没有任何结果时，MySQLi查询给出错误

Question

I am trying to find and echo one result from my database. I am using it like below

<?php
error_reporting(E_ALL);
include_once("includes/connection.php");
$input = "OUT";
$answer = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1")->fetch_object()->answer; 
echo $answer;
?>

Its working fine when there some result but giving error like below when there no any result

Notice: Trying to get property 'answer' of non-object in C:\xampp\htdocs\new\test.php on line 5

I want handle this error using if else. I do not know much PHP and specially does not know more about query. Let me know if someone can help me for get out from this. Thanks

Answer 1

Try this:

if ($answer) {
    echo $answer;
} else {
    die('No Answer Found');
}

Hope it will work.

Answer 2

It's caused by the overuse of method chaining, why do you need to put every call in one long line?

You should check if an object is retrieved first...

$query = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1");
if ( $query && ($row = $query->fetch_object()) ) {
    $answer = $row->answer; 
    echo $answer;
}

You should also look into prepared statements as this may lead to SQL injection and other problems. - How to create a secure mysql prepared statement in php?

Answer 3

You can also use try{} catch {}

try {
    $answer = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1")->fetch_object()->answer; 
}
catch (Exception $e) {
    $errormsg = $e->getMessage();
}