如何从下拉列表中选择值，如何从另一个表中获取相应的值，然后将其放入另一个下拉列表中？

Question

Ok, here's a database.

http://i.stack.imgur.com/j05AB.png

Say I've inserted values into the database for each of these tables, although the IDs would be auto incrementing. There are many BVALUES from each AVALUE, thus the AB table. I have all the AVALUEs from TABLE A in a drop-down list. A user selects an AVALUE which I put into a variable using

$AVALUE = $_POST['AVALUE']

Then I do an sql statement to get the AVALUEs from TABLE A that equal $AVALUE.

$sql = "SELECT AVALUE FROM TABLEA WHERE" . $AVALUE . " = AVALUE";

How do I then get the NAMEID from TABLEA for that AVALUE, then reference to AB where TABLEANAMEID = NAMEID from TABLEA? Then I want to get all the BVALUES by getting all the TABLEBNAMEIDs that correspond to the TABLEANAMEIDs.

I then want those BVALUES in a drop-down list on a seperate HTML page. After a bit of Googling the solution I think would be to do some sort of a loop putting the BVALUES into a variable as all the NAMEIDs from TABLE B increment where the variable would be in an $BVALUE loop and the list values would show with all the BVALUES.

I hope I explained that right. I think I know what I'm trying to do but I have no idea how to actually implement it. Please help guys.

Answer 1

You need to join those table together. What you are describing is an m:n relation. In this case you have do use 2 joins like this:

SELECT * FROM TableA AS a WHERE a.avalue = $AVALUE
JOIN TableAB AS a2b ON a.namevalue = a2b.id_a
JOIN TableB AS b ON a2b.id_b = b.id

Answer 2

Maybe u means, u want to get all BVALUE which has relation with selected AVALUE in table AB

$sql = "SELECT B.NAMEID as id, BVALUE as value FROM TABLEA A
        LEFT JOIN AB ON TABLEANAMEID=A.NAMEID
        LEFT JOIN TABLEB B ON TABLEBNAMEID=B.NAMEID
        WHERE AVALUE = $AVALUE";`

get mysql result

$result = mysql_query($sql);

iterate

echo "<select>";
foreach ($r = mysql_fetch_object($result)) {
   echo '<option value="'.$r->id.'">'.$r->value.'</option>';
}
echo "</select>";