PHP: Create dynamic (variable) file name- Undefined variable: 65

Question

I need to set a dynamic file name in PHP. So I wrote a small example script to represent the problems I am facing.

When I run the following script, I get the following erroneous output, and the file created is just named .csv while it should be named 0101.csv

OUTPUT:

Notice: Undefined variable: 65 in C:\xampp\htdocs\testsEight.php on line 5

Notice: Undefined variable: 65 in C:\xampp\htdocs\testsEight.php on line 7
Array ( [0] => BillClinton )

Why does it call the variable as 65 rather than $the_id? I am trying to follow these guidelines. In the following code, I also tried to replace it by ${$the_id}, no luck!

CODE:

<?php

$type = 'BillClinton';
$the_id = 0101;
file_put_contents ( 'C:/'.$$the_id.'.csv' , $type ."," , FILE_APPEND );

$file = fopen('C:/'.$$the_id.'.csv', 'r');
$line = fgetcsv($file);
array_pop($line);

if ($line !== FALSE) {
    //$line is an array of the csv elements in a line. The fgetcsv() function parses a line from an open file, checking for CSV fields.The fgetcsv() function stops returning on a new line, at the specified length, or at EOF, whichever comes first.
    print_r($line);//check
} else {echo 'FALSE';}

Please help me fix this.

Answer 1

You had extra $ in $$the_id which lead to call the reference of $the_id intead of variable name the_id. So you need to erase that. Code will be as follow;

<?php

$type = 'BillClinton';
$the_id = 0101;
file_put_contents ( 'C:/'.$the_id.'.csv' , $type ."," , FILE_APPEND );

$file = fopen('C:/'.$the_id.'.csv', 'r');
$line = fgetcsv($file);
array_pop($line);

if ($line !== FALSE) {
    //$line is an array of the csv elements in a line. The fgetcsv() function parses a line from an open file, checking for CSV fields.The fgetcsv() function stops returning on a new line, at the specified length, or at EOF, whichever comes first.
    print_r($line);//check
} else {echo 'FALSE';}

for more details, you can look in PHP documentation

Answer 2

You're using two $ in $$the_id and one in $the_id = 0101;

"Why does it call the variable as 65"

The leading zero is treating 0101 as an octal, so wrap it in quotes $the_id = "0101";

Answer 3

Firstly, your example is wrong. PHP will never allow you to define integer and not even "string-casted" integers as a variable name.

The only problem in your script is that you are using double dollar signs, which is a refference to $0101 (assuming the $the_id is 0101 string or integer, doesn't matter).

The simple solution is to remove your double dollar signs in:

file_put_contents ( 'C:/'.$the_id.'.csv' , $type ."," , FILE_APPEND );

$file = fopen('C:/'.$the_id.'.csv', 'r');

The idea behind this is that a variable's name can be variable. And that's how your problem rose.

$a = 'hello';
$$a = 'world';

echo $hello; // will output 'world';

Regards.