PHP: Trying to get property of non-object

Question

I am working on a function, what returns whether a table exists or not.

But it always notices:

Notice: Trying to get property of non-object [...] on line 10

in

1  function table_exists($table) {
2      
3      // get the database
4      global $mysqli;
5      
6      // look for tables named $table
7      $result = $mysqli->query("SHOW TABLES LIKE $table");
8      
9      // if the result has more than 0 rows
10     if($result->num_rows > 0) {
11         return true;
12     } else {
13         return false;
14     }
15 }

the $mysqli var is set like this:

$mysqli = new mysqli(mysqli_host, mysqli_user, mysqli_password, mysqli_database);

How to solve that?

Why do you want such a function? Don't you know your tables already? — Your Common Sense, Sep 03 '13 at 20:15
Do you really want to see if a table exists? or do you mean whether the table has data in it? Using num rows is not the way to see if the table exists — tommyd456, Sep 03 '13 at 20:18
you can change "if($result->num_rows > 0)" to "if($result && $result->num_rows > 0)" — Scott, Sep 03 '13 at 20:18
it should be something like `show tables like '$table'` (you need the single quotes). You need to make sure you sanitize and property escape $table variable as well. — Orangepill, Sep 03 '13 at 20:28
after debugging the code, @Orangepill is correct. You are missing the single quotes around table. And as he said, make sure to sanitize input $table. — Scott, Sep 03 '13 at 20:30
@YourCommonSense i am making a user system and every user gets an own table :) — Julius Rickert, Sep 05 '13 at 12:41
@tommyd456 yes, i want to check whether a table EXISTS. (look at my previous comment — Julius Rickert, Sep 05 '13 at 12:42
@JuliusRickert this is what you are doing wrong. Awfully wrong. — Your Common Sense, Sep 05 '13 at 12:44
@YourCommonSense i now know it's wrong. i now have one table for all users — Julius Rickert, Sep 12 '13 at 21:19

Flavio · Answer 1 · 2013-09-03T20:35:39.437

1

Your SQL syntax is wrong. Check the value of your variable $table. You should have something like

SHOW TABLES LIKE "%"

edited Sep 03 '13 at 20:35

answered Sep 03 '13 at 20:19

Flavio

846
1
9
21

I'm not the downvoter but % is not a regular expression and this code will return all the tables not just the one that the user is look for. – Orangepill Sep 03 '13 at 20:30
Well sure, I could have used "." or "*" for what matters... It's just a wildcard I've used instead of $table... If $table=% he will get the list of all tables, if $table=whatever he will get just that table. Ok nvm... – Flavio Sep 03 '13 at 20:35

score 0 · Accepted Answer · answered Sep 05 '13 at 12:47

I left out the quotes.

$result = $mysqli->query("SHOW TABLES LIKE \"$table\"");

or

$result = $mysqli->query("SHOW TABLES LIKE '$table'");

or

$result = $mysqli->query("SHOW TABLES LIKE \"" . $table . "\"");

or

$result = $mysqli->query("SHOW TABLES LIKE '" . $table . "'");

thanks for your help.

PHP: Trying to get property of non-object

2 Answers2