从表中提取随机数据，然后显示它

Question

I want to pull 1 random ID from a table and display the "text1" and the "text2" when they click a button. This is my current code, it does not have the button because I didn't find any help on google.

<?php select * from TEST order by rand() limit 1 ?>


<label for="exampleInputEmail1">Text1</label>
<input type="email" class="form-control" id="exampleInputEmail1" placeholder="Text1" value="
<?php 

while($row = $result->fetch_assoc()) {
echo "" . $row["text1"];
                                     }

?>">

Note that I have this code already at the top of the page in use for other things

<?php
session_start();
include 'config.php';

if(!isset($_SESSION['username'])){
header('location:index.php');
exit();
}
?>

Inside the config.php it has all of the MySQL login/database info.

Answer 1

Firstly this makes no sense and it not doing anything as it is now..

<?php select * from TEST order by rand() limit 1 ?>

Take that out, im amazed you're not getting php errors when you try to run.. ?

In order to do this you want to structure your array first so you can pull out your random item from it.

$arrayOfResults = $result->fetch_assoc();

Now we select a random item

$randomItem = $arrayOfResults[array_rand($arrayOfResults)];

and viola echo $randomItem; you should have what you're looking for.

the filler code ill leave to you!

As per your comments, your $random item will probably have keys.. so you can echo it as so : $randomItem['name'] or $randomItem['id'] where the KEY is your column ID in the database..

References :

Array_Rand

Fetch-assoc return details