php sql check value为每条正面记录打印一行

Question

I have been working on my "administrative" webpage for a while now, almost finished. Just need a little help with one last thing;

I want to get a value (ID) from a table posted as Yes if the ID is found, but no if it isn't.

Have managed to get it printet, but it prints all the records in the table on every person that displays on the website (http://gyazo.com/13f271bbb8c4e83ff9ecd9908545c854 where it says "Betalt" it should just be ONE correct for each one).

The code for that part is here:

if ($row[9] == $row[1]) {
$betalt = "Ja";
}

$row[9] is the value from table SI_PAYMENTS. And $row[1] is the value from table BUSS1.

These two values should be the same if the record exists in SI_PAYMENTS, and if it exists, I want the webpage to display "Yes". But if it don't find it, I want the webpage to display "No".

EDIT

Here's my SQL Query:

SELECT buss1.navn, buss1.plassnummer, buss1.telefon, buss1.epost, buss1.fodselsdato, buss1.pastigningssted, buss1.bilettype, buss1.ankommet, buss1.merknader, si_payment.ac_inv_id FROM buss1, si_payment ORDER BY buss1.plassnummer ASC

Answer 1

I think you need to JOIN the two tables. At the moment, you will get every row from table buss1 next to every row from si_payment. So if there are 6 rows in buss1 and 4 in si_payment, you'll end up with 24 rows. I suspect there is a column in the two tables that links them, so you need a JOIN: something like this

SELECT buss1.navn, buss1.plassnummer, buss1.telefon, buss1.epost, buss1.fodselsdato, buss1.pastigningssted, buss1.bilettype, buss1.ankommet, buss1.merknader, si_payment.ac_inv_id 
FROM buss1 INNER JOIN si_payment WHERE buss1.some_column = si_payment.another_column
ORDER BY buss1.plassnummer ASC

I can't tell what some_column and another_column should be, I'm afraid. But if you change those, that will give you only the matching rows from the two tables. If that's not quite right, you may need a LEFT JOIN, which will give you all rows from the first table and the matching rows from the same table.

If I've misunderstood, I apologise.