CODEWITHSUNDEEP
phpmysqlfunction
Joe Cool
Joe Cool

Reputation: 62

passing return value from a php function to a mysql_query

I am having a problem with passing a $query variable to a mysql_query command. So if I do this I get an update in the database (which I wanted):

$query = "UPDATE master SET finished_size = '$finished_size' WHERE id = $id";
mysql_query ($query, $db);

However, I need to use a function to do this. I'm trying to use this code, but I am no longer able to get the update to the database that I wanted:

if ( !function_exists("testQuery") ) {
    function testQuery($id) {
        return 'UPDATE master SET finished_size = "$finished_size" WHERE id = $id';
    }
}

$query = testQuery($id);

mysql_query ($query, $db);

I have tried many ways of doing this, but the $query variable which contains the string to pass to the mysql_query function doesn't seem to be recognized. But I don't know how else to do this the proper way.

Also, I realize mysql_query is an old function, but this is the code that I have to use because I'm working on very old software.

Upvotes: 1

Views: 186

Answers (3)

Joe Cool
Joe Cool

Reputation: 62

It turned out that the problem was that I was using $_POST variables, that had already been extracted, so that $finished_size was no longer treated as a global variable (like $_POST['finished_size'] would be). And so the extracted variable did not have scope access within the function.

So I just re-extracted $_POST inside the function. I had many $_POST variables coming in from a form, and so extracting them inside the function seems to be a fairly convenient way to pass them back to $query.

The answers that were given prior to this were helpful for me to realize that it was a scope issue.

Upvotes: 0

Martin Heralecký
Martin Heralecký

Reputation: 5779

If you're putting a variable to a string without connecting more strings, you have to use quotes, not apostrophes. So:

return "UPDATE master SET finished_size = '$finished_size' WHERE id = $id";

Upvotes: 1

Barmar
Barmar

Reputation: 780798

Add another parameter to the function.

Also, variables are only expanded inside double quotes, not single quotes. See What is the difference between single-quoted and double-quoted strings in PHP?

if ( !function_exists("testQuery") ) {
    function testQuery($id, $finished_size) {
        return "UPDATE master SET finished_size = '$finished_size' WHERE id = $id";
    }
}

$query = testQuery($id, $finished_size);

mysql_query ($query, $db);

Upvotes: 2

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