[RESOLVED] Changing One Line In File?

Channel5

Hi guys,

I'm trying to change only one certain line in a file without changing anything else, but so far, I'm wiping out the entire file.

Here's how I've been trying to do it:

while(!feof($file)){
    $line = fgets($file);
    if($line == "Old Line 3"){
        $string = "New Line 3";
        fwrite($file, $string);
    }
}
fclose($file);

Any suggestions?

NogDog

You'll have to read the whole file into a string or array, make the change there, then write it back over the file. One way:

$text = file($file, FILE_IGNORE_NEW_LINES)
{
   foreach($text as $key => $line)
   {
      if($line == "Old Line 3")
      {
         $text[$key] = "New Line 3";
         // if you only expect one instace, do a break here:
         // break;
      }
   }
}
file_put_contents($file, implode("\n", $text));

Channel5

Hey NogDog,

When I run that code, I get an "unexpected {" error.

Though I'm not really familiar with some of the functions used here, it seems to me that there's no need for { starting after the first line of code.

But when I try removing those {} I get other errors.

Any ideas?

NogDog

Everyone's a critic. :p

<?php
$text = file($file, FILE_IGNORE_NEW_LINES);
foreach ($text as $key => $line)
{
   if ($line == "Old Line 3")
   {
      $text[$key] = "New Line 3";
      // if you only expect one instace, do a break here:
      // break;
   }
}
file_put_contents($file, implode("\n", $text));

Channel5

Yes, but what I'm saying is that even when I use this code, I get errors:

Warning: file() expects parameter 1 to be string, resource given in /home/content/j/a/y/jay/html/1/chset.php on line 3

Warning: Invalid argument supplied for foreach() in /home/content/j/a/y/jay/html/1/chset.php on line 5

Warning: implode() [function.implode]: Invalid arguments passed in /home/content/j/a/y/jay/html/1/chset.php on line 12

Warning: file_put_contents() expects parameter 1 to be string, resource given in /home/content/j/a/y/jay/html/1/chset.php on line 12

The only thing I could think of was my $file, which is:

$file = fopen("num.txt", "w");

Ideas?

NogDog

The first parameter to file() is the path/name of the file. Change it to whatever you want it to be, or assign the string to $file. (There is no need for fopen() when using file() or file_put_contents().)

Channel5

Excellent! It's working now.

Thank-you.