Want to protect your cyber security and still get fast solutions? Ask a secure question today.Go Premium

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 184
  • Last Modified:

Why my array output is wrong

PHP

Why my array output is wrong

i have this code in code snippet

my result should be
supermarketname | Amount_of_candy | Amount_of_milo | AMount_of_milk |...
    coles       |    1.00000      |   5.00000         |    2.00000     |...
    target      |    5.00000      |  10.00000      |    3.00000     |...
...
   
but i get result as below
      
coles | "1.00000"|target| "5.00000"| parkson| "2.00000"... and so on

other words my amount become
coles | array[0] | target | array[1] | parkson | array[2] .. and so on

this is so wrong as you can see from my code GetAmount(()

coles should follow by array[0],array[1] and array[2]

but dont know why array[1] has go to target which is wrong
how to do this?

function GetAmount()
{
$ItemID = array(1234,6545,5468,9856,"no stock",4561,5466);
$arrayElement[3] = "no stock";
$supermarketID = 21;
 
 $i = 0;
 $j = 0;
 
 for ($j = 0; $j < sizeof($ItemID); $j++)
  {
 if (($ItemID[$j] != $arrayElement[3]) 
	{
	 $query = "SELECT getitem_std_amount( \"$supermarketID\",  \"ItemID[$j]\") AS AMOUNT";
     
	 $result = mysql_query($query);
     if(!$result) die("Query didn't work. " . mysql_error());
    
     $temp = "";   	 
     while($row = mysql_fetch_array($result))
	      {
            $temp = '"'.$row['AMOUNT'].'"';
              } 
              $this->GetAmountReturn[$i++] .= $temp ; 
              }	 
   	}  
	else
    {
      $this->GetAmountReturn[$i] .=  "\"Full Time Employees\" |";
  	  $i++;
	}
   }
 
 Wishing you a very happy birthday and a year full of things you love 
function joinSUPERMARKETNAMEnAMOUNT()
{
//SUPERMARKETNAME select value is from another function
 
 foreach(($this->GetSupermarketNameReturn_value as $i => $SupermarketName)
	{
         $data1 = "{$SupermarketName} | {$this->GetAmountReturn[$i]}  "; 
	  print_r( $data1);
	 }
        }  
}

Open in new window

0
firekiller15
Asked:
firekiller15
  • 4
  • 3
1 Solution
 
hernst42Commented:
Try replacing

              $this->GetAmountReturn[$i++] .= $temp ;
              }  
        }  
        else
    {
      $this->GetAmountReturn[$i] .=  "\"Full Time Employees\" |";
          $i++;
        }
with

              $this->GetAmountReturn[$j] .= $temp ;
              }  
        }   else    {
      $this->GetAmountReturn[$j] .=  "\"Full Time Employees\" |";
        }
0
 
firekiller15Author Commented:
change it to $j doesnt make any different?
0
 
hernst42Commented:
please post the modified code
0
Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
firekiller15Author Commented:
function GetAmount()
{
$ItemID = array(1234,6545,5468,9856,"no stock",4561,5466);
$arrayElement[3] = "no stock";
$supermarketID = 21;
 
 
 
 for ($j = 0; $j < sizeof($ItemID); $j++)
  {
 if (($ItemID[$j] != $arrayElement[3])
      {
       $query = "SELECT getitem_std_amount( \"$supermarketID\",  \"ItemID[$j]\") AS AMOUNT";
     
       $result = mysql_query($query);
     if(!$result) die("Query didn't work. " . mysql_error());
   
     $temp = "";         
     while($row = mysql_fetch_array($result))
            {
            $temp = '"'.$row['AMOUNT'].'"';
              }
              $this->GetAmountReturn[$j] .= $temp ;
              }       
         }  
      else
    {
      $this->GetAmountReturn[$j] .=  "\"Full Time Employees\" |";
          
      }
   }
 
 Wishing you a very happy birthday and a year full of things you love
function joinSUPERMARKETNAMEnAMOUNT()
{
//SUPERMARKETNAME select value is from another function
 
 foreach(($this->GetSupermarketNameReturn_value as $i => $SupermarketName)
      {
         $data1 = "{$SupermarketName} | {$this->GetAmountReturn[$i]}  ";
        print_r( $data1);
       }
        }  
}
 
0
 
hernst42Commented:
ok slowly I get the datastructure. The replace with $j is wrong, replace the $j in those two lines with $supermarketID
0
 
firekiller15Author Commented:
is still the same
0
 
hernst42Commented:
hm no clue and without knowing you datastructures it can get very complicated.
0
 
fiboCommented:
Looking at << $query = "SELECT getitem_std_amount( \"$supermarketID\",  \"ItemID[$j]\") AS AMOUNT";>>
and using the sample data you are testing with
$ItemID = array(1234,6545,5468,9856,"no stock",4561,5466);
$arrayElement[3] = "no stock";
$supermarketID = 21;

The first run will give SELECT getitem_std_amount( "21",  "1234") AS AMOUNT;
The second SELECT getitem_std_amount( "21",  "5454") AS AMOUNT;
and 5th one SELECT getitem_std_amount( "21",  "no stock") AS AMOUNT;

What is precisely the result of running $result = mysql_query($query) ?


0

Featured Post

What does it mean to be "Always On"?

Is your cloud always on? With an Always On cloud you won't have to worry about downtime for maintenance or software application code updates, ensuring that your bottom line isn't affected.

  • 4
  • 3
Tackle projects and never again get stuck behind a technical roadblock.
Join Now