PHP variable replacement

Question

Is there any function which replaces params in a string? Something like this:

Code:

$format_str = "My name is %name."; /* this was set in a 
                                      configuration file - config.php */

$str = xprintf($format_str, array('name' => 'Joe', 'age' => 150)); 
              /* above is somewhere in main code */

The expected value of $str after the operation is:

My name is Joe.

Update: I am aware of sprintf. But, it would not suffice in this case. I have modified the code to show what is the difference.

probably an error on the part of the author

Jonathan Fingland
– Jonathan Fingland

2009-06-09 04:12:40 +00:00
Commented Jun 9, 2009 at 4:12 — Jonathan Fingland
– Jonathan Fingland, Commented Jun 9, 2009 at 4:12

2 revs · Accepted Answer · 2023-11-14 22:25:35Z

10

seems like strtr is what is a builtin function which can do the same. (got this from going thru drupal code).

>> $format_str = "My name is %name.";
My name is %name.
>> strtr($format_str, array('%name' => 'Joe', '%age' => 150))
My name is Joe.

edited Nov 14, 2023 at 22:25

community wiki

2 revs
Shameem

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1 Comment

Christian Over a year ago

This looks to be exactly what is required.

3 revs · Accepted Answer · 2009-06-09 15:42:51Z

3

you could use this:

function xprintf($str, $array, $chr = '%') {

   foreach ($array as &$key => $val) {
       $key = $chr . $key;
   }

   return strtr($str, $array);
}

$str = xprintf('My name is %name', array('name' => 'Joe'));

edited Jun 9, 2009 at 15:42

community wiki

3 revs
Ozzy

4 Comments

Ozzy Over a year ago

Wtf, why did i get downvoted. I would have talked about sprintf but from the way he asked the question, it seems he wants UNIQUE identifiers not just %s. I get downvoted for reading the question -_-

Stobor Over a year ago

Similar to your answer, Something more like what he's after: au2.php.net/manual/en/function.sprintf.php#83779 see the example sprintf3

Ozzy Over a year ago

Thx for the link. Ill update my answer with something more like what he wants

rojoca Over a year ago

The problem with str_replace is that it replaces replacements. For example try this: $str = xprintf('My name is %first %last', array('first' => '%last', 'last' => 'Smith')); You should get 'My name is %last Smith' but instead you get 'My name is Smith Smith'. Using strtr avoids this problem.

Paige Ruten · Accepted Answer · 2009-06-09 03:00:21Z

2

Do you mean sprintf()?

$str = sprintf("My name is %s.", 'Joe');

answered Jun 9, 2009 at 3:00

community wiki

Paige Ruten

Comments

razzed · Accepted Answer · 2009-06-09 04:55:21Z

This function, "map" (and replace), is part of my web application system:

function replace($search, $replace, $mixed)
{
    if (is_string($mixed)) {
        return @str_replace($search, $replace, $mixed);
    } else if (is_array($mixed)) {
        foreach ($mixed as $k => $v) {
            $mixed[$k] = replace($search, $replace, $v);
        }
        return $mixed;
    } else {
        return $mixed;
    }
}


function map($a, $contents, $case_sensitive=false)
{
    if (!is_array($a)) {
        return $contents;
    }
    if (!$case_sensitive) {
        $a = array_change_key_case($a);
    }
    $s = array();
    $r = array();
    foreach ($a as $k => $v) {
        if (is_scalar($v) || empty($v)) {
            $s[] = "{".$k."}";
            $r[] = $v;
        }
    }
    if (!$case_sensitive) {
        $contents = preg_replace_mixed('/{([-_ =,.\/\'"A-Za-z0-9]+)}/ei', "'{'.strtolower('\\1').'}'", $contents);
    }
    return replace($s, $r, $contents);
}

Does a very good job. Supply any string with bracketed variables names, and it does the trick.

Syntax is different, but could be modified for your purposes:

$str = map(array('name' => 'Joe'), 'My name is {name}');

I prefer it over the % syntax.

2 revs · Accepted Answer · 2023-11-20 22:45:59Z

0

You're looking for vsprintf. https://www.php.net/vsprintf

It allows you to pass an array of arguments as you indicated.

edited Nov 20, 2023 at 22:45

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2 revs
dirtside

Comments

2 revs · Accepted Answer · 2023-11-20 22:45:59Z

-1

Try giving printf() a whirl. That website has a lot of examples for using arrays with printf.

printf("My name is %s", "Joe");

edited Nov 20, 2023 at 22:45

community wiki

2 revs
poundifdef

Comments

nickf · Accepted Answer · 2009-06-09 04:19:52Z

-1

You could always use eval... (gasp!)

$myStr = 'Hi my name is $name. I\'m $age years old.';
$myVars = array("name" => "Joe", "age" => 6);
$parsed = parseString($myStr, $myVars);

function parseString($str, $vars) {
    extract($vars);
    return eval('return "' . addslashes($str) . '";');
}

Before anyone gets cranky about using extract AND eval, as long as you have control over the input template string, I can't see how this could be a security problem.

answered Jun 9, 2009 at 4:19

community wiki

nickf

Comments

too much php · Accepted Answer · 2009-06-09 04:34:53Z

-1

PHP can pull in values from an array without using sprintf():

$data = array('name' => 'Joe', 'age' => 150);
print "My name is $data[name], my age is $data[age].";

answered Jun 9, 2009 at 4:34

community wiki

too much php

1 Comment

thomasrutter Over a year ago

Doesn't help - needs to be able to deal with an actual string containing % followed by a word, where the string value is set BEFORE the replacement values are known.

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