Re: php seems to be inconsistent in its handling of backslashes ... maybe?

The proper way to handle special control PCRE characters like "\" is to use the hex [e.g., \x5C] value. Then you won't have a problem. The engine knows you want the the object treated as a character and an a control. ufan100@gmail.com wrote:

-- or maybe it's just the PCRE extension -- or quite likely I have got something wrong Hello members, I'm hoping you could enlighten me. Using error_reporting = E_ALL | E_STRICT, I tested the following statements: <?php preg_match('#\\#','any-string'); => warning preg_match('#\\\#','any-string'); preg_match('#\\\\#','any-string'); preg_match('#\\\\\#','any-string'); => warning preg_match('#\\\\\\#','any-string'); => warning preg_match('#\\\\\\\#','any-string'); ?> This seemed strange: warnings with 2 and 6 backlashes no warnings with 3, 7 warning with 5 but not with 3 and 7. The warning related of course to no matching delimeter '#' being found. So I wrote a little test script (preg.php) to test up to 10 backslashes: <?php for($i=1; $i<=10; ++$i) {

    echo "\n---------------------------------\n";
    echo "Number of '\\' is $i \n";
    $bs = '#';
    $bs .=  str_repeat('\\',$i);
    $bs .= '#';
    echo 'Pattern is: ' . $bs . "\n";
    $php_errormsg = "";
    @preg_match($bs, "anystring") . "\n";
    if($php_errormsg != '')
        echo "error\n";
    else echo "ok\n";

} ?> Here is the output: $ php preg.php --------------------------------- Number of '\' is 1 Pattern is: #\# error --------------------------------- Number of '\' is 2 Pattern is: #\\# ok --------------------------------- Number of '\' is 3 Pattern is: #\\\# error --------------------------------- Number of '\' is 4 Pattern is: #\\\\# ok --------------------------------- Number of '\' is 5 Pattern is: #\\\\\# error --------------------------------- Number of '\' is 6 Pattern is: #\\\\\\# ok --------------------------------- Number of '\' is 7 Pattern is: #\\\\\\\# error --------------------------------- Number of '\' is 8 Pattern is: #\\\\\\\\# ok --------------------------------- Number of '\' is 9 Pattern is: #\\\\\\\\\# error --------------------------------- Number of '\' is 10 Pattern is: #\\\\\\\\\\# ok End of output. This agrees with my understanding of backslash escaping (I hope that's right) but now I can't understand why I got the results earlier (shown in my first script). Many thanks.