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[PHP]: Error -> Too few arguments in sprintf();

Answer 1

+3 A:

Well, your string specifies 12 placeholders and you only provide 9 values. The sprintf function requires that you pass as many values as the number of placeholders you specify in the format string. I actually think the error message is strikingly clear and is about as good an error message as you will ever see.

Andrew Hare 2009-10-25 06:19:30

Exactly the error is too clear. I am passing 9 arguments and their are 9 %s. Can you please clear where are 12 place holders?

RPK 2009-10-25 06:22:54

You are forgetting about the `%Y %m %d` - those are placeholders as well.

Andrew Hare 2009-10-25 06:23:39

Well, at least `%d` is - I am not sure whether or not PHP considers `%Y` and `%m` valid type specifiers.

Andrew Hare 2009-10-25 06:25:11

But those are part of MySQL Data_Format function. I forgot to end with '. That makes it '%Y %m %d', but that is not solving my problem.Is it that sprintf considers everything that starts with % as a placeholder?

RPK 2009-10-25 06:25:27

Not everything - only strings that are valid type specifiers. I believe that you should be okay with `%Y` and `%m` since they don't appear to be type specifiers (see http://php.net/manual/en/function.sprintf.php) but `%d` _is_ a type specifier so you will need to double up the percent symbol (like this: `%%d`) if you want `sprintf` to output a literal `%d`.

Andrew Hare 2009-10-25 06:28:31

Escaping any `%` you want to output literally is a good idea, whether what follows is a legal type specifier or not. It may become one in the future.

Matthew Scharley 2009-10-25 06:57:48

@Matthew - Good point!

Andrew Hare 2009-10-25 12:26:52

Answer 2

+2 A:

The arguments are the expressions, and they should match in number the % format specifiers. If you really just need a % char, use %%.

DigitalRoss 2009-10-25 06:28:20

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[PHP]: Error -> Too few arguments in sprintf();

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