PHP Get variable value out of Object

Question

I got following invalid code: (e.g. $column->Field == 'email')

echo $row[$column->Field];

With the Error:

Fatal error: Cannot use object of type stdClass as array

Thats the var_dump of $row:

object(stdClass)[17]
   public 'id' => string '1' (length=1)
   public 'email' => string 'master' (length=9)
   public 'Name' => string 'THE MASTER' (length=28)
   public 'reply' => string '1' (length=1)

I now what the error means i just can'T figure out how to work around it (i Might be too tired)

Im looking for something like that: What is the correct/working way to do so?

echo $row->$column->Field;

IDK how i didnt got to that earlier but i just defined a variable before hand

$field = $column->Field
echo $row->$field;

if you want to do it in one line, try $row->{$column->Field} — Abdelrhman Adel
– Abdelrhman Adel, Commented Jul 5, 2015 at 9:14
The error message is pretty clear: $row is not an array (it is an object), yet you try to access it like an array ($row[$column->Field]). That is not possible, it is invalid code. — arkascha
– arkascha, Commented Jul 5, 2015 at 9:16
The rest of the question is inclear. What is Field meant to be? Do you mean that as a placeholder for on of the property names? You don't say so... — arkascha
– arkascha, Commented Jul 5, 2015 at 9:19

j_s_stack · Accepted Answer · 2015-07-05 09:54:34Z

2

So 2 Solutios to this one:

1) Define a Variable:

$field = $column->Field;
echo $row->$field;

2) Credit to Abdo Adel:

If you want to do it in one line, try
$row->{$column->Field} 

answered Jul 5, 2015 at 9:54

j_s_stack

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