PHP variable contents not getting displayed - php

Question

As you can see from the code below im very new to PHP, so please excuse the newbie question but ive been struggling with this all afternoon and just cant figure out what exactly is the problem.
Basically Ive got 3 dropdown menus for team1, team2 and venue.
When user selects teams and venue an isset function is triggered and I extract the result from the game out of my database.
As you can see in the code below I use 3 echo statements to test if correct values from dropdown menus has been captured. When I open the page the echo statements confirm the correct results are captured from the drop-downs, I go on and query the database with the variables I echoed above.
The Problem
In the while loop, I want to echo the results from the query BUT nothing gets displayed. I tried doing a var_dump to see the contents of the variables but NOTHING gets displayed. What am I doing wrong?
foreach($_REQUEST["team1"] as $teamone){
$team1 = $teamone;
}
foreach($_REQUEST["team2"] as $teamtwo){
$team2 = $teamtwo;
}
foreach($_REQUEST['venue'] as $venue){
$venue = $venue;
}
//These echo statments are a test to see if the correct dropdown values has been captured
echo $team1.'<br>';
echo $team2.'<br>';
echo $venue;
//Use results from dropdown menu to query database
$result = mysql_query('SELECT *
FROM `results`
WHERE `hometeam` = "$team1" && `awayteam` = "$team2"') or
die(mysql_error());
while($row = mysql_fetch_array($result)){
echo $row['awayteamscore'];
echo $row['hometeamscore'];
}
If anyone can point me in the right direction it would be greatly appreciated.

Answer 1

I suspect you're receiving sql syntax error for two reasons at least
First when you want to use the variable value within string you must use double embracing quotes
$sql="select * from mytable where col1='$my_var'";
And second: SQL expects string values ebraced with simgle quotes as I pointed above. In your case you use "" quotes.
And one more recommendation. When debugging php app it might be useful to enable extended output by inserting
error_reporing(E_ALL);
somewhere in the beginning of php script

Answer 2

you mixed between double and single quotes . while variables shouldnt be rounded by double quotes.
$result = mysql_query(" SELECT *
FROM `results`
WHERE `hometeam` = '$team1' && `awayteam` = '$team2' ") or
die(mysql_error());
while($row = mysql_fetch_array($result)){
echo $row['awayteamscore'];
echo $row['hometeamscore'];
}

Answer 3

You put your SQL query in single quotes and variables do not work in there. Try this:
<?php
$var = "stuff";
echo '$var';
echo "$var";
?>
What you can do instead is to concatenate strings like this:
$result = mysql_query('SELECT *
FROM `results`
WHERE `hometeam` = "'.$team1.'" && `awayteam` = "'.$team2.'"')

Answer 4

My Friend Your code is just fine you just need to check that your submiting the dropboxes with the form so first check your html and when you var_dump($_REQUEST) it should be an array containing all the values that you trying to gather

Answer 5

Your foreach loops are overriding the values. It would be best to put it in an array like so
$team1 = array();
foreach($_REQUEST["team1"] as $teamone){
$team1[] = $teamone;
}
if you need to make the values of $team1 into a string then this would be more appropriate
$team1 = "";
foreach($_REQUEST["team1"] as $teamone){
$team1 .= $teamone;
}
Fixing the above first might help. If not then maybe the problem is in your db queries.
//Using double quotes
"SELECT * FROM `results` WHERE `hometeam` = '$team1' && `awayteam` = '$team2'"
//Using single quotes
'SELECT * FROM `results` WHERE `hometeam` = "'.$team1.'" && `awayteam` = "'.$team2.'"'
Then display the result
while($row = mysql_fetch_array($result)){
echo $row['awayteamscore'];
echo $row['hometeamscore'];
}