preg_replace（）在数组上不起作用

Question

I have the following code:

$pattern_dig_answer = array(); // Identify patterns to replace and their replacements
$pattern_dig_answer[0] = "/(.*)IN(\s+)A(.*)/";
$replacement_dig_answer = array();
$replacement_dig_answer[0] = "<span style='color:red;'>$1IN$2A$3</span>";

The first entry for the pattern_dig_answer array is some regex for the pattern I'm looking to match. The first entry for the replacement_dig_answer is what I am looking to replace what the regex finds with.

I then execute a dig like so:

$dig_answer = exec("dig {$_SESSION['domain']} $digrecords[2] $digops[0] $digops[1] $digops[2] $digops[3]", $dig_answer_raw);

Which, as you may guess, executes a dig command with session variables captured from user input and then saves the output to $dig_answer_raw.

What I now do is clean up the dig with the following:

$dig_answer_out = preg_grep('/IN/', $dig_answer_raw);

And then I echo the results:

echo implode("\n", $dig_answer_out);

This all works fine up until this point. It's only when I go to do the following that I get stuck:

echo preg_replace($pattern_dig_answer, $replacement_dig_answer, $dig_answer_out);

Now, from the investigating I've done, this should work. The documentation for preg_replace clearly says that it can search either in a normal string or via an array with string entries (which I'm assuming I have). What makes me think that I might not is as soon as I change the dig command to this:

$dig_answer = shell_exec("dig {$_SESSION['domain']} $digrecords[2] $digops[0] $digops[1] $digops[2] $digops[3]");

It works like a charm, however negates all the output cleaning that preg_grep did previously even after changing around all the variables.

Is there something painfully obvious that I'm missing here?

EDIT: To try and simplify my question, when my dig command is changed from exec(), which saves the output to an array, to shell_exec, which saves the output to a string, then preg_replace works. It doesn't seem to want to work in an array and I'm not sure why...

Answer 1

I reconstructed the script based on your original post:

<?php

$dig_answer = exec("dig www.google.com.au", $dig_answer_raw);

$dig_answer_out = preg_grep('/IN/', $dig_answer_raw);

$pattern_dig_answer = array(); // Identify patterns to replace and their replacements
$pattern_dig_answer[0] = "/(.*)IN(\s+)A(.*)/";
$replacement_dig_answer = array();
$replacement_dig_answer[0] = "<span style='color:red;'>$1IN$2A$3</span>";

$x = preg_replace($pattern_dig_answer, $replacement_dig_answer, $dig_answer_out);
print_r($x);

The output from this script looks like it is working for me:

Array
(
   [8] => <span style='color:red;'>;www.google.com.au.      IN  A</span>
   [11] => <span style='color:red;'>www.google.com.au.  125 IN  A   203.219.219.163</span>
   ...
)

Am I missing something? For reference, this is the output of php -v for me.

PHP 5.4.24 (cli) (built: Jan 19 2014 21:32:15)

Answer 2

I just noticed that this line may not produce what you are expecting:

echo preg_replace($pattern_dig_answer, $replacement_dig_answer, $dig_answer_out);

In this instance, preg_replace() will return an array as you have passed an array as the first and second arguments. Try this instead:

$foo = preg_replace($pattern_dig_answer, $replacement_dig_answer, $dig_answer_out); var_dump($foo);

Answer 3

You've split your code so it's tricky to tell what you're running. I think this is it, reconstructed from all of the parts as follows:

<?php
$pattern_dig_answer = array(); // Identify patterns to replace and their replacements
$pattern_dig_answer[0] = "/(.*)IN(\s+)A(.*)/";
$replacement_dig_answer = array();
$replacement_dig_answer[0] = "<span style='color:red;'>$1IN$2A$3</span>";

$dig_answer = exec("dig google.com", $dig_answer_raw);
$dig_answer_out = preg_grep('/IN/', $dig_answer_raw);
echo implode("\n", $dig_answer_out);
echo preg_replace($pattern_dig_answer, $replacement_dig_answer, $dig_answer_out);

When I run this over a CLI (PHP 5.5.11) I get a warning: PHP Notice: Array to string conversion on the line where you're calling echo preg_replace(...); .

You're (implicitly) converting the array that is returned by preg_replace() to a string with echo . Since you're passing in an array() for preg_replace for both parameters, you're getting back an array (as you assumed). The best way to dump those to view what is inside is with var_dump() . If you want to print it in a different way you could use a foreach loop and print that element.

<?php
$pattern_dig_answer = array(); // Identify patterns to replace and their replacements
$pattern_dig_answer[0] = "/(.*)IN(\s+)A(.*)/";
$replacement_dig_answer = array();
$replacement_dig_answer[0] = "<span style='color:red;'>$1IN$2A$3</span>";

$dig_answer = exec("dig google.com", $dig_answer_raw);
$dig_answer_out = preg_grep('/IN/', $dig_answer_raw);
implode("\n", $dig_answer_out);
$total_output = preg_replace($pattern_dig_answer, $replacement_dig_answer, $dig_answer_out);
foreach ($total_output as $output) {
    echo $output."<br />\n";
}