mysqli returns empty result - how to display

Question

I have the following PHP which basically gets the result of a MySQL query:

$q2 = "SELECT FIELD FROM TABLENAME WHERE ID = 1;";
$con = new mysqli($server, $user, $pass);
if (mysqli_connect_errno()) 
{
    $error = mysqli_connect_error();
    exit();
}
else
{   
    $res = mysqli_query($con, $q2);
    if ($res)
    {   
        while($row = mysqli_fetch_assoc($res))
        {
            PRINT "THERE WAS A RESULT HERE: "; 
        }
    }
    else
    {
        $error = mysqli_error();
        exit();
    }
    mysqli_free_result($res);   
};
mysqli_close($con);

But on occasions, it will return an empty value. This is valid, based on how the parent application works, but how can I detect that empty row and return "THIS WAS AN EMPTY ROW: "?

Please define "empty row". Does it mean that no row has been returned? Or that inside one row the *FIELD* was `NULL` or an empty string (`""`)? — hakre, Oct 29 '12 at 16:24

score 5 · Accepted Answer · answered Oct 29 '12 at 16:18

5

You might want to look up mysqli_num_rows() in the PHP manual. It lets you see how many rows are in a result set generated by the previous query. You can use the row count to determine whether to display results or the "no matching results" message.

answered Oct 29 '12 at 16:18

GordonM

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score 2 · Answer 2 · answered Oct 29 '12 at 17:02

I hope this helps.

if ($res)
{   
    if($res->num_rows) {
        while($row = mysqli_fetch_assoc($res))
        {
           PRINT "THERE WAS A RESULT HERE: "; 
        }
    }
    else {
          PRINT "THERE WAS A EMPTY ROW: "; 
    }
}

Refer: php.net

score 0 · Answer 3 · answered Oct 29 '12 at 17:19

Thanks for that I ended up with:

$q2 = "SELECT FIELD FROM TABLENAME WHERE ID = 1;";
$con = new mysqli($server, $user, $pass);
if (mysqli_connect_errno()) 
{
    $error = mysqli_connect_error();
    exit();
}
else
{   
$res = mysqli_query($con, $q2);

$row_cnt = mysqli_num_rows($res);

if ($row_cnt == 0)
{
        PRINT "THERE WAS NO RESULT: "; 
 }

else
{    
if ($res)
{   
    while($row = mysqli_fetch_assoc($res))
    {
        PRINT "THERE WAS A RESULT HERE: "; 
    }
}
else
{
    $error = mysqli_error();
    exit();
}
mysqli_free_result($res);   
}
};
 mysqli_close($con);

score 0 · Answer 4 · edited May 25 '15 at 11:56

My php function query handle:

private function query($sql_query){

        $result = $this->connection->query($sql_query);

        if(!$result){           //query fail
            throw new Exception($this->connection->error.$sql_query);
        }else {//                 "SUCCESS";
            if(!$result->num_rows)
            {
                throw new Exception("THERE WAS NO RESULT: "); 
             }
            for ($res = array(); $tmp = $result->fetch_array(MYSQLI_BOTH);){ 
                $res[] = $tmp;                
            }
           return $res;
        }
    }

mysqli returns empty result - how to display

4 Answers4

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