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up vote 0 down vote favorite

I have a function, and it returns an array.

class myarray
{
 public function getAr()
   {
      // mysql query
      while($dd= $database->fetch(PDO::FETCH_ASSOC))
           {
               $data[] = $dd; //there's values in the array
           }
               return $data;
   }
 public function get3()
   {
      // mysl query
      while($dd= $database->fetch(PDO::FETCH_ASSOC))
           {
               $data[] = $dd; //there's values in the array
           }
               return $data;
   }

}

How come I tried to merge together the array:

$get = new myarray();
$arrayAr = $get->getAr();
$array3 = $get->get3();
$new_array = array_merge($arrayAr ,$array3);

It says that its not an array? 

array_merge() [function.array-merge]: Argument #1 is not an array

But I can print_r($arrayAr); and its like an array?

Help?

Thanks

share|improve this question
    
I don't think you'd really name your class array - that could cause some confusion... Is this the code you are using or some pseudo code to simplify the problem? –  Lix Sep 3 '12 at 5:35
    
$d gets out of nowhere too. –  Shikiryu Sep 3 '12 at 5:36
1  
I'm not even sure how you could run this the class array should give you a PHP Parse error. But besides that your code returns an array containing NULL two times. Which is to be expected since $d, like @Shikiryu stated, comes out of nothing. –  Michael Sep 3 '12 at 5:40

2 Answers 2

up vote 1 down vote

array is the name of a builtin; you must change the name of your class. Additionally, you should define $d (live demo):

<?php
class MyArray {
  public function getAr() {
    $d = 42;
    $data = array($d);
    return $data;
  }
  public function get3() {
    $d = 43;
    $data = array($d);
    return $data;
  }
}
$get = new MyArray();
$arrayAr = $get->getAr();
$array3 = $get->get3();
var_export(array_merge($arrayAr ,$array3)); // [0 => 42, 1 => 43]

In response to the edited version: What if the query doesn't return anything? And note you $database is undefined. Moreover, it is not actually a database, but a PDOStatement object. Assuming that it is in fact a valid PDO object, you should simply use fetchAll:

class MyArray {
  protected $statement;
  public function query($pdo, $sql) {
    $this->statement = $pdo->query($sql);
  }

  // Don't forget to call query before this method.
  public function getAr() {
    return $this->statement->fetchAll(PDO::FETCH_ASSOC);
  }
}
share|improve this answer
    
I need to have $data[] instead of array($d); there's supposed to have populated values in there, I just left it blank for simplicity –  andrewliu Sep 3 '12 at 5:51
    
@andrewliu Well, I can't regard code or requirements you haven't included in the question. But sure, if you define $data right after $d, you can use $data[] = $d;. –  phihag Sep 3 '12 at 5:54
    
I edited my OP, I wonder if it makes sense what my problem is now? –  andrewliu Sep 3 '12 at 5:56
    
@andrewliu In the future, please don't make drastic edits to your question. That will leave all future visitors confused. Instead, simply ask a new question (and maybe link to it in the comments or so). Your problem is unrelated to array_merge. Instead, your query either returns an empty result or $database is not, well, a database (calling fetch strongly suggests so). I have amended this answer. –  phihag Sep 3 '12 at 7:50
up vote 0 down vote

It seems that the problem is that you are using a saved name for your class. You really shouldn't use keywords with other meanings in the language as variable names. Just like you wouldn't call a variable array_merge because there is already a function called that...

This code seems to work for me - 

class customArray{
  public function getAr(){
    $data[] = 'Stack';
    return $data;
  }

  public function get3(){
    $data[] = 'Overflow';
    return $data;
  } 
}

$get = new customArray();
$arrayAr = $get->getAr();
$array3 = $get->get3();
print_r(array_merge($arrayAr ,$array3))

Outputs - 

Array
(
    [0] => Stack
    [1] => Overflow
)

However if I use the class name array, I'll get this error -

PHP Parse error: syntax error, unexpected T_ARRAY, expecting T_STRING...

share|improve this answer
    
sorry, I guess I should've used other pseudo code names... –  andrewliu Sep 3 '12 at 5:50
    
I edited my OP, if it makes sense now? –  andrewliu Sep 3 '12 at 5:57

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