php - define sequential numbers

Question

$ext is an array

$noofformats = count($data['formats']);
for($x = 0; $x < $noofformats; $x++){ 
    echo $select = "<option value=\"" . $x . "\">" . $ext[$x] . "</option>";
}

When I run this script I get the desired result. Result 1:

<option value="0">webm</option><option value="1">m4a</option> and so on.....

This works because it echos the result everytime it runs, so I get the string with every value of $noofformat possible. But I wish to define $select as the result shown above. What happens is everytime function loops the newer result is defined instead of being added to the previous value, resulting in only the last value

<option value="33">mp4</option>

I wish

echo $select;

to give a result like Result 1

Kevin · Accepted Answer · 2015-04-18 03:36:38Z

2

What happens right now is that every iteration, $select gets overwritten. You could continually concatenate the markup string first, then finally echo in the end:

$noofformats = count($data['formats']);
for($x = 0; $x < $noofformats; $x++){ 
    $select .= "<option value=\"" . $x . "\">" . $ext[$x] . "</option>";
        //  ^ continually concatenate values while the loop is running
}

echo $select;

answered Apr 18, 2015 at 3:36

Kevin

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2 Comments

Kazi Lotus Over a year ago

Thanks for replying so quickly :D

Kevin Over a year ago

@KaziLotus yes you could just add the concatenation operator to append each string into it instead of overwriting it each iteration

Tamil Selvan C · Accepted Answer · 2015-04-18 03:37:22Z

1

Use

$noofformats = count($data['formats']);
for($x = 0; $x < $noofformats; $x++){ 
    echo $select .= "<option value=\"" . $x . "\">" . $ext[$x] . "</option>";
}

answered Apr 18, 2015 at 3:37

Tamil Selvan C

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