passing php variable into html

Question

I'm new to php so pardon the ignorance
I'm trying to include a php variable, it is and is not working.

$currentdata = file_get_contents("http://www.abr.business.gov.au/abnDetails.aspx?abn=$abn");

This works with $abn being the variable being passed from earlier being set.

$abntypedata = '/d$abn&amp;ResultListURL=">(.+?)</'; This does not work.
$abntypedata = '/d33051775556&amp;ResultListURL=">(.+?)</'; This does.

I need to be able to use the variable $abn to insert that number as it will be user defined. Why is this not working?

Tieson T. · Accepted Answer · 2011-05-17 01:29:50Z

4

Variable substitution does not occur with single quotes. Change those to double quotes and should work.

EDIT: FYI, here's the PHP manual for strings in PHP. It's useful to know what the various types of strings are: http://www.php.net/manual/en/language.types.string.php

answered May 17, 2011 at 1:29

Tieson T.

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Comments

Mohamed Nuur · Accepted Answer · 2011-05-17 01:30:52Z

1

Php is perticular about the quotations. You have to use double quotes to include variables:

$abntypedata = "/d$abn&amp;ResultListURL=\">(.+?)</"; This works

or

$abntypedata = '/d'.$abn.'&amp;ResultListURL=">(.+?)</'; This works

answered May 17, 2011 at 1:30

Mohamed Nuur

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1 Comment

jnrcoder Over a year ago

Great, #2 worked perfectly and I learnt something along the way. Thanks.

John Flatness · Accepted Answer · 2011-05-17 01:30:46Z

1

You can't use inline variables like that in single-quoted strings.

$abntypedata = "/d$abn&amp;ResultListURL=\">(.+?)</";

answered May 17, 2011 at 1:30

John Flatness

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Comments

brenjt · Accepted Answer · 2011-05-17 01:31:22Z

0

Try adding Double Quotes around the string " not ' that should render the variables other wise do this:

$abntypedata = '/d'.$abn.'&amp;ResultListURL=">(.+?)</';

answered May 17, 2011 at 1:31

brenjt

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