Only 2 options should get echo

PHP
1

hi all

there are 2 products inserted in my cart table.

  1. product 1094
  2. product 1095

i want to echo the “message” according to their existence in the cart table.

but my code echos all 4 messages.


<?
$qry = "select * from cart_table";
$result = mysql_query($qry);
while($row=mysql_fetch_array($result))
		{
							 
			if($row['product_id'] == 1094)
			{
			echo "1094 is already there";
			}
			
			if($row['product_id'] != 1094)
			{
			echo "1094 is not there";
			}
								 
			if($row['product_id'] == 1095)
			{
			echo "1095 is already there";
			}
			
			if($row['product_id'] != 1095)
			{
			echo "1095 is not there";
			}
							 
		}
?>

out of 4 only 2 message should be displayed

vineet

2

Why should the “____ is not there” message be displayed? There is nothing in this code that checks the WHY. Once that’s resolved (and the code updated correctly) it will only display 2/4 at any given time.

3

hi wonshikee

when a product is added in the cart table then its product_id gets added

so i am checking if the product_id=1094 doesnt exists then the message should say “1094 is not there”

vineet

4

Why not just select the fields you need from the outset?


$tgt = array(1094, 1095);
$qry = "select product_id from cart_table where product_id IN (".join(',', $tgt).")";


 // echo $qry; // debug

// spoofing your result set
$rows[0]['product_id'] = 1000;
$rows[1]['product_id'] = 1094;

foreach( $rows as $row ){
  if( in_array($row['product_id'], $tgt ) ) {
    echo $row['product_id'] . " is in the table";
  }else{
    echo $row['product_id'] . " is NOT in the table"; 
  }
}
5
Edit:

Seems several others have posted along these lines too, this just explains why your original code is flawed, with a bit of logical thinking

Look at your logic and think it over again.

First let’s consider the fact that you can simply your IF statements to

            if($row['product_id'] == 1094)
            {
            echo "1094 is already there";
            } else {
            echo "1094 is not there";
            }

            if($row['product_id'] == 1095)
            {
            echo "1095 is already there";
            } else {
            echo "1095 is not there";
            }

Now what if your cart_table contains the following product_ids: 1094 and 1095

When you loop through your records, you can easily see why it prints out all 4 messages. First record will produce 1094 is there and 1095 is not there, Second record will produce 1094 is not there and 1095 is there.

6

thanks cups for this solution.

i learned about in_array() from you and used it in my another code also.
its a good function

vineet

7

Good for you, just try and pass the knowledge on one day, and lets try and keep the PHP community awesome! :wink:

8

Cups, your… example doesnt work?

$tgt = array(1094, 1095);
$qry = “select product_id from cart_table where product_id IN (”.join(‘,’, $tgt).“)”;

// spoofing your result set
$rows[0][‘product_id’] = 1000; <– this row would NOT be returned by the query you just gave.
$rows[1][‘product_id’] = 1094;

If the table contained entries for 1094 and 1095, and you searched for 1000 and 1094, the code would only spit out “1094 is in the table”.
My understanding of the OP was that he wanted a response for all search items. (If this is incorrect, let me know lol)

Consider implementing array_intersect and array_diff.


//Assume table holds entries for 1094,1095.
$srch = array(1000,1094);
$qry = "select product_id from cart_table where product_id IN (".join(',', $srch).")";

//fetch_assoc the results, stripping the values into $rows.
//$res = $db->query($qry);
//while($row = $res->fetch_assoc()) {
// $rows[] = $row['product_id'];
//}
//Spoofing the result:
$rows[0] = 1094;
//Only one result, because 1000 isnt in the table.
$in_table = array_intersect($srch,$rows);
$out_table = array_diff($srch,$rows);
if(count($in_table) > 0) {
 echo implode(',',$in_table)." are in the table already.<br />";
}
if(count($out_table) > 0) {
 echo implode(',',$out_table)." are not in the table already.<br />";
}
9

Ok, thanks, I remember dashing that off – I should have created some better test cases for myself.

I think I just wanted to question the whole logic of “selecting * from the db”, and then got a bit carried away.

Using array_intersect() looks better.