php - How to simplify if isset $var else $var = ' '; -

if want repopulate form $_post values example (there other applications problem, that's easiest) have check if $_post index set, before can use it's value, or i'll notice php.

for example:

echo '<input type="text" name="somefield" value="'; if(isset($_post['somefield']))   { echo $_post['somefield']; } echo '">'; 

with complex forms, seems cumbersome , lot of repetition. thought, let's extract function:

function varcheck_isset($vartocheck) {   if(isset($vartocheck))     {return $vartocheck;}   else     {return '';} } 

and do

echo '<input type="text" name="somefield" value="'; echo varcheck_isset($_post['somefield']); echo '">'; 

makes code nicer.

but when , $_post['somefield'] not set, says

notice: undefined index: somefield

:-(

anybody got idea or suggestions how make work?

---

edit:

here's ended doing - accepted organgepill's answer, modified slightly:

function arraycheck_isset($arraytocheck, $indextocheck) {   if(isset($arraytocheck) && is_array($arraytocheck) && array_key_exists($indextocheck, $arraytocheck))     { return $arraytocheck[$vartocheck];}   else     { return '';} } 

the comment below elclanrs pretty good. write:

echo $_post['field'] ?: ''; 

personally non-shorthand version better though, because may have cases need check other things, besides isset() - example regex. way keep consistent going through function each time.

try this:

function postedval($vartocheck) {   if(array_key_exists($vartocheck, $_post))     return $_post["$vartocheck"];   return ''; } 

your function hitting first case because checking if parameter set have been if (isset($_post[$vartocheck]))