php - show the mysql Select query results -


i using php , mysql wanted bring results of select statement.

i tried use code implementation

<head> <meta http-equiv = "content-type" content = "text/html" charset = "utf-8"> </head> <?php $conn = mysqli_connect("hostname","user","password","dbname"); // check connection if (mysqli_connect_errno()){ echo "mysql error : " . mysqli_connect_error(); } $result = mysqli_query($conn,"select kind,abs(sum(money)) money account userid = 't@t.t' , date_format(adate,'%y/%c') ='2014/8'  , checkio  = 'out'  group kind;"); echo "<table border='1'> <tr> <th>no</th> <th>kind</th> <th>money</th> </tr>"; $no = 1;  while($row = mysqli_fetch_array($result)){ echo "<tr>"; echo "<td>" . $no . "</td>"; echo "<td>" . $row['kind'] . "</td>"; echo "<td>" . $row['money'] . "</td>"; echo "</tr>"; $no++;  } echo "</table>"; mysqli_close($conn); ?>

no data not come result. this

no  kind  money

enter on real mysql select statement

select kind,abs(sum(money)) money account userid = 't@t.t' ,   date_format(adate,'%y/%c') ='2014/8'  , checkio  = 'out'  group kind;

as result

kind money     1000  b   2000  c   1500

come way ...

why don't have output of code?

  need change in way?

no need change thing in code replace $row = mysqli_fetch_array($result) $row = mysqli_fetch_array($result, mysqli_assoc); fine.


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