PHP/Mysql offset error

Question

I am making a simple blog in php and getting this error -

Illegal string offset 'name' in C:\xampp\htdocs\simple\index.php on line 28 and 29

  <?php
    foreach ( $posts as $post ) {
    if ( ! category_exists('name', $post ['name'] ) )  {   //line 28
                $post['name'] = 'Uncategorised';   
                }
                ?>

I know the string offset questions have been asked here a lot but still I could not figure out how to fix this one. Please help.

Here is the function category_exists

 function category_exists($field,$value)
 {
  $field=mysql_real_escape_string($field);  
  $value=mysql_real_escape_string($value);  

  $query=mysql_query("SELECT COUNT(1) FROM categories WHERE {$field}= '{$value}'");
   echo mysql_error();

    return (mysql_result($query,0)=='0')? false : true;


   }

This part of code hopefully will help you understand the $posts variable:

$posts=get_posts();

  function get_posts($id=null,$cat_id=null){
   $posts=array();
   $query=mysql_query("SELECT posts.id AS post_id, categories.id AS      category_id, title, contents, date_posted, categories.name FROM posts INNER JOIN categories ON categories.id = posts.cat_id ORDER BY post_id DESC");
   while($row=mysql_fetch_assoc($query))
   {
   $posts=$row; 
   }
   return $posts;
   }

I have the following attributes of posts table -
id,title,cat_id,contents,date_posted.
For the categories table:
id and name.

show the full code. according to yours $posts is undefined and u should at first get another error like 'invalid ..somt.. for foreach' — M0rtiis
– M0rtiis, Commented Sep 13, 2015 at 4:36
It looks like there is a space between $post and ['name'], could that be causing issues? — Shane A. Darr
– Shane A. Darr, Commented Sep 13, 2015 at 4:37
its not solve your problem. but you certainly shoult replace $posts=$row; whith $posts[]=$row; in last code-block — M0rtiis
– M0rtiis, Commented Sep 13, 2015 at 5:06

zedfoxus · Accepted Answer · 2015-09-13 04:59:30Z

2

Your variable $post probably is not an array. Try this first:

foreach ( $posts as $post ) {
    var_dump($post);
}

If you see that it is an array, great! Use it as an array. It is very likely that $post will be a string. Therefore when you try to use $post['name'], PHP tells you that 'name' is not an offset for $post string. Had you used $post[0], that'd give you the first character and 0 would be legal offset...but 'name' wouldn't be a legal offset.

answered Sep 13, 2015 at 4:59

zedfoxus

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2 Comments

user43052 Over a year ago

I tried your way and It's giving me this output -string(1) "3" string(1) "1" string(5) "first" string(5) "fdsdf" //which means its an array

zedfoxus Over a year ago

No, it's not an array. Each of those 4 items is a string. string(1) means string of length 1. "first" is a string of 5 character length and therefore it shows string(5).

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