My insert statement (php to mysql) fails to use my variables - php

Question

It must be the simplest error, but I dont see nor find it.
I fill a variable $aa_minerid with value 7.
I use this variable in a insert.
The insert always inserts a 0 (zero) in the database never a 7
The field i put it in is a smallint(6)
I tried
VALUES ('$aa_productid')
VALUES ($aa_productid)
VALUES ("$aa_productid")
VALUES ('{$aa_productid}')
VALUES ("{$aa_productid}")
and all with use of ` aswell
into script placed hereafter.
If I put there : VALUES ( 7 )
It does work perfect.
So what do I do wrong in this script?
BTW the echo at the end DOES show the right value of the variable $aa_productid
<?php
/* This php script should transfer data from the aa to the sql database */
// Info coming from aa
$aa_productid = 7 ;
include ("dogs.inc");
$cxn=mysqli_connect($host,$user,$passwd,$dbname);
$query = 'SELECT * FROM `Price` WHERE '
. ' `Time_Stamp`=(select max(`Time_Stamp`) from `Price` where `Product_ID` = \'1\')';
$result=mysqli_query($cxn,$query) or
die("Couldn't execute select query");
$row = mysqli_fetch_row($result);
$aa_price=$row[3] ;
$aa_value = $aa_price * $aa_amount;
// Info ready to go to database
$sqlinsert = 'INSERT INTO Mining (Product_ID)'
. ' VALUES ( $aa_productid )' ;
echo $aa_productid;

Answer 1

Single quotes don't do variable expansion in PHP. But I would recommend you use prepared statements, such as:
$stmt = $cxn->prepare('INSERT INTO Mining (Product_ID) VALUES ( ? )');
$stmt->bind_param('i', $aa_productid);
$stmt->execute();
See the documentation at prepare and bind_param.
This will protect you from SQL injection.

Answer 2

Try
'.$aa_productid.'
or
".$aa_productid."
Depending on the type of apostrophe used to beging the string, use the same one.
Also, if You are using ", then You should be able to Just do
$insert="INSERT INTO $tablename;";

Answer 3

It's been a while since I have done any PHP but..
I think you need to have smartquotes turned on
Try this instead:
$sqlinsert = 'INSERT INTO Mining (Product_ID)'
. ' VALUES ('. $aa_productid .' )' ;
concatenate the variable into the query.

Answer 4

When you are using variables within quotes, you must use the double-quote if you want PHP to parse variables within it. So, this would work:
$sqlinsert = 'INSERT INTO Mining (Product_ID) VALUES ('.$aa_productid.')';
Or this would:
$sqlinsert = "INSERT INTO Mining (Product_ID) VALUES ($aa_productid)";

Answer 5

Try:
$query = "SELECT * FROM Price WHERE Time_Stamp=(select max(Time_Stamp) from Price where Product_ID = "1")";
$sqlinsert = "INSERT INTO Mining (Product_ID) VALUES ( '$aa_productid' )" ;
Also, its always a good idea to escape the strings before entering them in the db.

Answer 6

Try this syntax instead:
$sqlinsert = "INSERT INTO Mining (Product_ID) VALUES ("' . $aa_productid . '")";
no need to concatenate the two parts of the insert. Also double quoting the variable seems to avoid problems.