PHP Mysql Correctly Iterate Over Tables With Alias Names - php

Question

I have the following Mysql query:
$sql = "select r.brand AS brand, r.name AS name, r.cost AS cost, e.price AS price, d.shipping as shipping
FROM tab.rawproduct r
INNER JOIN price e ON r.housecode=e.housecode
INNER JOIN product d ON e.productid=d.productid
WHERE r.housecode='$housecode'";
The houscode is assigned to a variable which is then passed to the sql statement:
<label>Housecode:</label><input class="boxes" type="text" name="housecode" value="<?php echo $housecode; ?>"><br />
$housecode = $_POST['housecode'];
Housecode is submitted through a form with action set to $_SERVER[PHP_SELF]
I am trying to iterate over the results with PHP with the following:
$result = $con->query($sql);
if ($result->num_rows >0) {
while($row = mysqli_fetch_array($result)) {
$brand = $row['brand'];
$housecode = $row['housecode'];
$name = $row['name'];
$cost = $row['cost'];
$salesprice = $row['price'];
$shipraw = $row['shipping'];
}
} else {
echo "0 Results";
}
$con->close();
Nothing is getting returned when a user submits a housecode and the "0 Results" is echoed.
I have looked into this problem and read Lucas Knuth's post:
If two or more columns of the result have the same field names, the
last column will take precedence. To access the other column(s) of the
same name, you must use the numeric index of the column or make an
alias for the column. For aliased columns, you cannot access the
contents with the original column name. So, you can either use an AS
in your SQL-query to set other names for the doubled rows or use the
numbered indexes to access them.
So I have used the AS keyword in the above query but I still don't get any results. I have also tried changing to mysqli_fetch_row($result) and tried to assign the $row[0], 1 ... etc instead. Again no luck.
When I run Apache error_log I get the following:
Trying to get property of non-object on line 34. On line 34 and 35 I have:
if ($result->num_rows >0) {
while($row = mysqli_fetch_array($result)) {
Any help would be much appreciated.
Cheers

Answer 1

I have figured out the answer for myself.
The problem was that I was selecting just one database, ie tab.rawproduct BUT tables price e and procduct d are from a totally different database.
So the sql query should have been this:
$sql = "SELECT r.brand, r.name, r.cost, e.price, d.shipping FROM
tab.rawproduct r
INNER JOIN t1.price e on r.housecode=e.housecode
INNER JOIN t1.product d on e.productid=d.productid
WHERE r.housecode = '$housecode' ";
If found the error by checking the mysqli->errno:
if(!$result = $mysqli->query($sql)) {
echo "Error: Our query failed to execute and here is why: \n";
echo "Query: " . $sql . "\n";
echo "Errno: " . $mysqli->errno . "\n";
echo "Error: " . $mysqli->error . "\n";
exit;
}
Hopefully this may help someone else facing the same problem.
Thanks for all your comments and help.

Answer 2

I reproduced your example and I found a bug in your select query: you're missing the r.housecode.
$sql = "select r.brand AS brand, r.name AS name, r.cost AS cost, e.price AS price, d.shipping as shipping, r.housecode AS housecode
FROM tab.rawproduct r
INNER JOIN price e ON r.housecode=e.housecode
INNER JOIN product d ON e.productid=d.productid
WHERE r.housecode='$housecode'";
Check if the cross references match in your tables, because in my examples the rows are correctly returned
table price:
housecode productid price
HT0008 4 3400
HT0008 5 5400
table product:
shipping productid
64 4
78 5
table rawproduct:
brand name cost housecode
Cani bellaaa 63824 HT0008