Peter Ford
2014-09-03 09:23:34 UTC
I came across a bug in some of my code, where I do something like the following:
$obj = new MyObject();
foreach ($obj as $key=>$value)
{
echo $key;
}
Note that I don't use $value in the loop - I'm only doing processing based on
the field name.
Now my IDE flags this unused variable as a hint, so (on autopilot) I changed it
to be
$obj = new MyObject();
foreach (array_keys($obj) as $key)
{
echo $key;
}
Which removes the hint, and looks like the right sort of thing.
Of course this doesn't work: "array_keys() expects parameter 1 to be an array,
object found".
So what is the correct way to list the field names of an object?
Cheers
Pete
--
Peter Ford phone: 01580 893333 fax: 01580 893399
Justcroft International Ltd. www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17 1XS
$obj = new MyObject();
foreach ($obj as $key=>$value)
{
echo $key;
}
Note that I don't use $value in the loop - I'm only doing processing based on
the field name.
Now my IDE flags this unused variable as a hint, so (on autopilot) I changed it
to be
$obj = new MyObject();
foreach (array_keys($obj) as $key)
{
echo $key;
}
Which removes the hint, and looks like the right sort of thing.
Of course this doesn't work: "array_keys() expects parameter 1 to be an array,
object found".
So what is the correct way to list the field names of an object?
Cheers
Pete
--
Peter Ford phone: 01580 893333 fax: 01580 893399
Justcroft International Ltd. www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17 1XS
--
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