ID: 16786
Updated by: [EMAIL PROTECTED]
Reported By: [EMAIL PROTECTED]
Status: Feedback
Bug Type: Regexps related
Operating System: Linux 2.2.19
PHP Version: 4.2.0
New Comment:
This code should demonstrate the problem:
<?php
$str = "word1-word2";
$str = ereg_replace("^([^-]*)-(.*)", '\1+\2', $str);
print("1) Result = '$str'\n");
$str = "-word2";
$str = ereg_replace("^([^-]*)-(.*)", '\1+\2', $str);
print("2) Result = '$str'\n");
if ($str == "+word2") print ("No Bug Detected\n");
else print("Bug Detected (The result should be '+word2')\n");
?>
Previous Comments:
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[2002-04-24 06:30:42] [EMAIL PROTECTED]
Could you paste complete code?
(With complete code, we can test immediately and put the code
to test suite, if it's needed)
------------------------------------------------------------------------
[2002-04-24 05:12:16] [EMAIL PROTECTED]
Hi,
I have just downloaded PHP4.0.2 and several pieces of code broke
because ereg_replace() seemed to be working incorrectly:
if for example I did the following:
$str = "word1-word2";
$str = ereg_replace("^([^-]*)-(.*)", '\1+\2', $str);
it works as expected, ($str == "word1+word2")
The problem arises if one of the parenthasized subexpression is empty
the \x (where x is the number of the subexpression) in the replace
string does not get replaced at all, instead of with the empty string
eg:
$str = "-word2";
$str = ereg_replace("^([^-]*)-(.*)", '\1+\2', $str);
for this example I'd expect the result to be ($str == "+word2") but
instead you get ($str == "\1+word2") because \1 is empty.
I don't believe that this is meant to happen but if it is how are we
meant to deal with such a case?
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Edit this bug report at http://bugs.php.net/?id=16786&edit=1
