PHP variable in loop

Question

I have the following:

$textq1 = 'text';
$textq2 = 'more text';

I have a loop:

for ($i = 1; $i <= 70; $i++) {

    echo "<div>". $textq1." ";

}

Now....what I'd like to do is, within the loop have the 1 part of the $textq1 be the value of $i.

Is that possible?

Would be a whole lot easier if you had $textq = array('text','more text'); rather than $textq1 = 'text'; $textq2 = 'more text'; — Mark Baker
– Mark Baker, Commented Mar 1, 2012 at 17:11
@MarkBaker - yep, sure would be easier in an array but I've inherited a whole host of code... — Homer_J
– Homer_J, Commented Mar 1, 2012 at 18:45

Sarfraz · Accepted Answer · 2012-03-01 17:12:25Z

3

Change:

echo "<div>". $textq1." ";

To:

echo "<div>". ${'textq'.$i} ." ";

edited Mar 1, 2012 at 17:12

answered Mar 1, 2012 at 17:07

Sarfraz

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Comments

csanyigabor · Accepted Answer · 2012-03-01 17:06:37Z

1

for (...)
{
    echo "<div>{${textq$i}}</div>";
}

answered Mar 1, 2012 at 17:06

csanyigabor

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2 Comments

Sarfraz Over a year ago

That would result in Parse error: syntax error, unexpected T_VARIABLE error.

Travesty3 Over a year ago

Yes it would. Would work if it were echo "<div>{${"textq$i"}}</div>"; (need quotes around the variable name).

Travesty3 · Accepted Answer · 2012-03-01 17:07:36Z

1

$var = "textq{$i}";
echo "<div>". $$var." ";

answered Mar 1, 2012 at 17:07

Travesty3

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Comments

Andy · Accepted Answer · 2012-03-01 17:07:32Z

0

Take a look at this example about variable variables in the PHP docs: http://www.php.net/manual/en/language.variables.variable.php#105282

answered Mar 1, 2012 at 17:07

Andy♦

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