0
我有下面的代碼,應該從我的數據庫中選擇一行與當天的日期。SELECT where date =當前日期
$sql = "SELECT * FROM Forecast WHERE date(epoch) LIKE '". $mysqldate . "';";
echo $sql . "\n";
$result = mysqli_query($conn, $sql);
if (!$result || mysqli_num_rows($result) == 0) {
$row = mysqli_fetch_object($result);
$pop = 0;
$forecast = new DailyForecast($row->epoch, ...);
return $forecast;
}
else {
echo "no data found..\n";
}
在PHPmyAdmin中運行SELECT * FROM Forecast WHERE date(epoch) LIKE '2015-11-03'
工作正常。在我的腳本中,它返回假或無...
添加或死亡(mysqli_error($ DB));在mysqli_query的末尾,併發布錯誤。 :) –