Variable as part of other variable php

Question

I have a for script:

<?php for($i=1; $i<10; $i++):?>
<?php if($this->item->nuotrauka{$i}):?>
<div class="vobsmall">
<img src="<?php echo $this->item->nuotrauka{$i}; ?>"/>
<?php echo $this->item->nuotrauka{$i}; ?>
</div>
<?php endif; ?>
<?php endfor; ?>

There's 9 variables like $this->item->nuotrauka1, $this->item->nuotrauka2 etc.

As you can see here I'm trying to call these variables using {i} as number of object variable. Hovewer the code above returns single char from variable and not full variable as I want. How should I write {i} here to get what I want ?

Why are you using variables named like this instead of using an array? — Barmar, Sep 10 '14 at 16:04
Having to use dynamic variables can be a sign for bad design somewhere. As Barmar said, you probably want to use an array instead. — Felix Kling, Sep 10 '14 at 16:06
Just make a variable in each loop. Like so: $node = 'nuotrauka' . $i; and then when you display it: $this->item->$node. — ShayneStatzell, Sep 10 '14 at 16:08

score 3 · Accepted Answer · answered Sep 10 '14 at 16:06

3

Try this:

echo $this->item->{"nuotrauka".$i};

answered Sep 10 '14 at 16:06

CMPS

7,733
4
28
53

1

Doesn't `nuotrauka` have to be a string? – Felix Kling Sep 10 '14 at 16:06
1

(I mean, it does work, but more accidentally. You'd get a notice like `Notice: Use of undefined constant nuotrauka - assumed 'nuotrauka' in /foo.php on line xyz`) – Felix Kling Sep 10 '14 at 16:09

Variable as part of other variable php

1 Answers1