troubleshooting Question

% character in sprintf

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tr5 asked on
PHP
2 Comments1 Solution602 ViewsLast Modified:
Hi
This is crazy, but I can't get a % character to appear in a sprintf() statement.
What is wrong with this:
$str = sprintf("Percentage: %0.2f%", 53.0);
echo "str = $str<br>";
prints only "str = "
The % after the 0.2f causes the problem. I tried escaping it but it doesn't work. In c you can use a % in a printf. What's the deal here?

Thanks
tr5
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