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PHP - Manual: printf

2024-12-22

printf

(PHP 4, PHP 5, PHP 7, PHP 8)

printf输出格式化字符串

说明

printf(string $format, mixed $args = ?, mixed $... = ?): int

依据 format 格式参数产生输出。

参数

format

format 描述信息,请参见 sprintf()

args

...

返回值

返回输出字符串的长度。

参见

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User Contributed Notes 16 notes

up
13
dhosek at excite dot com
22 years ago
Be careful:
printf ("(9.95 * 100) = %d \n", (9.95 * 100));

'994'

First %d converts a float to an int by truncation.

Second floats are notorious for tiny little rounding errors.
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4
php at mole dot gnubb dot net
17 years ago
[Editor's Note: Or just use vprintf...]

If you want to do something like <?php printf('There is a difference between %s and %s', array('good', 'evil')); ?> (this doesn't work)  instead of <?php printf('There is a difference between %s and %s', 'good', 'evil'); ?> you can use this function:

<?php
function printf_array($format, $arr)
{
    return
call_user_func_array('printf', array_merge((array)$format, $arr));
}
?>

Use it the following way:
<?php
$goodevil
= array('good', 'evil');
printf_array('There is a difference between %s and %s', $goodevil);
?>
and it will print:
There is a difference between good and evil
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-3
maybird99 at yahoo dot com
20 years ago
instead of writing a function to round off a float (let's call it 'x') accurately, it's much easier to add a small number to x and then truncate it...
For example: if you want to round off to the nearest integer, just add 0.5 to x and then truncate it. if x=12.6, then it would calculate 13.1, and truncate it to 13. If x=14.4, it would calculate 14.9 and truncate it to 14.
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-6
deekayen at hotmail dot com
20 years ago
You can use this function to format the decimal places in a number:

$num = 2.12;
printf("%.1f",$num);

prints:

2.1

see also: number_format()
up
-15
spiffytech at gmail dot com
10 years ago
Be careful when relying on typecasting with printf(). For example,

    printf("%d", "17,999")

returns "17".
up
-11
shepard at ameth dot org
20 years ago
Be sure that the output channel is available to write on before executing printf()!  Some functions in classes available from various sources (in my case, DB_Sql::query() in PHPLIB) assume that printing will work, even after the default output stream has been closed. 

For me the issue was most notable in PHP4 session management when I was creating my own sess_write() handler.  Since I was unable to find any function that checks for the output stream that printf() uses, I just had to drop the crazy use of the printf(). (does some function already exist to check for the presence of an output stream?)
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-14
eugenew at starhub dot net dot sg
19 years ago
If anyone is looking for writing a quine using printf(),
this is my example:

<?php $f='<?php $f=%c%s%c; printf($f,39,$f,39); ?>'; printf($f,39,$f,39); ?>

This also helps those who are new to printf() see one way of using the 'mixed args' part, rather than just a single argument as in most examples I've seen.
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-12
RS
5 years ago
If your missing features such as "-"*100 to print a single character multiple times you can use the slightly longer and less readable PHP equivalent printf("%'-100s",""); and sprint("%'-100s","").
up
-16
dalu at uni SPAMHAM dot de
19 years ago
copypasted from msdn

A format specification, which consists of optional and required fields, has the following form:

%[flags] [width] [.precision] [{h | l | I64 | L}]type

Each field of the format specification is a single character or a number signifying a particular format option. The simplest format specification contains only the percent sign and a type character (for example, %s). If a percent sign is followed by a character that has no meaning as a format field, the character is copied to stdout. For example, to print a percent-sign character, use %%.

