MySQL with php, displaying 50 most recent - php

Question

I have a database which simply records words in a table in a single column (with RID beside). All I want to do is display the words in order with a space in between (which I have a working code for below)
<?php
$mysql_host = "mysql1.000webhost.com";
$mysql_database = "db name";
$mysql_user = "user";
$mysql_password = "pass";
// Create connection
$conn = new mysqli( "mysql1.000webhost.com","databaseuser","password","databasename");
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = 'SELECT Words FROM Poetry ';
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo $row["Words"]. " ";
}
} else {
echo "0 results";
}
?>
HOWEVER. I want to only display a set amount (say 50) of the MOST RECENT db entries (i.e) the 50 with the highest RID. Can somebody quickly tell me what code I need (in php) which will display the 50 most recent on a web page?
Thanks!

Answer 1

Don't do it with PHP, just add it to your query.
$sql = 'SELECT Words FROM Poetry order by RID desc limit 50';
This query will order the result set by RID descending, most recent first I'm assuming, and limit the result set to 50 records.

Answer 2

You can use
SELECT Words FROM Poetry order by RID desc limit 0,50
0 is the offset and 50 is limit.u can dynamically change those value if needed.
For the better security purpose You can use mysqli prepared statements.
You can see dynamically-bind_param-array-mysqli for Dynamically Bind Params

Answer 3

Try saving it in a variable and echoing it after your while statement.
Something like this
$all_words = '';
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$all_words = "$all_words, $row['Words']";
}
echo"$all_words";
} else {
echo "0 results";
}