Using a PHP Variable in a MySQL Query - php

Question

So I have the following MySQL query that I want to execute through a PHP script:
for($y=0;$y<$usercount;$y++){
$findweight = "replace into weight (user_id, column1)
select user_id,
0.4*column1_a+0.6*column1_bon as calc1
from history_summary where user_id=$y";
if (mysqli_query($link, $findweight)) {
echo "success <br>";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($link);
}
}
But somehow everytime I execute the script it fails to work. Everytime the script is executed, the text "success" is outputted, which means that the MySQL query should have been successfully executed, but when I check my database, nothing seems to have changed. Yet when I run the query directly such as the following:
replace into weight (user_id, column1)
select user_id,
0.4*column1_a+0.6*column1_bon as calc1
from history_summary where user_id=1
It works just fine. I'm assuming the issue is that the variable $y failed to be passed on properly to the query, but I'm not quite sure. I've tried the other solutions posted on this site but none of them seem to work. If anyone can help out that'd be great!

Answer 1

The DB part of me is crying because you should really be submitting one query instead of multiple queries. Specifically, if $y is a safe variable, you should have something like this:
$sql = "replace into weight (user_id, column1)
select user_id,
0.4*column1_a+0.6*column1_bon as calc1
from history_summary where user_id < $y";
There are other questions I have, such as if you really want REPLACE or you want UPDATE but that is a problem for another day.
BUT, your question was more about debugging, and it seems that the command(s) are executing successfully. Try running your query on many different values of $y. Perhaps you have encountered an edge case

Answer 2

The question is about debugging, so I'll respond to that, but I double-down on #mankowitz points below.
Debugging
Do you have access to the DB server? Can you watch the query log on the server to confirm that it received what you think the code is sending?
Try the old standby of println style debugging and echo out the SQL string before sending. The <pre> and --- help to make sure you are able to see any extraneous characters you might have in your SQL string:
...
echo "About to send SQL: <pre>--- $findweight ---</pre>\n"
if (mysqli_query($link, $findweight)) {
...
Is this code wrapped inside of a SQL transaction that is not (auto)committed when the script ends?
Do other, perhaps simpler queries work? Have you tried the simple SELECT 1 and echoed out the result?
More arcane and improbable given the context of the question: is fsync disabled on the server? If so, then each mysql_query does not guarantee the DB has actually written the contents to disk yet, and you may eventually see the rows get updated.
What does mysqli_query return? The code currently ignores the result other than checking if it's falsey. Store the result and introspect it.
An edge case of mysqli_query. Ten plus years ago, I encountered cases where mysql_query (note the lack of i) would not return FALSE despite a failing query. It was an oddball case of the connected user not having access to manipulate or view the rows in question, highlighting a bug in mysql_query as well as my connection string.
Database interaction
Be smart about DB queries, minimize the communication, and use bulk operations wherever possible. Your code, as structured could be rewritten in a couple of ways, both of which require only a single call to the database:
The WHERE-IN construct allows you to be very precise about which rows you address:
$ids = [];
for ( $y = 0; $y < $usercount; ++$y )
$ids[] = $y;
$ids = implode(',', $ids);
$findweight = "REPLACE INTO weight (user_id, column1)
SELECT user_id, 0.4*column1_a + 0.6*column1_bon
FROM history_summary WHERE user_id IN ($ids)";
Or, since the IDs the code addresses are contained within a contiguous range, do as #mankowitz suggested, removing the need for a for-loop entirely:
$findweight = "REPLACE INTO weight (user_id, column1)
SELECT user_id, 0.4*column1_a + 0.6*column1_bon
FROM history_summary WHERE user_id BETWEEN 0 AND $usercount";
# alternate form
$findweight = "REPLACE INTO weight (user_id, column1)
SELECT user_id, 0.4*column1_a + 0.6*column1_bon
FROM history_summary WHERE 0 <= user_id AND user_id <= $usercount";
If you absolutely must use individual per-row statements, consider a prepared statement so the DB need only parse and plan the statement once:
$findweight = 'REPLACE INTO weight (user_id, column1)
SELECT user_id, 0.4*column1_a + 0.6*column1_bon
FROM history_summary WHERE user_id = ?';
if ($psth = mysqli_prepare($link, $findweight)) {
$y = 0;
mysqli_stmt_bind_param($psth, 'i', $y);
while ( ++$y < $usercount ) {
if ( mysqli_stmt_execute( $psth ) ) {
echo "Successful REPLACE operation for user_id = $y<br>\n";
}
else {
echo "Error in REPLACE operation for user_id = $y<br>\n";
break;
}
}
}