Why isn’t printf() in PHP working when I use double quotation marks?

4 comments

1. Because when you are using double quotes the $s is treated as a variable

   As in:

   $x = "World";
   echo "Hello $x"; // Will print: "Hello World
   

   where as when using:

   $x = "World";
   echo 'Hello $x'; // Will just print "Hello $x"
   

   For a more detailed explanation you can check the manual:

   Strings in General

   Single quoted vs Double quoted

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2. It is very important to realize php is treating single quoted and double quoted strings differently.

   You can read more in official php docs, but let me give you a highlight:

   $t = 'bla';
   echo '$t';
   

   will output $t, where

   $t = 'bla';
   echo "$t";
   

   will output bla

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3. That’s because you have the ‘$s’ bit in your string. When using double quotation marks PHP interprets it as a variable and tries to parse it. You probably used double quotations without the $ in it earlier.

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4. As the other answers say, it treats $s as a variable, you could always escape the $

   printf("by %1$s on %2$s", 'string1', 'string2');
   

   I would however use single quotation marks, as php doesn’t need to parse the string and is therefore faster.

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