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我需要使用內部聯接來顯示來自另一個表的數據。SQL內部聯接
$pid=intval($_SESSION["Patient_id"]); $query = "SELECT Appointment_id, Doctor_id, Patient_id, Appointment_time, Appointment_date FROM Appointment where Patient_id=$pid";
SELECT Doctor_ID
FROM Appointment
INNER JOIN Doctor
ON Appointment.Doctor_id=Doctor.Doctor_id
目前我有一個表顯示數據,但我也需要使用內部連接顯示來自另一個表的數據。我如何將創建的SELECT代碼插入到我當前的編碼中?醫生的詳細信息是我正在嘗試使用醫生ID在我的頁面中輸出的內容。我是新來的PHP。
謝謝
完整的PHP代碼appointment.php
<?php
{
mysql_connect("localhost" , "" , "") or die (mysql_error());
mysql_select_db("") or die(mysql_error());
$pid=intval($_SESSION["Patient_id"]); $query = "SELECT Appointment_id, Doctor_id, Patient_id, Appointment_time, Appointment_date FROM Appointment where Patient_id=$pid";
SELECT Doctor_id
FROM Appointment
INNER JOIN Doctor
ON Appointment.Doctor_id=Doctor.Doctor_id
//executes query on the database
$result = mysql_query ($query) or die ("didn't query");
//this selects the results as rows
$num = mysql_num_rows ($result);
//if there is only 1 result returned than the data is ok
if ($num == 1) {}
{
$row=mysql_fetch_array($result);
$_SESSION['Appointment_date'] = $row['Appointment_date'];
$_SESSION['Appointment_time'] = $row['Appointment_time'];
}
}
?>
<strong>Dates available</strong>
<select id="Availability" name="Availability">
<option value="0">--Select date--</option>
<option value="1"><?php echo $_SESSION['Appointment_date'];?></option>
</select>
<br />
<br />
<strong>Times available</strong>
<select id="Availability" name="Availability">
<option value="0">--Select time--</option>
<option value="2"><?php echo $_SESSION['Appointment_time'];?></option>>
</select>
全部代碼行從57至98
看起來好像你有你需要的所有信息在你的問題。你卡在哪裏? – SomeSillyName
它顯示一個語法錯誤。我有沒有留下什麼? – user2216325
這是我得到的錯誤。解析錯誤:語法錯誤,/web/stud中出現意外的T_STRING // 64行上的PHP/appointment1.php – user2216325