PHP sprintf escaping %

Question

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?

@Col. Shrapnel My question is about vsprintf not printf, I am using this for the first time and could not assume the similarity between the two. However, searching escape or escaping in both php.net/printf and php.net/vsprintf both does not show the answer immediately. When I search for %% it shows the answer in php.net/printf but I didn't know about %%!!! Did you search for the answer there before downvoting? — Sandeepan Nath
– Sandeepan Nath, Commented Sep 8, 2010 at 10:40
@sandeepan: vsprintf belongs in the same family of functions as printf. The correct documentation to find the format, though, is php.net/sprintf. Both pages even point to it: "See sprintf() for a description of format." Didn't you at least click it? — BoltClock
– BoltClock, Commented Sep 8, 2010 at 10:43
@Col. Shrapnel ok fine let's take php.net/sprintf, where is the answer? It is halfway down the page With printf() and sprintf() functions, escape character is not backslash '\' but rather '%'. What is there to downvote here? It was just not that obvious to me as it was to you. If you find a duplicate question you can better write the link. But I am sure many will find this question helpful. But you won't accept that and you will still say something, I know. — Sandeepan Nath
– Sandeepan Nath, Commented Sep 8, 2010 at 10:49
oh I thought the second comment was by Col. Shrapnel , sorry — Sandeepan Nath
– Sandeepan Nath, Commented Sep 8, 2010 at 10:51
SO should have a flag for RTFM responses. It's almost like people troll just so they can tell people to read the docs. He needed help and asked a question and then someone answered helpfully and got points for it. The world went on and the internet was used to someone's benefit. Meanwhile I'm getting heated over a two year old argument. — rob5408
– rob5408, Commented Jul 2, 2013 at 3:22

BoltClock · Accepted Answer · 2010-09-08 10:26:43Z

441

Escape it with another %:

$stringWithVariables = 'About to deduct 50%% of %s %s from your Top-Up account.';

answered Sep 8, 2010 at 10:26

BoltClock

728k165 gold badges1.4k silver badges1.4k bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Jan Hettich Over a year ago

sprintf("SELECT * FROM ... WHERE name LIKE '%%%s%%%s%%'", $fname, $lname); -- Ugly but it works!

Lukas Liesis Over a year ago

you can add that part with another "%s": sprintf('from %s you can get %s', 'something', '50%')

budji Over a year ago

How do you escape it if the string is dynamic? Let's say, sprintf('This is %s.', the_title())

Ivanka Todorova Over a year ago

@madastro str_replace('%', '%%', the_title())

Collectives™ on Stack Overflow

PHP sprintf escaping %

1 Answer 1

4 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

4 Comments

Linked

Related