var_dump displays text, but echo does not

Question

I have an array called $worker
the array consists of only strings, each of which has multiple lines.

if I do

var_dump($worker);

it displays all the information, but do

for($i=0,$size=sizeof($worker);$i<$size;++$i)
{
    echo $worker[i];
}

I end up with nothing on the page.

I'm very new to php, so sorry if this is a noob question: how do I get the information in the array to print to the screen correctly?

Enable on-screen display of warnings in your development environment, and you'll notice these mistakes right away. — Matthew, Jun 05 '11 at 19:57

score 5 · Answer 1 · edited Jun 06 '11 at 06:55

5

You're missing the '$' for your '$i' variable inside the for loop.

It's a good idea to turn on error reporting while developing in PHP: http://php.net/manual/en/function.error-reporting.php

This is the conventional syntax for for loops in PHP:

for ($i=0, $c=count($worker); $i<$c; $i++) {
    echo $worker[$i];
}

edited Jun 06 '11 at 06:55

Jordon Bedwell

answered Jun 05 '11 at 19:45

asdf

1

A little more detail on what is actually going wrong with the missing $. When PHP sees `$worker[i]` it thinks that `i` is a bare string instead of a variable. Since there is no constant named `i` it converts `i` into a string and tries to access `$worker['i']` since PHP arrays work like associative arrays. Since `$worker['i']` hasn't been set it returns `null`. `echo null` prints an empty string. Also, as the code @asdf posted points out in PHP it is customary to use `count` instead of `sizeof`, `sizeof` is just an alias of `count`. – Useless Code Jun 05 '11 at 20:34

score 3 · Answer 2 · edited Jun 06 '11 at 06:55

3

for($i=0,$size=count($worker);$i<$size;++$i)
{
echo $worker[$i];
}

You forgot '$' int echo $worker[$i];

edited Jun 06 '11 at 06:55

Jordon Bedwell

answered Jun 05 '11 at 19:39

therealszaka

@motsu: coming from c++ you find PHP as the paradise. – dynamic Jun 05 '11 at 19:47

patapizza · Answer 3 · 2011-06-05T19:51:46.983

1

You forgot the dollar sign before i in $worker[$i].

-edit-: Removed the second part, maybe I'm too tired :)

edited Jun 05 '11 at 19:51

answered Jun 05 '11 at 19:39

patapizza

3

The second part is nonsense as its return value is not used. – Gumbo Jun 05 '11 at 19:43

3 Answers3