CODEWITHSUNDEEP
php
Javacadabra
Javacadabra

Reputation: 5758

Trying to wrap a php variable in single quotes

I am trying to wrap a variable in single quotes in PHP. This is my code: 

$my_post = array(
            'post_title' => $orderRequestCode.' - ' . $customerName,
            'post_content' => $orderContent,
            'post_status' => 'draft',
            'post_author' => "'".$user_ID."'",
            'post_type' => 'orders'
            );

Currently var_dump($my_post) outputs:

array (size=5)
  'post_title' => string '2014-06-13-15-13-52 - xxxxxxx' (length=35)
  'post_content' => string 'Order Code: 2014-06-13-15-13-52
   Customer Name: xxxxxxxxxxxx
   Customer Email: [email protected]
   Order Items: 
   Stock Code: Q20-50-6101 Quantity: 12
   Comments: 
' (length=162)
  'post_status' => string 'draft' (length=5)
  'post_author' => string ''1'' (length=3)     <--------------- should be '1'
  'post_type' => string 'orders' (length=6)

This line:

'post_author' => string ''1'' (length=3)

Needs to be:

'post_author' => string '1' (length=3)

Upvotes: 2

Views: 6387

Answers (2)

Marc B
Marc B

Reputation: 360672

No, you're telling PHP to create a 3-character string:

$x = 1;

$y = "'" . $x . "'";
$y = ' 1 '
     1 2 3

Since it's a string, it will get wrapped by OTHER ' as well when you do your dump. If your ID is an integer, and you want it to be treated as a string, then you could something as simple as:

 'post_author' => (string)$user_ID // cast to string
or
 'post_author' => '' . $user_ID; // concatentate with empty string

Upvotes: 3

Ray
Ray

Reputation: 41428

The outside quotes in your var_dump aren't really part of the string, hence the length being 3, not 5. If you echo your string it will be '1'

Upvotes: 4

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