JSON输出使用PHP ajax返回null
当我从数据库查看记录时,结果显示为JSON输出使用PHP ajax返回null,php,ajax,Php,Ajax,当我从数据库查看记录时,结果显示为null 这就是我迄今为止所尝试的: <?php include("db.php"); $fetchqry = "SELECT mid, planid, paid_date, expire_date FROM payment"; $result = mysqli_query($conn, $fetchqry); $num = mysqli_num_rows($result);
null
这就是我迄今为止所尝试的:
<?php
include("db.php");
$fetchqry = "SELECT mid, planid, paid_date, expire_date FROM payment";
$result = mysqli_query($conn, $fetchqry);
$num = mysqli_num_rows($result);
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
$mid = $row['mid'];
$planid = $row['planid'];
$date1 = $row['expire_date'];
if (strtotime(date("Y/m/d")) < strtotime($date1))
{
$status = "Active";
}
else
{
$status = "Expired";
}
}
echo json_encode($row);
?>
将从数据库读取的值添加到数组中,然后对其进行编码
$output = [];
$today = strtotime(date("Y/m/d"));
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
$mid = $row['mid'];
$planid = $row['planid'];
$date1 = $row['expire_date'];
if($today < strtotime($date1))
{
$status = "Active";
}
else
{
$status = "Expired";
}
$output[] = ['mid' => $mid, 'planid' => $planid, 'status' => $status];
}
echo json_encode($output);
$output=[];
$today=标准时间(日期(“Y/m/d”);
而($row=mysqli\u fetch\u数组($result,mysqli\u ASSOC))
{
$mid=$row['mid'];
$planid=$row['planid'];
$date1=$row['expire_date'];
如果($今天<标准时间($日期1))
{
$status=“活动”;
}
其他的
{
$status=“过期”;
}
$output[]=['mid'=>$mid,'planid'=>$planid,'status'=>$status];
}
echo json_编码($output);
循环在$row
为NULL
时结束。循环完成后,您希望json_encode($row)是什么?while($row=mysqli_fetch_数组($result,mysqli_ASSOC)){…}echo json_encode($row)代码>。获取结果后,mysqli\u fetch\u array
返回null。这就是为什么会得到null
在循环中设置的所有变量是什么?记录的设置应类似于以下格式$output1[]=array(“mid”=>$mid);像wise as json数组一样,您应该使用json\u encode($output1)
谢谢您,先生。