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Home > Forums > Programmers > Web Development > PHP Forum

preg_replace with backreference AND curly bracesHelpful Member! 

thread434-1685408
Read More Threads Like This One
OsakaWebbie (Programmer)
11 Jun 12 3:18
Trying to use PHP to construct TeX code, and TeX loves curly braces. But in PHP, curly braces are interpreted as part of parsing variables. I have strings like "Dm7 G" (yes, they are guitar chords), and want them to become like "Dm\Sup{7} G".

If I do this:

CODE --> PHP

preg_replace("/([0-9]+)/","\\Sup{$1}",$chord)
I get a syntax error.

But if I do this:

CODE --> PHP

preg_replace("/([0-9]+)/","\\Sup\{$1\}",$chord)
the backslashes actually end up in the resulting string, which is not desired.

Suggestions?
Helpful Member!  feherke (Programmer)
11 Jun 12 4:24
Hi

Quote (OsakaWebbie)

But in PHP, curly braces are interpreted as part of parsing variables.
Correct. So you have to stop the variable parsing :

CODE --> PHP

preg_replace("/([0-9]+)/","\\Sup{\$1}",$chord);

May be caused by my affinity for shell scripting, but I generally prefer to use the most restrictive quoting possible in each case. In this case I would use single quotes ( ' ) :

CODE --> PHP

preg_replace('/([0-9]+)/','\\Sup{$1}',$chord);

Feherke.
http://feherke.github.com/

OsakaWebbie (Programmer)
11 Jun 12 11:13
Thanks! I must be losing my mind - I thought I tried that second version (among various things I tried in desperation) and got an error. Anyway, naturally you know what you're doing, and both solutions work.

Quote (feherke)

So you have to stop the variable parsing...
I'm a little confused - when I stop the parsing (either by escaping the $ or using single quotes), why don't I end up with a literal "Dm\Sup{$1} G"?
feherke (Programmer)
11 Jun 12 11:25
Hi

Quote (OsakaWebbie)

when I stop the parsing (...) why don't I end up with a literal "Dm\Sup{$1} G"?
In this case $1 is not a variable, it is a placeholder. The preg_replace() function replaces such placeholders after her finds a substring matching the regular expression and extracted the captured groups.

Variable parsing had to be stopped to avoid the PHP interpreter trying to interpret $1 as variable before the preg_replace() function receives its string parameters.

Feherke.
http://feherke.github.com/

OsakaWebbie (Programmer)
11 Jun 12 19:53
Okay, thanks for the clarification. smile
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