".$row['AntwB']."<br>";
echo "AntwC :".$row['AntwC']."<br>";
echo "AntwD :".$row['AntwD']."<br>";
}
} else {
echo "no results found";
}
mysqli_close($conn);
?>

Answer 3

For select function mysql_query() returns a resource on success, or FALSE on error.
so your assignment statement
$test = mysql_query($sql);
assign the resource to $test.
if you want the data inside the resource you can do
while($row= mysql_fetch_assoc($test)):
print_r($row);
endwhile;
also you can access the $row['column_name']
If you want to return only one row you can do this limit in query
$sql = 'SELECT id, vraag, AntwA, AntwB, AntwC, AntwD FROM vraag1 limit 1';

Answer 4

You need to add something like the following:
while($row = mysql_fetch_array($result)){
echo $row['id'];
echo $row['vraag'];
echo $row['AntwA'];
echo $row['AntwB'];
echo $row['AntwC'];
echo $row['AntwD'];
}

Answer 5

use mysqli instead of mysql
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$conn = mysqli_connect($dbhost, $dbuser, $dbpass);
mysqli_select_db('lotto');
$sql = 'SELECT id, vraag, AntwA, AntwB, AntwC, AntwD FROM vraag1';
$test = mysqli_query($sql);
echo $test;
?>