some problems with in_array function

Question

Hi I am trying to use in_array, I think my syntax is correct, but it says "Wrong datatype for second argument"

My code is

$result = mysqli_query($con, "SELECT * FROM Products WHERE Quantity_On_Hand < Min_Stock");
$filter = mysqli_query($con, "SELECT ProductID FROM Orders");

while($row = mysqli_fetch_array($result))
    {
    if (in_array($row['ProductID'], $filter))
        {
        }
    }

My idea is to find out if the ProductID from Products Table is in the Order Table.

Could someone helps me, Thanks :-)

Michael · Accepted Answer · 2013-03-30 23:47:08Z

3

$filter isn't an array; it's a mysqli_result object:

$filter = mysqli_query($con, "SELECT ProductID FROM Orders");

I think you want to iterate over that, add each ProductID to a new array, and then pass that array to the in_array function like so:

$filter = mysqli_query($con, "SELECT ProductID FROM Orders");

$product_ids = array();

while ($row = $filter->fetch_assoc())
{
    $product_ids[] = $row['ProductID'];
}

$result = mysqli_query($con, "SELECT * FROM Products WHERE Quantity_On_Hand < Min_Stock");

while($row = mysqli_fetch_array($result))
{

    if (in_array($row['ProductID'], $product_ids))
    {

    }

}

edited Mar 30, 2013 at 23:47

answered Mar 30, 2013 at 22:44

Michael

12.9k6 gold badges57 silver badges69 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

James Chen Over a year ago

Hi thanks for helping, but is seems does not work. The table does not display anything inside...

Michael Over a year ago

You are putting this outside (before) the while($row = mysqli_fetch_array($result)), right?

James Chen Over a year ago

no, do you mean put these two loops before the while($row = mysqli_fetch_array($result)), and then I am going to have two while loops, and the second one is inside the if loop? Sorry I am not quite fimilar with PHP...

James Chen Over a year ago

oh thanks so much, it solved. Actually the answer shouldn't use "$row" twice, and I correct it. Anyway, THanks for helping :-)

Michael Berkowski · Accepted Answer · 2013-03-30 23:58:16Z

3

Your code is failing because $filter is a MySQLi result resource, not an array. Really, this is better accomplished with a simple inner join between the two tables. If a ProductID does not exist in Orders, the INNER JOIN will exclude it from the result set in the first place.

$sql = "
SELECT Products.* 
FROM
  Products
  INNER JOIN Orders ON Products.ProductID = Orders.ProductID
WHERE Quantity_on_Hand < Min_stock";

$result = mysqli_query($con, $sql);
if ($result) {
  $results = array();
  while ($row = mysqli_fetch_array($result)) {
    $results[] = $row;
  }
}
// Now $results is a 2D array of all your Products

If instead you want to retrieve all the Products, and simply have an indication of whether it has an active order, use a LEFT JOIN and test if Orders.ProductID is null in the SELECT list:

$sql = "
SELECT
  Products.* ,
  /* No orders will print 'no-orders' in a pseudo column called has_orders */
  CASE WHEN Orders.ProductID IS NULL THEN 'no-orders' ELSE 'has-orders' AS has_orders
FROM
  Products
  LEFT JOIN Orders ON Products.ProductID = Orders.ProductID
WHERE Quantity_on_Hand < Min_stock";

$result = mysqli_query($con, $sql);
if ($result) {
  $results = array();
  while ($row = mysqli_fetch_array($result)) {
    $results[] = $row;
  }
}
// Now $results is a 2D array of all your Products
// And the column $row['has_orders'] will tell you if it has any...

In this case, you may test in a loop over your rowset whether it has orders:

foreach ($results as $r) {
  if ($r['has_orders'] == 'has-orders') {
    // this has orders
  }
  else {
    // it doesn't...
  }
}

edited Mar 30, 2013 at 23:58

answered Mar 30, 2013 at 22:44

Michael Berkowski

271k47 gold badges450 silver badges395 bronze badges

2 Comments

James Chen Over a year ago

Hi thanks for helping, but I am have some questions, bc I will still have some actions in "else statement", if i put "while" statement inside the "if" statement, how could I process to "else" statement which required to process the result that does not in the array? Thanks :-)

Michael Berkowski Over a year ago

@JamesChen The rows have all been stored into the array $results, so you can check while looping over it. This isn't much different than your original attempt. Added above.

Collectives™ on Stack Overflow

some problems with in_array function

2 Answers 2

4 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

4 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related