PHP MySQL : inserting Json response values into database by iterating through loops

Question

I have a JSON response which I want to store in database. But here I am unable to loop through the values within the inner loops.

JSON Twitter Data:

Array
(
  [0] => stdClass Object
  (
    [created_at] => Thu Apr 23 05:23:39 +0000 2015
    [id_str] => 591110161239945216
    [entities] => stdClass Object
    (
      [hashtags] => Array
      (
        [0] => stdClass Object
        (
          [text] => HelloWorld
          [indices] => Array
          (
            [0] => 0
            [1] => 11
          )
        )
      )
     )
   )       
 )

Here i am trying to access [text]=>helloworld field and insert it to database. I want to iterate through loop using while. But this returns me an error if there is no hashtags posted.

Code :

i=0
while($i<3)
{
   $tweet_id     = $json[$i]->id_str;
   $created_at   = $json[$i]->created_at;
   $hashtag_text = $json[$i]->entities->hashtags[$i]->text;

   $this->sql="INSERT INTO twitter_scan('tweetid','created_at','hashtag_text').                        VALUES('$tweet_id','$created_at','$hashtag_text')";
   $this->query=mysqli_query($this->conn,$this->sql);
   $i++;
}

i have 3 tweets in my account out of which 1 tweet has a #hashtag other two are plain text. When I run my code, for the 1st iteration it works well, but for the 2nd and 3rd iteration it returns me an error PHP Notice:Undefined offset: 1 , PHP Notice:Undefined offset: 2.

This is because my other two tweets contain no hashtags and still my code is trying to access it. Here I want to store NULL value to database when there is no value for this field. How can I achieve this ?

Thank you for any help.

use foreach instead of while.

Sougata Bose
– Sougata Bose

2015-04-29 05:56:36 +00:00
Commented Apr 29, 2015 at 5:56 — Sougata Bose
– Sougata Bose, Commented Apr 29, 2015 at 5:56

Sougata Bose · Accepted Answer · 2015-04-29 05:57:40Z

3

There is only one hashtag so it will be on 0, 2nd & 3rd loop $i will be 1, 2 which is not present. Replace $i with 0 & add a check for it else store blank to $hashtag_text. Try with -

$hashtag_text = !empty($json[$i]->entities->hashtags[0]->text) ? $json[$i]->entities->hashtags[0]->text : '';

answered Apr 29, 2015 at 5:57

Sougata Bose

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