php mysql+ Notice: Array to string conversion in C

Question

I keep getting this error when I try to explode the array by a comma, How to fix this error can anyone help me ???

friend_request.php

 <?php

     if(isset($_POST['acceptrequest'.$user_from]))
     {
         //select the friend array  row  from the logged in user
         $get_friend_check = mysql_query("SELECT friend_array FROM user WHERE user_name = '$login_user'") or die(mysql_error());
         $get_friend_row = mysql_fetch_assoc($get_friend_check);
         $friend_array = $get_friend_row['friend_array'];
         $friendArray_explode = explode(",", $friend_array);
         echo $friendArray_explode;
     }
     ?>

The last line of code produce this error how to fix it ??

imgs.xkcd.com/comics/exploits_of_a_mom.png

imulsion
– imulsion

2013-04-21 09:27:06 +00:00
Commented Apr 21, 2013 at 9:27 — imulsion
– imulsion, Commented Apr 21, 2013 at 9:27

Nir Alfasi · Accepted Answer · 2013-04-21 09:28:45Z

1

This is a NOTICE (not an error!) - you're trying to print an array as if it was a string. Use print_r or var_dump instead of echo:

print_r($friendArray_explode);

edited Apr 21, 2013 at 9:28

answered Apr 21, 2013 at 9:22

Nir Alfasi

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Comments

Community · Accepted Answer · 2017-05-23 11:50:15Z

0

this is because you are trying to pinrt an array as it was a string only using echo. For arrays you can use both print_r or var_dump

I would also sugeest you to stop using mysql_ api since they are depecrated, please switch to PDO or mysqli

Furthermore you are ready to mysql injection. there is a nice tutorial here which explain you everything about that -> How can I prevent SQL injection in PHP?

edited May 23, 2017 at 11:50

CommunityBot

11 silver badge

answered Apr 21, 2013 at 9:28

Fabio

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