Accessing records with same name but different ID - php

Question

I am handling a database table, where I have a quantity field in it. The quantity is manipulated through a form where it is reduced. The record gets inserted in the table with the new value as a new record with new quantity(reduced) and through this table it is being fetched in the next form. Now the issue is that whenever I fetch the quantity the older record(record with original quantity) is fetched where as the record with new quantity should be fetched.
I tried updating the same record with new value but it is not updating.
Here is the code for the same :
$haha="SELECT quantity FROM ready_for_delivery WHERE joborderid='".$data['joborderno']."'";
$haharesult = mysqli_query($link,$haha);
if(mysqli_num_rows($haharesult)>0)
{
$sql1="UPDATE ready_for_delivery SET quantity='".$_POST['rp_qty']."' WHERE joborderid='".$_data['joborderno']."'";
$sq1result=mysqli_query($link,$sql1);
//echo "I am here";
}
else
{
$quantity="INSERT INTO `ready_for_delivery` (`joborderid`,`joborderdetailsid`,`datetime`, `quantity`) VALUES ('".$data['joborderno']."',' ', now(), '".$_POST['rp_qty']."');";
$res1 = mysqli_query($link,$quantity);
echo "done";
}
I am checking if there is any data in the table with specified job order no, if yes, update it else insert as a new record. Every time it inserts as new record. Please guide me. Thanks.

Answer 1

Remove ; from your first line, hope this will work .
$haha="SELECT quantity FROM ready_for_delivery WHERE joborderid='".$data['joborderno']."'"; //remove ;
//echo $haha;
$haharesult = mysqli_query($link,$haha);
if(mysqli_num_rows($haharesult>0))
{
$sql1="UPDATE ready_for_delivery SET quantity='".$_POST['rp_qty']."' WHERE joborderid='".$_data['joborderno']."'";
$sq1result=mysqli_query($link,$sql1);
//echo "I am here";
}
else
{
$quantity="INSERT INTO `ready_for_delivery` (`joborderid`,`joborderdetailsid`,`datetime`, `quantity`) VALUES ('".$data['joborderno']."',' ', now(), '".$_POST['rp_qty']."');";
$res1 = mysqli_query($link,$quantity);
echo "done";
}

Answer 2

In the mysqli_query you have $sql a query variable while the actual query is in the $haha var.
Also, did you check if variable $data['joborderno'] is not empty?

Answer 3

The error lies in this line of code:
if(mysqli_num_rows($haharesult>0))
the closing brackets is in the wrong place. Correct way:
if(mysqli_num_rows($haharesult)>0)