PHP：使用表达式初始化数组

I have a simple (in my opinion) question. I try to initialize an array in php as follows:

array(
  'type' => 'hidden',
  'id' => "request_params$suffix",
  'name' => "request_params$suffix",
  'value' => "?rule_id=$rule_id&cur_instr=$cur_instr&dev_id=$dev_id&cmd_id=$cmd_id" + ($disabled? '&disabled' : ''))

This construction is passed as a parameter in a function call. All of the variables are defined. And as a result I get the type, id and name are initialized well, but the value is initialized with 0. If I comment out the + ($disabled? '&disabled' : '') then the value initialized too. I had tried to enclose all the expression in parenthesis with the same result -- initializing with 0.

Has anybody any idea?

You are using arithmetic operator for concatination of two strings i.e. + instead of .

array(
  'type' => 'hidden',
  'id' => "request_params$suffix",
  'name' => "request_params$suffix",
  'value' => "?rule_id=$rule_id&cur_instr=$cur_instr&dev_id=$dev_id&cmd_id=$cmd_id" . ($disabled? '&disabled' : ''))

Because you are using+ its trying to add numbers and because it is unable to find that its assuming both variables as 0 so 0+0=0.

To concatenate strings use the dot

array(
  'type' => 'hidden',
  'id' => "request_params$suffix",
  'name' => "request_params$suffix",
  'value' => "?rule_id=$rule_id&cur_instr=$cur_instr&dev_id=$dev_id&cmd_id=$cmd_id" . ($disabled? '&disabled' : ''))

I think that code will help you alot...

$disabled='';
$suffix='suffix';
$array=array(
  'type' => 'hidden',
  'id' => 'request_params'.$suffix,
  'name' => 'request_params'.$suffix,
  'value' => '?rule_id=$rule_id&cur_instr=$cur_instr&dev_id=$dev_id&cmd_id=$cmd_id'.($disabled? '&disabled' : '')
);

var_dump($array);