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

Question

This line is giving error: "Too few arguments". What is the problem?

$InsertQuery = sprintf("INSERT INTO listing (ldate, places, company, designation, projectdetails, desiredcandidate, hrname, hrcontact, email) VALUES (DATE_FORMAT(%s,'%Y %m %d),%s,%s,%s,%s,%s,%s,%s,%s)", $ldate,$places,$company,$designation, htmlentities($projectdetails), htmlentities($desiredcandidate),$hrname,$hrcontact,$email);

Andrew Hare · Accepted Answer · 2009-10-25 06:19:30Z

6

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.

answered Oct 25, 2009 at 6:19

Andrew Hare

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6 Comments

RKh Over a year ago

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?

Andrew Hare Over a year ago

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

Andrew Hare Over a year ago

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

RKh Over a year ago

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?

Matthew Scharley Over a year ago

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.

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DigitalRoss · Accepted Answer · 2009-10-25 06:28:20Z

5

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

answered Oct 25, 2009 at 6:28

DigitalRoss

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