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POEscaping reference look-alikes in the replacement parameter of PHP preg_replace
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19579445
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2013-10-25T01:09:01.310
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2013-11-04T21:28:44.270
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2013-10-25T07:21:04.020
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text
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I would like to replace a pattern string '/r/' within a subject string 'r r' with predetermined replacements like '"\\23"' and '[4]'. So (showing you only a subset of my replacements array) I try: <pre><code>$subject = 'r r'; $replacements = array('"\\23"', '[4]'); $replacements = array_map('preg_quote', $replacements); print_r($replacements); foreach($replacements as $replacement){ $subject = preg_replace('/r/', $replacement, $subject, 1); } echo $subject; </code></pre> The string I want is '"\\23" [4]', but I get back '"\23" \[4\]', which is both missing characters and has additional characters. Does PHP have an escape function that would work here? How do I do this? Note: Without the preg_quote line the result is '"" [4]'.
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<php><reference><escaping><preg-replace>
Title
Escaping reference look-alikes in the replacement parameter of PHP preg_replace
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1. PTQuestion
UserLastEditorUserId
1. USzylstra
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1. USzylstra
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