-1
此刻，我在一張表格中顯示了所有註冊到健康中心的患者。我添加了一個刪除按鈕或刪除鏈接，將病人從桌面上移開。當我點擊刪除按鈕時，我收到一條錯誤消息，並且之前顯示的所有患者都消失了，現在「所有患者查看」頁面都會回顯「0結果」。
如果有人可以幫助我解決問題，以便我可以從表中移除病人，這將非常感激。試圖從表格中刪除一行
錯誤消息
警告：主（）：於E無法提取mysqli_result：\ WebProgrammingAssignment \視圖\ AllPatientsView.php在線路48上
最新登記患者模型
<?php
require_once('DAO.php');
class RegisterPatientModel extends DAO
{
protected $target = "patient";
public function __construct()
{
parent::__construct();
}
public function insertPatient($firstname, $lastname, $patstreet, $patcity, $patpostcode, $patphone, $doctorid, $dob)
{
$firstname = parent::escape($firstname);
$empnin = parent::escape($lastname);
$patstreet = parent::escape($patstreet);
$patcity = parent::escape($patcity);
$patpostcode = parent::escape($patpostcode);
$sql = "INSERT INTO {$this->target} (`firstname`, `lastname`, `patstreet`, `patcity`, `patpostcode`, `patphone`, `doctorid`, `dob`) VALUES ('{$firstname}', '{$lastname}', '{$patstreet}', '{$patcity}', '{$patpostcode}', '{$patphone}', '{$doctorid}', '{$dob}');";
return parent::query($sql);
}
public function deletePatient($patientid)
{
$sql = "DELETE
FROM {$this->target}
WHERE patientid='{$patientid}'";
return parent::query($sql);
}
function getAllPatients()
{
$sql = "SELECT a.patientid,
concat(d.firstname, ' ', d.lastname) as fullname_doctor,
a.firstname, a.lastname, a.patstreet, a.patcity, a.patpostcode, a.patphone, a.dob
FROM patient as a
INNER JOIN doctor as d
on a.doctorid = d.doctorid;";
return parent::query($sql);
}
}
?>
更新所有的患者視圖
<html>
<tr>
<td colspan="5" align="center">
<div id="boxalign2" class="boxalign2">
<div class="inputwrap">
<br>
<table id="customers" width="900" border="1" cellspacing="0" cellpadding="1">
<tr align="center">
<td bgcolor="#008000">Patient ID</td>
<td bgcolor="#008000">Doctor Name</td>
<td bgcolor="#008000">First Name</td>
<td bgcolor="#008000">Last Name</td>
<td bgcolor="#008000">Street</td>
<td bgcolor="#008000">City</td>
<td bgcolor="#008000">Post Code</td>
<td bgcolor="#008000">Telephone</td>
<td bgcolor="#008000">DOB</td>
</tr>
<?php
$allpatients = $_SESSION['patients'];
if ($allpatients->num_rows > 0) {
while ($row = $allpatients->fetch_assoc()) {
echo "<td>" . $row["patientid"] . "</td>";
echo "<td>" . $row["fullname_doctor"] . "</td>";
echo "<td>" . $row["firstname"] . "</td>";
echo "<td>" . $row["lastname"] . "</td>";
echo "<td>" . $row["patstreet"] . "</td>";
echo "<td>" . $row["patcity"] . "</td>";
echo "<td>" . $row["patpostcode"] . "</td>";
echo "<td>" . $row["patphone"] . "</td>";
echo "<td>" . $row["dob"] . "</td>";
echo "<td><a href='../controllers/ViewAllPatientsController.php?patientid=" . $row['patientid'] . "'>Delete</a></td>";
echo "</tr>";
}
} else {
echo "0 results.";
}
?>
更新。查看病人控制器
<?php
session_start();
require_once("../models/RegisterPatientModel.php");
$vapc = new RegisterPatientModel;
if (isset($_GET['patientid'])) {
$vapc->deletePatient($_GET['patientid']);
}
$allpatients = $vapc->getAllPatients();
require_once("../views/AllPatientsView.php");
您希望爲患者刪除多少個數據行？ 1？ 2？ 3？ – Martin
如果你的代碼是實時代碼，那麼它就是視圖，你有數據庫操作的風險，並且在把它們放入SQL之前不檢查這些值，你應該查找[Prepared Statements]（http://stackoverflow.com/questions/60174/how-can-i-prevent-sql -injection-in-php）並實際使用它們。 – Martin
現在只是一行 – Matthew