参数上未定义的索引，无法定义

Question

Hello I receive the following error:

Notice: Undefined index: $db_field.

It's because I have not declared the parameter: $db_field - when this function is called:

if (checkInField($webmailUserName,$db_connection,$db_con_table,"userName") == false)

The fourth bit is a clear string, PHP is not like Java where your declare the variables int, long, string, etc. So I used echo $db_field and sure enough the echo onscreen shows 'userName' so I think its recognised as a string?

I have tried echo $row[$db_field]; (does not work)

I have tried echo $row['db_field']; (indicating a string using ' does not work / if I understand right)

I have tried echo $row['userName']; (this string works)

I can however check with isset() which I do not understand how to do. Other posts with my error description do not seem to match what I'm trying to achieve.

Please can someone explain to me why echo $row[] does not recognise '$db_field' as a string and what is needed to fix it. Hopefully this will help other beginners too.

I'm using PHP5.3 and cannot check any server ini's as I'm using a domain and webserver company for my script. (Another Undefined Index (this one's defined though))

function checkInField($value,$db_connection,$db_con_table, $db_field)  
// Checks a value is not within a database field
{       
    $query = "SELECT $db_field FROM $db_con_table WHERE $db_field='$value'"; 
    $result = $db_connection->query($query) or die($mysqli->error());

    // GOING THROUGH THE DATA
    if($result->num_rows > 0) 
    {
        while($row = $result->fetch_assoc()) 
        {
            echo $row['$db_field']; 
        }
    }
    else
    {
        echo 'NO RESULTS';  
    }

Answer 1

The warning (aka Notice) about Undefined index: $db_field should indicate that you're trying to perform an array-index (like Map.get() in Java) with the string $db_field. (When you typed in '$db_field', you expected it to evaluate to be userName, but it isn't.)

This is because in PHP, single-quoted strings are different than double-quoted strings. (PHP Docs)

So, instead of this:

echo $row['$db_field']; // indexing with the literal string $db_field

... use:

echo $row[$db_field]; // if $db_field is numeric, acts like List.get(int); if $db_field is a string, acts like Map.get(Object)

or

echo $row["$db_field"]; // indexing with the string-representation of the variable named $db_field

---

About forward-declaring variables: One of your method parameters is named $db_field, so it's perfectly safe to refer to the variable of that name inside your method (but not outside it). I think you assumed the Undefined... had to do with the variable itself, but that is not the case.