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Fatal error: Only variables can be passed by reference

<?php
function readParams( $source ) {
    $params = array();
    $fh = fopen( $source, 'r' ) or die("problem with $source");
    while ( ! feof( $fh ) ) {
        $line = trim( fgets( $fh ) );
        if ( ! preg_match( "/:/", $line ) ) {
            continue;
        }
        list( $key, $val ) = explode( ':', $line );
        if ( ! empty( $key ) ) {
            $params[$key]=$val;
        }
    }
    fclose( $fh );
    //$params.='extra';
    $params=array_push(4);
    return $params;
}

function writeParams( $params, $source ) {
    $fh = fopen( $source, 'w' ) or die("problem with $source");
    foreach ( $params as $key=>$val ) {
        fputs( $fh, "$key:$val\n" );
    }
    fclose( $fh );
}


$file = "./params.txt"; 
$array['key1'] = "val1j";
$array['key2'] = "val2j";
$array['key3'] = "val3j";
writeParams( $array, $file );
$output = readParams( $file );
print_r( $output ); 


?>

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in order to learn (I think)
I am trying to add to the $params variable

I tried adding a string but there was array to string error

so now I am trying to push a value to the end of array


Fatal error: Only variables can be passed by reference in C:\wamp\www\POPP-edition4-code\9781430260318_Chapter_06_Code\listing06.01.php on line 17

and what does this error mean
what is difference pass by value, pass by reference
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rgb192
Asked:
rgb192
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2 Solutions
 
Ray PaseurCommented:
On line 17, the number 4 is not a variable, hence the fatal error.

On any occasion that you encounter a function that you're not 100% sure you understand, you should look up the function definition on PHP.net.  Example:
http://php.net/manual/en/function.array-push.php

If you read the man page and still do not understand how to write the statement on line 17, please post back and I'll be glad to help.

If you want to learn the difference between variables and references, this article explains it.  But I think the misunderstanding is really about how (or whether) to call array_push().
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_12310-PHP-Variables-and-References.html
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rgb192Author Commented:
not only did I read
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_12310-PHP-Variables-and-References.html

but I also ran all your code samples in my ide


but I do not understand the php error (how did I pass by reference???)

$stack = array("orange", "banana");
array_push($stack, "apple", "raspberry");
print_r($stack);


//$params = array("orange", "banana");
array_push($params, "apple", "raspberry");
//print_r($stack);


Array ( [key1] => val1j [key2] => val2j [key3] => val3j [0] => apple [1] => raspberry )

this does not add to the keys
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Ray PaseurCommented:
I think the error occurred on this line, according to the line number in the error message:

$params=array_push(4);

That does not make sense because the function expects array &$array for the first argument of the function.
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SurranoSystem EngineerCommented:
Try this:

    array_push($params, 4);

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SurranoSystem EngineerCommented:
in the fruity example, do you expect that new elements are created with keys "key4" and "key5"? Well, that won't work. What array_push does is to take the lowest non-negative integer that doesn't exist in the array yet, and uses it as a key. You'll have to implement your own function to achieve some custom numbering.
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rgb192Author Commented:
Array ( [key1] => val1j [key2] => val2j [key3] => val3j [0] => 4 )

array_push($params, 4);
added
key 0, value 4

 lowest non-negative integer that doesn't exist in the array yet, and uses it as a key
so key=0



That does not make sense because the function expects array &$array for the first argument of the function.
variable is $array
reference is &$array

I do not understand a practical example
array_push($params, 4);

$four=4;
array_push($params, &$four);
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SurranoSystem EngineerCommented:
No, this ref thing is in the definition of the array_push function. When you call it, you don't use the "&" sign.  Also, only the first parameter (the array) is what must be a reference.

To understand reference: In C analogy, a left-value, i.e. whatever may appear on the left side of an assignment operator. This makes sense:
$four=3+1;

This doesn't:
4=3+1;

Since array_push wants to modify the array you specify as the first parameter, you may invoke it like this:

array_push($myarray, 1);

but you can't use a literal:

array_push(array(11,22,33), 44);

Okay, I could imagine it to run without errors, but makes no sense and can't test it right now.
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Ray PaseurCommented:
Let's not overcomplicate...

In the original code snippet, we found a fatal error on line 17 where it said this:

$params=array_push(4);

Please step back from the technical details and just tell us in plain language what you want to accomplish when you wrote that instruction.  For example, "I wanted to put the number 4 into the first (or last) position of the $params array."  Or "I wanted to add 4 to one of the values in the $params array."  Or whatever explains why you wrote the instruction.  Once we understand that, we can show you the correct PHP statement to get it done.

Thanks, ~Ray
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rgb192Author Commented:
I am getting
Array ( [key1] => val1j [key2] => val2j [key3] => val3j [0] => 4 )

but output I would like is
Array ( [key1] => val1j [key2] => val2j [key3] => val3j [key4] => 'secret' )

When you call it, you don't use the "&" sign.  Also, only the first parameter (the array) is what must be a reference.
so the array is the reference and the second parameter is the value which is the value to be added to array

array_push($myarray, 1);
$myarray is reference
1 is value


is this correct?
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Ray PaseurCommented:
You would get the desired result with an assignment statement, something like this:

$myarray['key4'] = 'secret';

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SurranoSystem EngineerCommented:
The catch here is that you probably want to generate the key.

How about this?

$myarray['key'.(count($myarray)+1)] = 'secret';

Open in new window


But I still feel like the real question being, why do you want to enumerate the keys with anything other than 0..(n-1)? Are these keys coming from somewhere? (e.g. from a database query) In that case you should get the next key from an iteration (e.g. fetch row #n from database query above)
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rgb192Author Commented:
okay, this can not be done with push

Are these keys coming from somewhere?
no this is a book php code example

thanks
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