php "replace" help please.

[Jazzy_Girl]

Hi, Getting tired of much reading and getting nowhere with php replace. Whats the best way to replace whole functions with symbols, numbers, and text. I tried this, just to test this one with no luck along with several others. Could use some php wizards advice. Thank you very much.

<?php
$patterns = array ('I am just seeing if this will be replaced. ad;alkj646546///');
$replace = array ('All this is what I am trying to replace it with a;dfjk64646588800//, with no luck.');
echo preg_replace($patterns, $replace);
?>

[stolzyboy]

<?php
$string = 'The quick brown fox jumped over the lazy dog.';
$patterns = array('/quick/', '/brown/', '/fox/');
$replacements = array('slow', 'black', 'bear');
echo preg_replace($patterns, $replacements, $string);
?> 

[NogDog]

Note that if you just want to replace one string with another without using any regex features (wildcards, character classes, etc.) it's both faster and simpler to use str_replace().

[Jazzy_Girl]

I'm still having trouble with number and symbols and even php strings with variables in it. What php "replace" functions will pass php variables. Thank you very much.

[Weedpacket]

What do you mean "pass php variables"? Perhaps some real code will help, instead of a vague random mash.

$search = array('$a', '$b', '$c');
$replace = array('$z', '$y', '$x');
$string = 'function foo($a,$b){$c = $a+$b; return $c;}';

echo str_replace($search, $replace, $string);

[Jazzy_Girl]

Sorry about that, but that was a pretty impressive little algebra or whatever string you just did. I mean all important php variables. For instance if I try to replace this whole thing, something like it, or one elements at a time of it, with any of the replace commands I have trouble.

<?php
$getfile = rand(1,5);
$imagefile = "file" . $getfile . ".jpg";
echo "<img src='/$imagefile'>";
?>

I always get an "unexpected T variable" message. There must be a way for Php to replace even its own code and not just $functions and straight text, but even php code itself. Like this "<?php" and the just above and other items.

[liquorvicar]

I'm assuming you're treating the PHP code you posted as a string and then trying to replace various bits of it. If so, you need to make sure it's properly escaped. As your PHP code string has both single and double quotes in it, enclosing it in either single of double quotes won't work.
Either escape one type of quotes:

$string = '<?php
$getfile = rand(1,5);
$imagefile = "file" . $getfile . ".jpg";
echo "<img src=\'/$imagefile\'>";
?>';

Or use the heredoc syntax:

$string = <<<STRING
<?php
$getfile = rand(1,5);
$imagefile = "file" . $getfile . ".jpg";
echo "<img src='/$imagefile'>";
?>
STRING;

[Jazzy_Girl]

Yes, I was having trouble trying to figure out what to do about those differring quotation marks, thank you very much. I think I already tried that but not positive, I'll test it tomorrow and see how it goes.

[Jazzy_Girl]

I don't know, I tried simplifying it even further to see if I could get it working with only an echo command first. All I am trying to do is echo some php. Once I can get that mastered then I can move on to the replace functions. I even studied this page in detail
http://www.php.net/manual/en/language.types.string.php
and some of their instructions don't work. Putting slashes in front of the \$ variables etc. don't seem to take. Am I reading their string page correctly? Doesn't their table named:
Table 11-1. Escaped characters Just mean in order to parse php properly some of the php code needs its variables altered? So a $ would instead be \$ etc? Because none of it works for me. I could use some help steering me further in the right direction. If anyone knows how to echo the following php code, or get me to the right pages for help please let me know, I am getting tired of going in circles. Thank you very much.

<?php
echo "<?php
$getfile = rand(1,15);
$imagefile = "file" . $getfile . ".jpg";
echo "<img src=\'/$imagefile\'>";
?>";
?>

[liquorvicar]

Have you tried my first example?

<?php
 $string = '<?php
$getfile = rand(1,5);
$imagefile = "file" . $getfile . ".jpg";
echo "<img src=\'/$imagefile\'>";
?>';
 echo($string);
 ?>

outputs

<?php
$getfile = rand(1,5);
$imagefile = "file" . $getfile . ".jpg";
echo "<img src='/$imagefile'>";
?>

[Jazzy_Girl]

You are so right and thank you for helping me with this. I did see that it put it in the source code soon after you first gave it to me. Then I got sidetracked and used it often in other other circumstances to test it. For example it wouldn't work in replace and so many other situations even with your altered code and I couldn't understand why since it worked in the echo. But thanks for bringing me back to it again to evaluate it. How would I make it execute? For instance even though I now have it in the source code the server isn't reading it. Is it because the server is reading it between the other php tags and its confusing the server? So I took out the php tags on the echo and then I get this.
$getfile = rand(1,5); $imagefile = "file" . $getfile . ".jpg"; echo " "; Between the echo quotations it shows a spot for an image but no image appears. When I look in the image location properties, it only shows the image as /$imagefile instead of file04.jpg etc. So either way its not showing an image. Can you shed some light on why? You might be wondering why I just don't put it in the regular way instead of going through all this.
Its only because I'm trying to learn how to use echo, replace, and eval, properly with php code itelf, instead of just text. Thank you very much.

[liquorvicar]

Can't say I'm an expert on eval() but:

<?php
$str = '$getfile = rand(1,5);
$imagefile = "file" . $getfile . ".jpg";
echo "<img src=\'/$imagefile\'>";';
eval("\$str = $str;");
echo($str);
?>

outputs:

<img src='/file2.jpg'>2

which is almost what you want...

[Jazzy_Girl]

You just made my month, I've been working on that for weeks! Now I'm feeling much better. Thanks a bunch!!!!! A number is appearing next to the image, any way to get rid of it?

[liquorvicar]

<?php
$str = '$getfile = rand(1,5);
$imagefile = "file" . $getfile . ".jpg";
echo "<img src=\'/$imagefile\'>";';
eval($str);
?> 

[Jazzy_Girl]

Thats it! I'm sending you something, thank you very much!!

[Weedpacket]

I take it you're planning on storing that string somewhere for later use?

[Jazzy_Girl]

Yes, experimentation. I'll let you know if anything else comes of it.