PHP check if record exist in SQL

Question

I am trying to check if record exist in SQL and get a return -> True or False. It returns a notice: Notice: Trying to get property of non-object in.... if($result->num_rows > 0)

$connection = new mysqli('localhost', 'user', 'pass','db_name');
$query = "SELECT * FROM order WHERE telephone = '".$telephone."' AND order_status_id='0' ";

$result = $connection->query($query);
            if($result->num_rows > 0) {
                    echo 'found'; // The record(s) do exist
                }
            else{
                    echo 'not found'; // No record found
                }

    $connection->close();

check first `$result` value. use `print_r($result)` to check the value. — Yash, Mar 30 '16 at 10:16
You query fails because `order` is `reserved keyword` in mysql it must be in backtick https://dev.mysql.com/doc/refman/5.7/en/keywords.html — Saty, Mar 30 '16 at 10:18
Apart from using the reserved word, why are you closing the connection at the end of your script? — Mjh, Mar 30 '16 at 10:41

score 2 · Answer 1 · edited May 23 '17 at 11:45

2

ORDER is a mysql reserved word. Using it as field name, table name or whatever except ORDER BY fieldname requires using backticks:

SELECT * FROM `order` WHERE ....

Otherwise, you will get a query error, which will turn $result into boolean false.

Also, your code is vulnerable to sql-injection. I advise you to use prepared statements.

edited May 23 '17 at 11:45

Community

1
1

answered Mar 30 '16 at 10:18

u_mulder

54,101
5
48
64

1

As a small tip to add, if you want to still use "order" as tablename but don't want to "escape" it each time, you can follow name convention and user plural : "orders" – Robin Choffardet Mar 30 '16 at 10:24

score 0 · Answer 2 · answered Mar 30 '16 at 10:48

0

You are checking the rows of the query object. The query object does not contain the rows.

What you want to do is something like

$data = $result->fetch_array();
if ($data->num_rows > 0)
{
    //Rows exist
} else {
    //No rows exist
}

answered Mar 30 '16 at 10:48

Petter Pettersson

377
7
20

Ashwani Goyal · Answer 3 · 2016-03-30T10:38:04.697

-2

It happens in case query do not injects object in your $result variable. You can make the following changes to proceed.

if(is_object($result) && $result->num_rows > 0) {
     echo 'found'; // The record(s) do exist
} else{
     echo 'not found'; // No record found
}

OR

if(!empty($result->num_rows)) {
         echo 'found'; // The record(s) do exist
    } else{
         echo 'not found'; // No record found
    }

Also keep the thing simple:

$query = "SELECT * FROM `order` WHERE telephone = '$telephone' AND order_status_id='0' ";

edited Mar 30 '16 at 10:38

answered Mar 30 '16 at 10:16

Ashwani Goyal

616
4
18

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Your query fails because order is reserved keyword and must be in backtick – Saty Mar 30 '16 at 10:24
Great! You are great! No error, but returns only result of ELSE. Why? – Mar 30 '16 at 10:48
Are you sure your `$result` gets the expected results? I do not think so, please share your `$result` contents? – Ashwani Goyal Mar 30 '16 at 10:58

Gorakh Yadav · Answer 4 · 2016-03-30T10:48:31.160

-3

Use like this

 $sqlReq = " SELECT * FROM order WHERE telephone = '".$telephone."' AND order_status_id='0' ";

                $queryReq = mysql_query($sqlReq) or die(mysql_error());

                if(mysql_num_rows($queryReq) > 0)
                {
                    echo "found";
                 }else{
                     echo "not found";
                       }

edited Mar 30 '16 at 10:48

answered Mar 30 '16 at 10:34

Gorakh Yadav

304
5
19

PHP check if record exist in SQL

4 Answers4