MySQL PHP printf into $variable adding additional value

Question

I'm experiencing something strange, and I can't figure out why it's happening. I'm running a query to pull data from a column in a mysql table, and when I do a straight printf(), the data comes out as expected. However, when I do $variable = printf(), I'm getting an additional value in the string.

The following prints something like: 22611,21435,23327,22876,22986,23692,21581,21832,22337,24313,22174,24368,

$query  = "SELECT column FROM table WHERE year in (2012)";
if ($result = mysqli_query($connect, $query)) {
    while ($row = mysqli_fetch_assoc($result)) {
        printf ("%s,", $row["column"]);
    }
}

But if I try to put the result into a variable like so:

$data = printf ("%s,", $row["column"]);

I get an output of 22611,21435,23327,22876,22986,23692,21581,21832,22337,24313,22174,24368,6

Why is it adding this extra value? Am I adding the result to the variable incorrectly? FYI, this is just a snippet from the code, I have error handling in place.

You should have a look at the printf() manual. By the way, why are you using "in" in your request ? — mimipc
– mimipc, Commented Feb 6, 2013 at 18:30
Because the year is actually a variable, and sometimes its value is "2008,2009,2010" or something to that effect. — Jesse Schoff
– Jesse Schoff, Commented Feb 6, 2013 at 18:37

David Harris · Accepted Answer · 2013-02-06 18:26:34Z

4

Use sprintf() instead (it works exactly like printf()). This 'silences' it and gives a return value.

answered Feb 6, 2013 at 18:26

David Harris

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2 Comments

Jesse Schoff Over a year ago

Hmm, I tried that but when I do sprintf(), I'm only getting one value returned. In this instance, it just returns 24368,

David Harris Over a year ago

It's not sprintf(), rather, you're not assigning your first sprintf() to a variable and appending the rest to it.

Tim Fountain · Accepted Answer · 2013-02-06 18:28:50Z

2

printf() outputs the data and returns its length. So that's where the extra value is coming from. You want sprintf() which just returns the value.

answered Feb 6, 2013 at 18:28

Tim Fountain

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Comments

Orbling · Accepted Answer · 2013-02-06 18:43:46Z

2

Just out of curiosity, why are you using sprintf() at all?

You can assign values to variables directly or with string concatenation.

$data = $row["column"];
$data = $row["column"] . ',';

3 Comments

David Harris Over a year ago

Totally agree with this OP, sprint() is just using more (albeit little) resources.

Jesse Schoff Over a year ago

This is much better, thanks. I am using this solution. Accepting the first response, though, as I think it answers the question more directly.

Jesse Schoff Over a year ago

I'm having the same issue mentioned here. It's adding the array twice even though I'm using mysql_fetch_assoc(). Any idea why? That thread that I linked to suggests that mysql_fetch_assoc shouldn't do that.

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