Php How to put a variable with in a variable

Question

I have this code:

echo $b1[1][wood];   // It would say 100

But i want to change the 1 in $b1, for example:

$id = 1;
echo $b(The $id here)[1][wood];

I tried

echo $b'.$id.'[1][wood];

But it didnt work. Does any one have any suggestions ?

Thanks

Ick. Try to avoid "variable variables" - not many languages have such a construct, but work fine anyway. Don't confuse "convenient" (per the documentation's wording) with "appropriate". — user2246674
– user2246674, Commented Jun 2, 2013 at 0:15
Milan's answer below is perfect for this. however, I suggest you would want to try out using array instead. — Aldee
– Aldee, Commented Jun 2, 2013 at 4:59

zerkms · Accepted Answer · 2013-06-02 00:07:05Z

5

Try this:

$id = 1;
echo ${'b'.$id}[1]['wood'];

zerkms

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answered Jun 2, 2013 at 0:04

Milan Babuškov

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Comments

Oussaki · Accepted Answer · 2013-06-02 00:05:31Z

2

why don't you use an array() type ?

answered Jun 2, 2013 at 0:05

Oussaki

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jcsanyi · Accepted Answer · 2013-06-02 00:38:00Z

2

Try this:

$varName = 'b' . $id;
echo ${$varName}[1]['wood'];

(Notice I also put quotes around 'wood' - PHP treats wood as an undefined constant, which will work, but generates a warning)

A better solution would be to avoid variable variables altogether. Couldn't you just make $id another dimension of the array? $b[$id][1]['wood'] ?

answered Jun 2, 2013 at 0:04

jcsanyi

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