Variable in array value

[webdev007]

Is there a way to achieve this
(Which of course does not work!)
I try to set a var in an array 'value'=>
for example:
array('value'=> '/aa/$my_var/cc/', 'text' => 'My TEST',)'
and more ......
);
so how can I write an array in order to include such a var in the array?

I do not figure my way around that litle problem

thanks

Regards

 

[hos2]

$test=array()

$test['value']="aa" . $my_var . "cc";
$test['text']="My TEST";

and to put the values in an var with the corresponding value you do

while ( list($field, $value) = each ($test))
{
$$field=$value;
}

something like this ??

 

[webdev007]

Absolutely!

Thank you for your swift answer

that will do fine.

 

[hos2]

you're welcome

 

[webdev007]

I came with the following
but I cannot find a way to make this next line working:
$lib_dir_2['value']=($lib_dir),
as is ($lib_dir), does not get its value.


using the variable variable structure

// Disregarding beginning of script
// query is fine and verified

while($data = mysql_fetch_array($result)){

$lib_dir=$data['lib_dir'];
} // echoing $lib_dir works fine.

$cfg['ilibs_dir']=array(
$lib_dir_2['value']=($lib_dir),// no data passing in $lib_dir
$lib_dir_2['text']=(' "My TEST" ')
);
while ( list($field, $value) = each ($lib_dir_2))
{
$$field=$value;
}

array was never my forte; understatement!

Thanks

Regards

 

[hos2]

$lib_dir_2['value']=$lib_dir perhaps ??

 

[webdev007]

Hello hos2,
Well I tried a whole bunch of variations
and still cannot get the value out of the $lib_dir
in:
$lib_dir_2['value']=($lib_dir),// no data passing in $lib_dir

your suggestion does not work for $lib_dir could not hang without surrounding parentheses or comma or something else?

I guess the structure of the array has an error
(not only that line but possibly the array construction?)
but cannot figure it out.
thanks