$sql = "SELECT lnumber,violation,datetime FROM violators WHERE lnumber='".$lnumber."'";
选择从我的数据库三列,并将结果解析成JSONArray和我得到这个结果
{"lnumber":"2","violation":"Beating the red light","datetime":"2017-10-15 13:02:34"}
现在我想做的是如何解析“datetime”字段没有秒?
下面是完整的代码:
$sql = "SELECT lnumber,violation,datetime FROM violators WHERE lnumber='".$lnumber."'";
$stmt = $con->prepare($sql);
$stmt->execute();
$stmt->bind_result($lnumber, $violation, $datetime);
while($stmt->fetch())
{
$temp = [
'lnumber'=>$lnumber,
'violation'=>$violation,
'datetime'=>$datetime
];
array_push($result, $temp);
}
echo json_encode($result);
如何实现这一目标?通过做这个? '$ sql =“SELECT lnumber,violation,DATE_FORMAT(”2017-10-15 13:52:35“,”%Y-%m-%d%H:%i“)违规者WHERE lnumber ='”。$ lnumber。“'”;'?这是正确的吗? –
是的,但用日期时间表格列替换日期文本,如下所示: DATE_FORMAT(datetime,“%Y-%m-%d%H:%i”) – Keith