PHP select不返回值
I have this code and I want to get the info in medication table and display it where acc_id in account table is = to acc_id in medication table and where med_timeoftheday='morning'
$postdata = file_get_contents("php://input");
if (isset($postdata)) {
$request = json_decode($postdata);
$User_ID = $request->acccid;
$sql = sprintf("SELECT * FROM account_info
join medication on account_info.acc_id = medication.acc_id
where account_info.acc_id='%s'",
mysqli_real_escape_string($conn,$User_ID));
$result=$conn->query($sql);
if ($result->num_rows>0)
{
while($row=$result->fetch_assoc())
{$data[]=$row;
}
echo json_encode($data);
}
}
this is my ts :
how can I do that ?
Thank you in advance!
我有这段代码，我想在 这是我的问题： p>
我该怎么做？ p>
提前谢谢！ p>
div>药物表 code>中获取信息并显示 其中帐户表 code>中的 acc_id code>是 = code>到 accicid code> drug table code>和 med_timeoftheday ='morning' code> p>
$ postdata = file_get_contents（“php：// input”）;
if（isset（$ postdata））{\ n $ request = json_decode（$ postdata）;
$ User_ID = $ request-＆gt; acccid;
$ sql = sprintf（“SELECT * FROM account_info
join drug on account_info.acc_id = medication.acc_id
其中account_info .acc_id ='％s'“，
mysqli_real_escape_string（$ conn，$ User_ID））;
$ result = $ conn-＆gt; query（$ sql）;
if（$ result-＆gt; num_rows＆gt; 0）
{
while（$ row = $ result-＆gt; fetch_assoc（））
{$ data [] = $ row;
}
echo json_encode（$ data）;
}
\ n}
code> pre>
Try somehting like this:
SELECT * FROM medication
INNER JOIN account_info ON account_info.acc_id = medication.acc_id
WHERE medication.med_timeoftheday='morning'
firstly if you selected data from medication table so select first medication table and then using join with account table .
$sql = "SELECT * FROM medication JOIN account_info ON account_info.acc_id = medication.acc_id WHERE medication.med_timeoftheday='morning'";