No output from mysql query in PHP - php

Question

I am troubleshooting this query that I was previously trying to use to count rows but now I have discovered that 0 rows are being found in the database even though they exist. I have checked previous queries I have used and they work and look the same. The $db_conx definitely works to connect to the database as I am using it on other web pages. Does any one have any suggestions? Thanks
$result =("SELECT * FROM Request WHERE user2='joey' AND accepted='0'");
$query = mysqli_query($db_conx, $result);
$rows = mysqli_num_rows($query);
if (rows < 1) {
echo '<strong style="color:#F00;">No rows</strong>';
exit();
} else {
echo '<strong style="color:#F00;">Rows exist</strong>';
exit();
}

Answer 1

You are missing an important step to select something from the database and that is what are you trying to select. And, you are also missing the $ in the rows.
$result =("SELECT something FROM Request WHERE user2='joey' AND accepted='0'");
$query = mysqli_query($db_conx, $result);
$rows = mysqli_num_rows($query);
if ($rows < 1) {
echo '<strong style="color:#F00;">No rows</strong>';
exit();
} else {
echo '<strong style="color:#F00;">Rows exist</strong>';
exit();
}

Answer 2

You edited your original post https://stackoverflow.com/revisions/33307758/1
being SELECT FROM Request
You have selected "nothing" from your query
SELECT FROM ...
Read the manual on SELECT:
https://dev.mysql.com/doc/refman/5.0/en/select.html
and this if (rows < 1) { - rows is treated as a constant instead of a variable.
Error reporting would have told you that if it was setup to catch and display on your system.
if ($rows < 1) {
and make sure you are successfully connected to your db using the same MySQL API as your query.
Check for errors on your query
http://php.net/manual/en/mysqli.error.php
Look into using a prepared statement also.
https://en.wikipedia.org/wiki/Prepared_statement
Nota:
You will also need to escape your data should it be coming from user input at one point.
I.e.:
$var = $_POST['var'];
and if $_POST['var'] is equivalent to joey's bistro.
MySQL will see that as WHERE user2='joey's bistro' if it were passed as a variable in the query WHERE user2='$var' resulting in a syntax error.
Escaping it will render it as joey\'s bistro being valid syntax.
$var = mysqli_real_escape_string($db_conx,$_POST['var']);
This would be beneficial for just that as well as helping to protect against an SQL injection.
https://en.wikipedia.org/wiki/SQL_injection