CODEWITHSUNDEEP
phpmysqlfunctionvariables
beaverjam
beaverjam

Reputation: 43

PHP Returning a string from a function to be used later on?

function foo() {
    $foo1= mysql_query();
    $db_field = mysql_fetch_assoc($foo1);
    $word= $db_field['word'];
    $lower=strtolower($word);
}
foo();
print $lower;

The function above executes a block of code with the endgame being the $lower variable, which is the lowercase of a word from a database. As the code is written above, I get an error saying that $lower is not defined.

I tried returning $lower with return $lower, but that didn't work. How do I keep a variable from a function for later use?

Upvotes: 0

Views: 41

Answers (5)

Stone Lasley
Stone Lasley

Reputation: 76

put your return statement back 

...
return $lower;
}

$result = foo();
print $result;

Upvotes: 0

kjames
kjames

Reputation: 656

You could actually use 'global' Like This.

    $lower = '';

    function foo() {

        global $lower;

        $foo1= mysql_query();
        $db_field = mysql_fetch_assoc($foo1);
        $word= $db_field['word'];
        $lower=strtolower($word);

    }

    foo();
    print $lower;

Upvotes: 0

Ranjith
Ranjith

Reputation: 2819

$lower variable used only in the function. it can't work out of the function. Do like this, 

function foo() {
    $foo1= mysql_query();
    $db_field = mysql_fetch_assoc($foo1);
    $word= $db_field['word'];
    $lower=strtolower($word);
  return $lower;
}

$lower = foo();
print $lower;

Upvotes: 0

Jenz
Jenz

Reputation: 8369

$lower is defined inside the function which have scope only within that function. So trying to access it outside the function will definitely cause proble. You have to return it in the function .Try with :

function foo() {
    $foo1= mysql_query();
    $db_field = mysql_fetch_assoc($foo1);
    $word= $db_field['word'];
    return strtolower($word);  // return the value
}
$lower=foo();   // this will contain the return value of foo()
print $lower;

Upvotes: 0

Realitätsverlust
Realitätsverlust

Reputation: 3953

First of all, return $lower at the end of the function is correct. Then you use:

$var = foo();

You just discoverd the scope of a variable. $lower is only availible inside the function, so you cant use it outside of it.

Upvotes: 1

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