dongmiao4733 2016-06-28 03:28
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MySQL Query使用JSON

Straight forward question - but for the life of me, can't work out why below shows NO results!!

Was working - all I did was add the WHERE...

$query="select distinct c.id, c.FirstName, c.LastName, c.EmailAddress, c.MemberStatus from details c where c.MemberStatus = 'ACTIVE' order by c.LastName";
$result = $mysqli->query($query) or die($mysqli->error.__LINE__);

$arr = array();
if($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {
    $arr[] = $row;  
}
}
# JSON-encode the response
$json_response = json_encode($arr);

// # Return the response
echo $json_response;

I have do doubt it's something ridiculously simple and I'll look like a goose...

  • 写回答

1条回答 默认 最新

  • doubo6658 2016-06-28 03:48
    关注

    Silly question but are you certain that "ACTIVE" is actually a value that exists in the c.MemberStatus column? A case issue possibly?

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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