This is a PHP error that will happen if you attempt to use a scalar value as an array.
In PHP, a scalar value is one unit of data. In other words, it is a variable that contains a string, an integer, or a float value.
More often than not, this warning happens because the developer has forgotten about the existence of a previously-declared scalar variable.
$var = 1; //a lot of code goes here $var['name'] = 'John';
As you can see, I completely forgot about the fact that I already had an integer variable called $var.
If I were to run the code above, my script would display an unappealing “Cannot use a scalar” warning.
It is extremely important to note that the code above will not work. In other words, $var will remain as an integer. The data that I tried to add to the “array” will be lost.
If you’re unsure about whether a certain variable is a scalar value or not, then you can use the is_scalar function like so:
$test = 1; $isScalar = is_scalar($test); var_dump($isScalar);
If you run the code above, you’ll find that it returns a boolean TRUE value.
However, if you run the following:
$test = array(1, 2, 3); $isScalar = is_scalar($test); var_dump($isScalar);
You will see that the result is FALSE.
It is also worth noting that null and non-existent variables will also return FALSE.