If Statement proper formula - php

Question

I have looked on here about if statements. I have found a few things but I am having issues figuring out the proper statement formula.
I have 2 tables in the database with the following 2 fields
table 1
rct_app_id
table 2
uid
now if the uid field matches the rct_app_id field I want it to
echo "Green Light";
if they don't match
echo "No Go"
this is my formula
<?php
$user_id = $_SESSION['uid'];
$sql = "SELECT * FROM recruits WHERE rct_app_uid = {$user_id}";
$result = query($sql);
$rct_app_id = ['rct_app_id'];
if ($rct_app_id == 'uid') {
echo "Green Light";
} else {
echo "No Go";
}
?>
function query($query)
{
global $connection;
return mysqli_query($connection, $query);
}

Answer 1

Try this. but keep in mind its hard for people to figure out whats going on by bits and pieces and it makes it harder to help you.
<?php
$user_id = $_SESSION['uid'];
$sql = "SELECT * FROM recruits WHERE rct_app_uid = {$user_id}";
$result = query($sql);
while(($row = mysqli_fetch_assoc($result))!=false){
$rct_app_id = $row['rct_app_id'];
if ($rct_app_id == $user_id) {
echo "Green Light";
} else {
echo "No Go";
}
}
}
?>

Answer 2

You need to fix two lines. $result has the results from the database, so that's the source for the rct_app_id data. Then, when you do the comparison, you need to compare the two variables.
$rct_app_id = $result['rct_app_id'];
if ($rct_app_id == $user_id) {
The way you have it, you're comparing an array to a string.
When you do this:
$rct_app_id = ['rct_app_id'];
You're actually setting the variable $rct_app_id equal to an array with one element, although the syntax is incorrect. Instead, you need to get one element of the array that is returned from the database. This assumes that you have a function called query() that is working properly and returning an array.
Instead, we need to set the variable equal to one element of the array like so:
$rct_app_id = $result['rct_app_id'];
Then, when you do a comparison like this:
if ($rct_app_id == 'uid') {
you're saying if the variable $rct_app_id is equal to the string uid, which it's not. Variables always start with $ in php, strings are quoted. The variable set earlier in the script is $user_id (from SESSION), so we need to compare to that:
if ($rct_app_id == $user_id)

Answer 3

UPDATE: You've specified your sql lib, I've edited the answer below to work with your updated answer.
Since you didn't specify the library, I'm making the answer and the code edits with the assumption that you're using mysql. Though all queries and return functions use similar syntax, ie: mysql_fetch_assoc() = mysqli_fetch_assoc(), pg_fetch_assoc(postgres).
<?php
$user_id = $_SESSION['uid'];
$sql = "SELECT * FROM recruits WHERE rct_app_uid = {$user_id}";
$result = query($sql); //What type of query runs as just query()? mysql_query would go here if this was mysql. Some Libraries offer this as a function, but since you didn't specify the library, I'm going to change it to mysql_query and proceed as if you're using mysql.
//$rct_app_id = ['rct_app_id'];//This will never work.
//You need this:
while($row=mysqli_fetch_assoc($result)){
//We only expect one result
$rct_app_id=$row['rct_app_id'];
}
if ($rct_app_id == 'uid') {
echo "Green Light";
} else {
echo "No Go";
}
function query($query)
{
global $connection;
return mysqli_query($connection, $query);
}
?>