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Java Java Basics Getting Started with Java Strings, Variables, and Formatting

Posted by Ryan Galle'
Ryan Galle'
4,784 Points

what does: java.util.MissingFormatArgumentException: Format specifier '%s' () mean?

It's saying it, I have no clue what it means and I've watched that part of the videos twice...

Name.java 
// I have setup a java.io.Console object for you named console
String firstName = "Ryan";
console.printf("%s can code in Java!");

3 Answers

Ryan Galle'
Ryan Galle'
4,784 Points

Wut

Daniel Hartin
Daniel Hartin
18,106 Points
Daniel Hartin
Daniel Hartin
18,106 Points

Hey Ryan

Sorry but what don't you understand? simply putting 'wut' is neither proper English or helpful to anyone trying to expand on something which you don't understand.

The string formatters %s for a string or %d for an integer etc are intended to be used as a placeholder inside the string for you to slot in some other information, but how does the program know what to slot it? This is why you pass in as many arguments after the initial string into the method as you have placholders, that way the program can slot in these values like

String myName = "Daniel";
int myAge = 30;

console.printf("my name is %s and I am %d years old",myName,myAge);

Hopefully this expands a little further and you can make sense of what was in the video.

Daniel

Daniel Hartin
Daniel Hartin
18,106 Points
Daniel Hartin
Daniel Hartin
18,106 Points

Hi Ryan

When you declare a string formatter i.e %s you must have an equal number of arguments passed into the method like the below

// I have setup a java.io.Console object for you named console
String firstName = "Ryan";
console.printf("%s can code in Java!",firstName);

This replaces the %s with the first argument passed in which in this case is the firstName variable.

Hope this helps Daniel

Ryan Galle'
Ryan Galle'
4,784 Points
Ryan Galle'
Ryan Galle'
4,784 Points

I no speaka da chinese