Best way to check if MySQL results returned in PHP?

Question

I\'m looking for the best way to check and see if any results were returned in a query. I feel like I write this part of code a lot and sometimes I get errors, and sometimes

Answer 1

Usually I use the === (triple equals) and __LINE__ , __CLASS__ to locate the error in my code:

$query=mysql_query('SELECT champ FROM table')
or die("SQL Error line  ".__LINE__ ." class ".__CLASS__." : ".mysql_error());

mysql_close();

if(mysql_num_rows($query)===0)
{
    PERFORM ACTION;
}
else
{
    while($r=mysql_fetch_row($query))
    {
          PERFORM ACTION;
    }
}

Answer 2

$connect = new mysqli('localhost', 'user', 'password', 'db');
$result = $connect->query("select * from ...");
$count=$result->num_rows;
if(empty($count)){
echo"Query returned nothing";
}
else{
echo"query returned results";
} 

Answer 3

One way to do it is to check what mysql_num_rows returns. A minimal complete example would be the following:

if ($result = mysql_query($sql) && mysql_num_rows($result) > 0) {
    // there are results in $result
} else {
    // no results
}

But it's recommended that you check the return value of mysql_query and handle it properly in the case it's false (which would be caused by an error); probably by also calling mysql_error and logging the error somewhere.

Answer 4

If you would still like to perform the action if the $result is invalid:

if(!mysql_num_rows($result))
    // Do stuff

This will account for a 0 and the false that is returned by mysql_num_rows() on failure.

Answer 5

mysqli_fetch_array() returns NULL if there is no row.

In procedural style:

if ( ! $row = mysqli_fetch_array( $result ) ) {
    ... no result ...
}
else {
    ... get the first result in $row ...
}

In Object oriented style:

if ( ! $row = $result->fetch_array() ) {
    ...
}
else {
    ... get the first result in $row ...
}