PHP sprintf escaping %
PhpEscapingPrintfPhp Problem Overview
I want the following output:-
> About to deduct 50% of € 27.59 from your Top-Up account.
when I do something like this:-
$variablesArray[0] = '€';
$variablesArray[1] = 27.59;
$stringWithVariables = 'About to deduct 50% of %s %s from your Top-Up account.';
echo vsprintf($stringWithVariables, $variablesArray);
But it gives me this error vsprintf() [function.vsprintf]: Too few arguments in ... because it considers the % in 50% also for replacement. How do I escape it?
Php Solutions
Solution 1 - Php
Escape it with another %:
$stringWithVariables = 'About to deduct 50%% of %s %s from your Top-Up account.';
Solution 2 - Php
It is very easy.
Put another % in front of the original % to escape it.
For example,
$num=23;
printf("%%d of 23 = %d",$num);
Output:
%d of 23 = 23
Solution 3 - Php
For add % in your language string, you just need to add double percent %% instead of one
Solution 4 - Php
This works for me:
sprintf(
'%s (Cash Discount: %%%s, Deferred Discount: %%%s)',
$segment->name,
$segment->discount_cash,
$segment->discount_deferred,
)
// Gold (Cash Discount: %25, Deferred Discount: %20)
Solution 5 - Php
What about this:
$variablesArray[0] = '%';
$variablesArray[1] = '€';
$variablesArray[2] = 27.59;
$stringWithVariables = 'About to deduct 50%s of %s %s from your Top-Up account.';
echo vsprintf($stringWithVariables, $variablesArray);
Just add your percent sign in your variables array