The optional fields, which appear before the type character, control other aspects of the formatting, as follows:

type
Required character that determines whether the associated argument is interpreted as a character, a string, or a number (see the printf Type Field Characters table.
flags
Optional character or characters that control justification of output and printing of signs, blanks, decimal points, and octal and hexadecimal prefixes (see the Flag Characters table). More than one flag can appear in a format specification.
width
Optional number that specifies the minimum number of characters output (see printf Width Specification).
precision
Optional number that specifies the maximum number of characters printed for all or part of the output field, or the minimum number of digits printed for integer values (see the How Precision Values Affect Type table).
h | l | I64 | L
Optional prefixes to type-that specify the size of argument (see the Size Prefixes for printf and wprintf Format-Type Specifiers table).
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-17
ezislis at mail dot ru
20 years ago
be careful with integers, they cant hold large values.

printf("%d",10023123553.45634663);
will print out: 1433188961
and
printf("%.0f",10023123553.45634663);
will print out: 10023123553
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-8
steve at myschoolsystems dot com
11 months ago
To format a dollar value as in $123.00 that may otherwise look like $123  use this

print ('$');                    // the dollar sign in front of our answer
printf ('%.2f',$price);
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-20
codeslinger at compsalot dot com
17 years ago
Several people have commented about problems with aligning numbers.  I just wanted to add a bit of clarification.

According to the spec all True Type Fonts (especially porpotional spaced fonts) use a fixed width for numeric digits.  All digits have the same width which is equal to the width of "0".

Where things go haywire when trying to align numbers is that the space character does not have the same width as a digit when using a porportional spaced font.

Therefore if you want to line up a column of numbers, you can not use leading spaces to position them unless you use the same quantity of space characters for each row. e.g. your numbers have leading zeros etc.

The simplest solution is to switch to a monospaced font.  Alternativly you can use positioning tags.

Summary: Your numbers will always line up properly regardless of the font used, if you get the starting position to be consistant.
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-25
creating dot www at gmail dot com
7 years ago
Why rounding is not same as for round()?

Try this code:

<?php
printf
("%.02lf\n", 1.035);
printf("%.02lf\n", round(1.035, 2));
?>
Result:
1.03
1.04

In my opion it should be:
1.04
1.04

Why is that?
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-20
kalai_msc29 at rediffmail dot com
15 years ago
//If you want to make many Hidden fields you can use the function
//You can pass the values as array value,This will help you, when you are going to post many hiddend fields:-
function MakeHidden($ArrValues)
    {
       global $dearvar;
       echo $dearvar;
       if(is_array($ArrValues)){
             foreach($ArrValues as $key=>$values)
              {
                 echo $MakeHTML = "<input type='hidden' name='$values' value='$values'>";
               }
      }   
     
     
      else {
            echo $MakeHTML = "<input type='hidden' name='$ArrValues' value='$ArrValues'>";
         }
   return $MakeHTML;
}

//example:-

MakeHidden(array("value1","value2","value3"));

//OutPut :-

<input type="hidden" value="value1" name="value1"/>
<input type="hidden" value="value2" name="value2"/>
<input type="hidden" value="value3" name="value3"/>
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-22
sam[NOSPAM] at [NOSPAM]kingdomfaith dot com
19 years ago
I don't know if this is useful to anyone, but here goes! Example for using the printf function to output an object.

class person
{
    var $name = "";
    function name($newname = NULL)
    {
        if(! is_null($newname))
        {
            $this->name=$newname;
        }
        return $this->name;
    }
    var $surname = "";
    function surname($newsurname = NULL)
    {
        if(! is_null($newsurname))
        {
            $this->surname=$newsurname;
        }
        return $this->surname;
    }
    var $age = "";
    function age($newage = NULL)
    {
        if(! is_null($newage))
        {
            $this->age=$newage;
        }
        return $this->age;
    }
}

$bob = new person;
$bob->name('Bob');
$bob->surname('Builder');
$bob->age('50');

printf("Hi %s, your surname is %s and you are %s years old", $bob->name, $bob->surname, $bob->age);

Outputs:

Hi Bob, your surname is Builder and you are 50 years old
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-25
lordfarquaad at notredomaine dot net
18 years ago
In response to chris dot breen at accurate dot com :

This will work, but i didn't try:
<?php
$format
= 'The %2$s contains %1$d monkeys.
         That is a nice %2$s full of %1$d monkeys.'
;
printf($format, $num, $location);
?>

Your problem came from the fact that in the string "\$s", the $ is simply escaped by the \, but but you must not do that with single quoted strings. Try to echo your strings to test it, or just go to http://www.php.net/manual/en/language.types.string.php

官方地址:https://www.php.net/manual/en/function.printf.php

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