Nothing is wrong with my query, but doesn't retern data in php

[TomatoLover]

I am trying to run the SELECT statement which is supposed to return data that matches with the specific condition.
I ran this query in MySQL and it works fine.
Now I put the same query in php and tried to execute the query. But it doesn't return anything. I checked if $year and $app have values and they did. I printed out $result and it returned "Resource ID #4". I know it is php internal code but I don't know what #4 indicates. I printed out $row, and it returned a word "Array". And furthermore, I tried to print out what is in array, using $row[0], $row[1], etc..., but return anything, because it is empty.

if($year && $app){

$query = "SELECT * FROM table_status WHERE Program='$app' AND ProgAY='$year'";
$result = @($query); //run the query
$row = mysql_fetch_array($result, MYSQL_NUM);
}else{ //If everything wasn't OK
$message .= '<br>Please try again.</br>';
}

Can anyone tell me what is wrong with this code?

[TomatoLover]

Sorry, I made a mistake.
What I wanted to say is that it returns NOTHING when I tried to print out the contents of an array.

Thanks.

[barand]

Easiest way to print contents of an array is [man]print_r/man

if($year && $app){ 
$query = "SELECT * FROM table_status WHERE Program='$app' AND ProgAY='$year'"; 
$result = @mysql_query($query); //run the query 
$row = mysql_fetch_array($result, MYSQL_NUM);

if ($row) {
    echo '<pre>'; // to give more readable output
    print_r($row);
    echo '</pre>';
} else {
    echo "No record found<br>";
}

}else{ //If everything wasn't OK 
$message .= '<br>Please try again.</br>'; 
} 

hth

[joelvb]

Hi tomatolover,

if the error is in your SQL statement try changing it from:
$query = "SELECT * FROM table_status WHERE Program='$app' AND ProgAY='$year'";

to:
$query = "SELECT * FROM table_status WHERE (Program='$app' AND ProgAY='$year')";

the close & open parenthesis might help

also, i notice this statement in your code
$result = @($query); //run the query

maybe you can get rid of the @ sign, i usually code like this
$result = mysql_query($query); //run the query

if it doesn't work yet try removing MYSQL_NUM from this statement
$row = mysql_fetch_array($result, MYSQL_NUM);

make it simpler like this

$row = mysql_fetch_array($result);

hope this helps

[helz]

are you expecting more than 1 row for the query? if so:
you need to use a while loop, also you need to get the info from the array using the $row['field_name'] so try this:

$query = mysql_query("SELECT * FROM table_status WHERE (Program='$app' AND ProgAY='$year')");
while ($row = mysql_fetch_array($query)) {
   $field1 = $row['field1'];
   $field2 = $row['field2'];
   echo "Field 1 = $field1, Field 2 = $field2";
}

with the loop above the echo will display 1 row, then as the loop goes to the next row then the fields for that row will be shown...

[TomatoLover]

Thank you all who gave me an advice.

Actually, I found out that the problem wasn't query.
The query returns the 1 row in a table, but
for some reason, if($row){ ... wasn't working.
But I fixed it, so it is working fine.

But advices you gave me was very helpful!
I can try them when I encounter another problems.

Thanks again.