Unable to echo the real value of a variable - php

Question

Iam given the following variable: %4$s that outputs the text "a test" in the code.
I am trying to echo strlen('%4$s'); but it always returns 4 , while adding it as echo strlen("%4$s"); is not returning again the real value (2), which in my opinion means that for some reason, is uses the number of the variable.
My primary scope is to check if %4$s contains two or more words or to count the characters of a returned string.
At this time, %4$s , returns 'a test' in my HTML, so I m expecting an echo of %4$s to return 'a test' in PHP and a strlen of %4$s to return number 6

Answer 1

It looks like you are using a string format suitable for something like printf() or sprintf() and you are wanting to know the length of the 4th entered value.
Sample code:
$format = '%4$s';
$val1 = 'one';
$val2 = 'two';
$val3 = 'three';
$val4 = 'a test';
echo sprintf($format,$val1,$val2,$val3,$val4);
Which will display:
a test
And you want to know the length of that 4th value. Instead of strlen('%4$s'), you should be doing strlen($val4), for example:
echo strlen($val4);
Which will show:
6
A full example would be:
$format = '%5$d is the strlen of "%4$s"';
$val1 = 'one';
$val2 = 'two';
$val3 = 'three';
$val4 = 'a test';
echo sprintf($format,$val1,$val2,$val3,$val4,strlen($val4));
Which will display:
6 is the strlen of "a test"
Edit: It is still not very clear what you are doing even after looking at the pastebin link you posted. That said, the following is a guess that works. It uses the vsprintf() method:
$format = '%6$s is the strlen of "%4$s"';
$retArr[0] = array('post_id' => 'one',
'icon' => 'two',
'title' => 'three',
'permalink' => 'a test',
'image' => '/path/to/img.png');
$retArr[0]['len'] = strlen($retArr[0]['permalink']);
echo vsprintf($format,$retArr[0]);
And still outputs:
6 is the strlen of "a test"

Answer 2

Referring : http://php.net/manual/en/language.types.string.php
Note: Unlike the double-quoted and heredoc syntaxes, variables and escape sequences for special characters will not be expanded when they occur in single quoted strings.
Double quoted ¶
If the string is enclosed in double-quotes ("), PHP will interpret the following escape sequences for special characters:
$
is an escape character in php and if you don't escape it correctly using "\" like "\$s" "s" will be treated as a variable "$s".
Hence you are getting count as 2

Answer 3

strlen returns the amount of letters in a string so that's why you are getting the number 4 instead of your string.
try : '%4'.$s

Answer 4

Here's an answer that will count string length (by char) and do a word count.
Note: This isn't a copy-and-paste solution, so you will need to change your code to fit.
You can get the word count easily by doing the generic rule that words are separated by spaces. Here's where a PHP function called explode() comes in. We can use that combined with count() to get the word count:
$foo = 'a test';
$wordCount = count(explode(' ', $foo));
$strLength = strlen($foo);
echo $wordCount; # will output 2
echo $strLength; # will output 6
I've seen your pastebin for your array data, you can just loop through the array to get the individual values or use $array[$key] to specify a certain value to target.