php syntax - why does making my code a function stop it working?

Question

Probably a simple one so please excuse me.

At the top of my page I include a file:

<?
    require("common.php");
    ?>

Later on in my code I do the following:

  $stmt = $db->prepare($query);
  $result = $stmt->execute();
  $result = $stmt->fetchAll();

(the db connection files being in my include file).

This works fine, until I try and make the code a function For example: If I do

    function doAQuery(){
 $stmt = $db->prepare($query);
                    $result = $stmt->execute();
                    $result = $stmt->fetchAll();
}

It can't find the db details. Please can someone explain why?

There is no `$query` available in your function. (And PHP would have told you that, if you only had set your error reporting to a sensible value – so please go do that _now_!) [Basics, Baby!](http://www.php.net/manual/en/language.variables.scope.php) — CBroe, Nov 13 '13 at 11:22
Because `$db` is not in the scope of your function. Use `global` or make a DB class that does this for you while keeping `$db` as a local class variable. — h2ooooooo, Nov 13 '13 at 11:23

score 2 · Answer 1 · answered Nov 13 '13 at 11:23

2

It's not a syntax issue. Your $db and $query variables are not available to your function scope. You can make them available by declaring global $db, $query as the first line of your function.

answered Nov 13 '13 at 11:23

Tim

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This is horrible practice, you should be using dependency injection. – cryptic ツ Nov 13 '13 at 11:25
2

the question asked why it doesn't work?, not how can I be a better programer? – Tim Nov 13 '13 at 11:26
I agree with Tim, he answered the question correctly. Of course there are better design patterns than this, but that's a way more broader question. – Guilherme Sehn Nov 13 '13 at 11:30
I am of course interested in best practice for this. Any advice will be taken gratefully. – Programatt Nov 13 '13 at 11:35
2

@Programatt see http://stackoverflow.com/a/5166527/1592648 don't use `global` use dependency injection instead. – cryptic ツ Nov 13 '13 at 11:40

score 2 · Answer 2 · answered Nov 13 '13 at 11:24

It's because $db is a global variable. PHP ignores it in the function scope unless you explicity declare it should be used.

function doAQuery() {
    global $db;
    $stmt = $db->prepare($query);
    $result = $stmt->execute();
    $result = $stmt->fetchAll();
}

Take a look at how variables scope works in PHP.

score 0 · Answer 3 · answered Nov 13 '13 at 11:24

0

Pass Query string to function,

function doAQuery($query){

answered Nov 13 '13 at 11:24

Krish R

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Please read the question. – Programatt Nov 13 '13 at 11:24

php syntax - why does making my code a function stop it working?

3 Answers3