[SOLVED] Correct My Code?

[herghost]

Well, using the little php I know I have come up with this:

 

<?php
		function changelink($change)
            {
		$find = "htmlspecialchars($row["source"])";
  $replace = "<a href="http://www.amazon.co.uk/dp/($row["source"])?tag=speciainvita-21&camp=1406&creative=6394&linkCode=as1&creativeASIN=($row["source"])" />";
  
  


  $change = preg_replace($find, $replace, $change);
  return $change;
}


             <?php echo $change; ?>

 

Which obviously doesn't work!

 

What I am trying to do is insert "source" into various places in the bottom line. I am guessing there is another way of doing this apart from the find and replace method as I am guessing trying to find a string will not work. I just wanna know how to do this! I welcome anyone who can show me how to do this and explain how it works as I am really trying to learn php and not just copy and paste!

 

This is the line originally:

 

<?php echo htmlspecialchars($row["source"]); ?>

 

and "source" is gathered using:

 

<input name="source" type="text" value="<?php echo htmlspecialchars(stripslashes($source)); ?>" maxlength="255" size="50"/>

 

Thanks

 

 

[PersonPerson]

Escape all double quotes except the initial and final ones.

[teng84]

this should work

$find = htmlspecialchars($row["source"]);
  $replace = '<a href="http://www.amazon.co.uk/dp/('.$row["source"].')?tag=speciainvita-21&camp=1406&creative=6394&linkCode=as1&creativeASIN=('.$row["source"].')" />';

 

[cooldude832]

that will never work ever regardless of if you fix quotes

 

 

go to php.net and look at variable scope you are trying to use global variables (or what looks to be global) in a function and that isn't going to work with your syntax.

 

$row['source'] is undefined in the function thus it can't do anything about it.

 

 

Your even trying to use the local variable $change outside the function in the global area that won't work also.

[teng84]

that will never work ever regardless of if you fix quotes

 

 

go to php.net and look at variable scope you are trying to use global variables (or what looks to be global) in a function and that isn't going to work with your syntax.

 

$row['source'] is undefined in the function thus it can't do anything about it.

 

 

Your even trying to use the local variable $change outside the function in the global area that won't work also.

oh sorry i don't see that i though its not inside the function and i dont see this function changelink($change)

[herghost]

Thanks for your help so far, so how do i get this to work? can I do it without creating a function, my knowledge of php is very limited, im learning as I go!

[herghost]

Is there anyway to get this to work?

 

Thanks

 

dave

[herghost]

ok I have been looking at this for a few days now and have got no closer, any ideas people?

[cooldude832]

explain clearly what u trying to do

[atlanta]

its because your using preg_replace and not using the correct syntax for it .

[herghost]

Ok,

 

The user inserts a product number, for example : 1000 and it is stored using this:

 

<input name="source" type="text" value="<?php echo htmlspecialchars(stripslashes($source)); ?>" maxlength="255" size="50"/>

 

I then want to display "source" inside a line:

 

i.e

 

http://www.amazon.co.uk/dp/"source"?tag=speciainvita-21&camp=1406&creative=6394&linkCode=as1&creativeASIN="source") using the echo

 

Is this possible?

 

Thanks dude

 

sorry post updates : I know I am using the wrong syntax, i just have no idea what to use or what it does, it is outside my limited knowledge of php

 

[atlanta]

try

 

 

<?php
		function changelink($change)
            {
		$find = "source";
		$replace = htmlspecialchars($change);
  $change = '<a href="http://www.amazon.co.uk/dp/(source)?tag=speciainvita-21&camp=1406&creative=6394&linkCode=as1&creativeASIN=(source)" > test</a>';
  

$change2 = str_replace($find, $replace, $change);
  
echo $change2;
}
changelink($row["source"]);
?>

[herghost]

Thanks for this, I think I understand it! I am I correct in thinking that I can load the function by saving it as say functions.php and calling it by typing:

 

<?php echo $row["source"]; ?>

 

making sure I include functions.php

 

thanks

 

 

[herghost]

One last question, I have it working, almost! The only thing is that when the link is echo'ed it is placing the brackets around the code, so the url is not working, how do I get it to display the code without the brackets?

 

Thanks

 

Dave

[herghost]

nevermind got it!

 

 

Thanks to everyone that helped