Help with SQL query... PLEASE!

[feinhorn]

Hi...

I am a newbie so please... HELP ME... I am a bit desperate because I did 1000 changes and nothing works...

I have this code working fine:

mysql_select_db("tlibros_shop", $link);

$ventas = mysql_query("SELECT orders_id, customers_name, customers_country FROM orders WHERE orig_reference LIKE '%" . $url . "%'", $link);

if ($row = mysql_fetch_array($ventas)){

	echo "<table border = '1'> \n";

	echo "<tr> \n";

	echo "<td><b>Nº de Venta</b></td> \n";

	echo "<td><b>Nombre del Cliente</b></td> \n";

	echo "<td><b>Total de la Venta</b></td> \n";

	echo "</tr> \n";

	do {


		echo "<tr> \n";

		echo "<td align=center>".$row["orders_id"]."</td> \n";

		echo "<td>".$row["customers_name"]."</td>\n";

		echo "<td>".******."</td>\n";

		echo "</tr> \n";

	} while ($row = mysql_fetch_array($ventas));

		echo "</table>";

		echo "</table> \n";

} else {

	echo '<div class="Estilo3"><strong>¡ No se ha registrado ninguna venta proveniente de tu URL !</strong></div>';

}

Now, I want to show where i placed the "*****" the price of the sale that its save in another table, the query should be like this:

 $result = mysql_query("SELECT text FROM orders_total WHERE orders_id = '" . $row["orders_id"] . "', title = 'Total:', $link); 

and place the ONLY one result that it will get in the first table where i placed *****...

But I don't know how to do this...

PLEASE HELP ME

Fede

[ragedigital]

Are you getting an error?

Are you getting more than 1 result per Query?

What is the title = 'Total:' for?

$result = mysql_query("SELECT text FROM orders_total WHERE orders_id = '" . $row["orders_id"] . "', title = 'Total:', $link);

[feinhorn]

ragedigital wrote:

Are you getting an error?

Are you getting more than 1 result per Query?

What is the title = 'Total:' for?

$result = mysql_query("SELECT text FROM orders_total WHERE orders_id = '" . $row["orders_id"] . "', title = 'Total:', $link);

Thanks for your answer ragedigital!!

Let me explain it a little bit more

(first, sorry for my english, Im from a spanish country)

I have to show in one table, results from 2 diferent queries. But the second query use the results of the first query to select the results.

The firs query, selects from table orders, orders_id and orders_customers where referingurl is ****** (depends on user). But order total ($$), is not in the same table... so I need to do a second query to select the total amount of the order, as in the table orders_total are different lines for the same order the query needs to select the amount (text) from the table orders_total where orders_id (from the first query) AND title (is the type of data sorted in "text") = "Total:"...

I tried many diferent ways to do this but I cant get any result. I just get the table resultant from the first query... but I dont know where to put the second or how to do that query...

If you can help me I would be very appreciate!!

Thanks again,

Feinhorn

[ragedigital]

One issue I see is that you aren't pulling the array of data from the 2nd query - you may need to add:

$row_text = mysql_fetch_array($result);

after

$result = mysql_query("SELECT text FROM orders_total WHERE orders_id = '" . $row["orders_id"] . "', title = 'Total:', $link);

[feinhorn]

Hey again!

Well I tried this:

mysql_select_db("tlibros_shop", $link);

$ventas = mysql_query("SELECT orders_id, customers_name, customers_country FROM orders WHERE orig_reference LIKE '%" . $url . "%'", $link);

if ($row = mysql_fetch_array($ventas)){

	echo "<table border = '1'> \n";

	echo "<tr> \n";

	echo "<td><b>Nº de Venta</b></td> \n";

	echo "<td><b>Nombre del Cliente</b></td> \n";

	echo "<td><b>Total de la Venta</b></td> \n";

	echo "</tr> \n";

	do {

	$monto = mysql_query("SELECT text FROM orders_total WHERE text = 'Total:', orders_id = '".$row["orders_id"].", title = 'Total:'", $link);

	$row_text = mysql_fetch_array($monto);

		echo "<tr> \n";

		echo "<td align=center>".$row["orders_id"]."</td> \n";

		echo "<td>".$row["customers_name"]."</td>\n";

		echo "<td>".$row_text["text"]."</td>\n";

		echo "</tr> \n";

	} while ($row = mysql_fetch_array($ventas));

		echo "</table>";

		echo "</table> \n";

} else {

	echo '<div class="Estilo3"><strong>¡ No se ha registrado ninguna venta proveniente de tu URL !</strong></div>';

}

But I get this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/tlibros/public_html/referidos/members/index.php on line 131

and line 131 is this: $row_text = mysql_fetch_array($monto);

I dont know where is the error... may be I need to add an if at the sql_fetch_array.. I dont know... may be the query is made wrong...

Well... Thanks for your help ragedigital!!

Feinhorn

[tomhath]

It looks like the query failed. Check what mysql_error returns after the mysql_query.

[feinhorn]

tomhath wrote:

It looks like the query failed. Check what mysql_error returns after the mysql_query.

I Did it... !!!

I change 2 thinks! but now its working!! 🙂

here is the final code... I don't think it will be usefull for anybody... but maybe... who knows... 😃

mysql_select_db("tlibros_shop", $link);

$ventas = mysql_query("SELECT orders_id, customers_name, customers_country FROM orders WHERE orig_reference LIKE '%" . $url . "%'", $link);

if ($row = mysql_fetch_array($ventas)){

	echo "<table border = '1'> \n";

	echo "<tr> \n";

	echo "<td><b>Nº de Venta</b></td> \n";

	echo "<td><b>Nombre del Cliente</b></td> \n";

	echo "<td><b>Total de la Venta</b></td> \n";

	echo "</tr> \n";

	do {

		echo "<tr> \n";

		echo "<td align=center>".$row["orders_id"]."</td> \n";

		echo "<td>".$row["customers_name"]."</td>\n";

		$monto = mysql_query("SELECT text FROM orders_total WHERE orders_id = '".$row["orders_id"]."' AND title = 'Total:'", $link);

		while ($row_text = mysql_fetch_array($monto)){

			echo "<td>".$row_text["text"]."</td>\n";

		}

		echo "</tr> \n";

	} while ($row = mysql_fetch_array($ventas));

		echo "</table>";

		echo "</table> \n";

} else {

	echo '<div class="Estilo3"><strong>¡ No se ha registrado ninguna venta proveniente de tu URL !</strong></div>';

}

Thank you tomhath!!!