PHP - Variable inside a loop and another one outside

Question

as the title said, i've this block of code and i want to know if i can write a variable that takes the name of another one similar to it but outside the loop.

is that makes a problem ?!

<?php 
   $query = "SELECT * FROM subjects";
   $subjects_set = mysql_query($query);
   errors_of($subjects_set);

    while ($db_subjects_rows = mysql_fetch_array($subjects_set)){
           echo "<li>{$db_subjects_rows["menu_name"]}</li>";

           $pages_set = mysql_query("SELECT * FROM pages WHERE subject_id = {$db_subjects_rows["id"]}");


        errors_of($pages_set);
        echo "<ul class=\"pages\">";
        while ($db_pages_rows = mysql_fetch_array($pages_set)){
        echo "<li>{$db_pages_rows["menu_name"]}</li>";
        }
            echo "</ul>";                       
    }

?>

You can access all variables that were declared outside of the loop inside it — kero
– kero, Commented Mar 31, 2013 at 13:48
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — Madara's Ghost
– Madara's Ghost, Commented Mar 31, 2013 at 13:56

omars · Accepted Answer · 2013-03-31 13:49:18Z

2

Using local variables inside a loop does not create a problem, if thats what you are asking, you should read about variable scopes

The function has a scope, the class has a scope and there is a global scope

answered Mar 31, 2013 at 13:49

omars

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5 Comments

Kareem Over a year ago

A.Shabaan but what i'm asking for the variable inside the loop. the similarity between it and the one outside could make an error if i use the outside one in a function for example ?!

omars Over a year ago

What is the name of the variable you are talking about ?

Kareem Over a year ago

i think i get it permanently. but if u donn mind i've another question :) i've this function that i want to pass a value on it "function get_all_subjects ($all_subjects){ $query = "SELECT * FROM subjects"; $subjects_set = mysql_query($query); errors_of($subjects_set); }" and when i recall it in the page this error appear "Notice: Undefined variable: all_subjects in C:\wamp\www\test_of_test\content.php" although that in the content.php page i put the function as it get_all_subjects($all_subjects); why its undefined ?!

omars Over a year ago

The variables inside the function are read only inside the function to change the value of the variable passed in the global scope you should pass it by reference, hence function get_all_subjects(& $all_subjects){// do some coding}

Kareem Over a year ago

i sent you a message on your web page please find it out and send me back :)

Ritesh Kumar Gupta · Accepted Answer · 2013-03-31 13:53:36Z

1

All the variables that is used outside the loop can be accessed inside the loop.Refer this manual to know more about the PHP Variable Scope.

answered Mar 31, 2013 at 13:53

Ritesh Kumar Gupta

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