CODEWITHSUNDEEP
phpmysql
hurnhu
hurnhu

Reputation: 936

Error produced from mysql select

I'm currently learning MySql, but ive hit this problem. the following code should just query the db for everything in the users table. but it returns this error. Error: SELECT * FROM users which helps me not at all. I am able to successfully insert an item into the database, but I am unable to select from it. I've also tried $sql = "SELECT * FROM ama.users"; my DB structure is 

 ama
 |-users

any help would be much appreciated.

  $conn = new mysqli($_ENV['OPENSHIFT_MYSQL_DB_HOST'],$_ENV['OPENSHIFT_MYSQL_DB_USERNAME'], $_ENV['OPENSHIFT_MYSQL_DB_PASSWORD'], 'ama');
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$username = "Doe";
$password = "johnexample";
$sql = "SELECT * FROM users";
if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();

Upvotes: 5

Views: 99

Answers (2)

devpro
devpro

Reputation: 16117

From the PHP Manual:

mysqli::query will return object in success and return false in failure.

So you can use it without checking data type (===):

if ($conn->query($sql)) { 
     echo "New record created successfully"; 
} else { 
     echo "Error: " . $sql . "<br>" . $conn->error; 
}

For more better understanding you can use var_dump() and check what are you getting like:

var_dump($conn->query($sql));

Upvotes: 5

warzon
warzon

Reputation: 187

Documentation says:

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

So, do something like 

$result= $db->query($sql);

and then check the rows in $result

Upvotes: 1

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