SQL Where Help

[DudleyRay]

Hey all, I am just learning how to do SQL etc. I have a php script that used to work, but now it doesn't. Can someone tell me if this is the correct code? $sql_Eq = "SELECT Code1 AS Code FROM CodeTable WHERE Code2 = '$computercode'"; Names changed of course.

[Don E]

Looks right to me, but if you want to be more precise, you can post the actual code?

[DudleyRay]

function check_existing($eq,$exists,$ac,$link) { $sql_Eq = "SELECT Code1 AS Code FROM CodeTable WHERE Code2 = '$computercode'"; $typeRow = doQuery($link, $sql_Eq); if ($exists) { // (eg if there is any existing route) echo "Comparing : is $typeRow->CODE inside $eq ?<br>"; if (strpos(" ".$eq,$typeRow->CODE) > 0) { // the ac is there already. do nothing. $need = false; } else { // the ac isn't there. the list must be updated $need = true; } } else { // no route found : must add the route $need = true; } if ($need) { // prepare new ac to be added //echo "Need to add this : $typeRow->CODE<br>"; return $typeRow->CODE.","; } else { return false; } sqlsrv_free_result($types); }

[justsomeguy]

What happens when you run that? What do you mean when you say it isn't working? Is there an error message?

[DudleyRay]

It should fail out if I enter a value that isn't in the table. It should display values that are in the table and it isn't doing that.

[justsomeguy]

Add this to the top of whatever page you're running that function on: ini_set('display_errors', 1);error_reporting(E_ALL); Does the doQuery function check for database errors? On the last line you're using a variable called $types that I don't see defined, and the function sqlsrv_free_result isn't a built-in function if you're trying to use one there.

[DudleyRay]

Notice: Trying to get property of non-object in URL on line 42 This is line 42 return $typeRow->CODE.","; There is stuff I am leaving out I know is working because the information I want added is getting added. Just a one column of a table isn't have data entered. Even though the information is getting added to different columns within the table.

[justsomeguy]

That error means that $typeRow is not an object, so doQuery is not returning what you think it is. You can use var_dump if you want to figure out what it's returning. It's fine if you want to leave out code, but I can only answer questions based on what I see. If something that you think is working or is not important is actually important then I can't take that into account if you leave it out. I can tell you that $typeRow is not an object, but I don't know why because I don't know what the doQuery function does for example.

[DudleyRay]

Thanks for the help, I am going to contact the person who wrote this code. It used to work, and most of the file is. Just one field fails to get entered.