Pass Array Value through $_POST

jlpulido

I am in desperate need. I'm hoping that this is enough information to obtain the answer I'm searching for.

I'm trying to find the easiest method to pass the value from this code to another form. Here's a snapshot of the code that retrieves the fields from the db.

$res = @($sql, $link) or die(mysql_error());
$num = @mysql_num_rows($res);
$c = 0;
while ($row = @mysql_fetch_array($res)) {
$field_1[] = $row['field_1'];
$field_2[] = $row['field_2'];
$field_3[] = $row['field_3'];
$field_4[] = $row['field_4'];
$field_5[] = $row['field_5'];

		echo "<table border=1>";
		echo "<tr><td>"."$field_1[$c]"."$field_2[$c]"."</td><td>"."<input type=\"submit\" 			name=\"submit\"	value=\"Test\">"."</td></tr>";
		$c++;
		echo "</table>";
	}

I'm aware of the code inside the <table> tags doesn't reflect all the fields pulled from the db, but I'm sure everyone will see what the end result would be.

I hope this is enough to get some guidance.

Weedpacket

I think you might be in the wrong forum for this one. Whatever the question is.

jtp51

jlpulido: by having the line:

"<input type=\"submit\" name=\"submit\" value=\"Test\">"

in your loop, you'll have created the same number of submit buttons as you have records. I don't believe you want that.

If you have the submit button on the outside of the loop; then put your results in input text fields so you can submit the information by clicking on a single submit button.

Right now your just writing the results directly to the page and your unable to post or get the returned MySQL records.