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好的，我正在尝试将 php 表单中的值提交到多个表中。我的 php 代码工作正常，但是诸如patientID插入到“ patients”中的值例如：PatientID; 100很好，但是相同的值PatientID没有将相同的唯一值插入到另一个表中，例如：“ Disease”表。难道我做错了什么？

**修改后的问题

我不确定我是否正确分配了表之间的关系。以下是表格和它们之间的关系。

Patient Attends Accident & Emergency 
Patient seen_by Nurse
Nurse assesses disease of patient 
{{nurse assigns priority to patient}} Priority linked to patient and nurse
{{nurse gives patient waiting time}} Time linked to nurse and patient 
{{doctor will see patient based on their waiting time and priority}} Doctor linked to both time and priority. 
Accident & Emergency; (ID(PK), PatientID(FK) Address, City, Postcode, Telephone)
Patient (ID(PK), Forename, Surname, Gender, Dateofbirth, Address, Patienthistory, illness, 
Nurse(ID(PK) Forename, surname)
Assesses(ID(PK)NurseID(FK), PatientID(FK))
Disease(ID(PK), illness, symptoms, diagnosis, treatment) {{nurse assesses disease of patient (these tables should all be linked}}
Priority (ID, NurseID(FK), PatientID(FK), DoctorID(FK), Priority)
Time(ID,NurseID, PatientID, DoctorID, Arrival Time, Expected waiting time, Discharge time)
Doctor (ID,Firstname, Surname)

修改后的 PHP 代码。ID 没有插入到表中；例如：PatientID 未插入到疾病表中。

<?php
$con = mysql_connect("localhost","root","") or die('Could not connect: ' . mysql_error());
mysql_select_db("a&e", $con) or die('Could not select database.');

//get NURSE values from form
$nurse_ID = $_POST['nurse_ID'];
$nurse_name = $_POST['nurse_name'];
$nurse_lastname = $_POST['nurse_lastname'];

//get Disease values from form
$disease_ID = $_POST['disease_ID'];
$symptoms = $_POST['symptoms'];
 $diagnosis = $_POST['diagnosis'];
$treatment = $_POST['treatment'];

//get Patient values from form 
$patient_id = $_POST['patient_id'];
$patient_name = $_POST['patient_name'];
$patient_lastname = $_POST['patient_lastname'];
$gender = $_POST['gender'];
 $dateOfBirth = $_POST['dateOfBirth'];
$monthOfBirth = $_POST['monthOfBirth'];
$yearOfBirth = $_POST['yearOfBirth'];
$address = $_POST['address'];
$history = $_POST['history'];
$illness = $_POST['illness'];
$priority = $_POST['priority'];
$priority_id = $_POST['priority_id'];

// Validate
$date = $dateOfBirth.'-'.$monthOfBirth.'-'.$yearOfBirth;

$sql ="INSERT INTO Nurse(Forename, Surname)
VALUES('$nurse_name', '$nurse_lastname')";
mysql_query($sql,$con) or die('Error: ' . mysql_error());
echo "$nurse_ID"; mysql_insert_id(); //get the assigned id for a nurse

$sql ="INSERT INTO Disease(Illness, Symptoms, Diagnosis, Treatment, PatientID)
   VALUES('$illness', '$symptoms', '$diagnosis', '$treatment', '$patient_id')";
mysql_query($sql,$con) or die('Error: ' . mysql_error());
echo "$patient_id"; mysql_insert_id(); //get the assigned id for a patient 

//use nurse_id and patient_id
$sql ="INSERT INTO Priority(NurseID, PatientID, Priority)
   VALUES('$nurse_ID', '$patient_id', '$priority')";
mysql_query($sql,$con) or die('Error: ' . mysql_error());
echo "$priority_id"; mysql_insert_id(); //get the assigned id for priority
echo "$patient_id"; mysql_insert_id(); //get the assigned id for a patient

$sql="INSERT INTO Patient(Forename, Surname, Gender, Date_Of_Birth, Address, Patient_History, Illness, Priority)
  VALUES     ('$patient_name', '$patient_lastname', '$gender', '$date', '$address', '$history', '$illness', '$priority')";
 mysql_query($sql,$con) or die('Error: ' . mysql_error());
 echo "$patient_id"; mysql_insert_id(); //get the assigned id for a patient

echo "1 record added";
 // close connection 
 mysql_close($con);
 ?>
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2 回答 2

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虽然我不完全理解您的系统应该如何工作，但您可以在以下代码中看到它永远不会为疾病 ID 和患者 ID 插入不同的 ID：

$sql ="INSERT INTO Disease(ID, Illness, Symptoms, Diagnosis, Treatment, PatientID)
VALUES('$id', '$illness', '$symptoms', '$diagnosis', '$treatment', '$id')";

基本上，您插入的疾病 ID 与患者 ID 完全相同。您可能希望为这些变量设置不同的变量。

关于我上面的评论：

您可以像这样过滤：

$id = intval($_POST['ID']);
$name = filter_input(INPUT_GET | INPUT_POST, $_POST['name']); // works in PHP 5.2.x and above

关于 MySQL，请参阅这篇文章： 为什么我不应该在 PHP 中使用 mysql_* 函数？

于 2013-02-02T17:49:47.333 回答
0
  1. 您需要为不同的实体（护士、患者、疾病等）使用唯一的 ID、姓名和姓氏。INSERT然后在语句中适当地使用它们。请参阅下面的修订代码。
  2. 在脚本的开头只选择一次你的数据库mysql_select_db（如果你打算坚持使用 mysql_*）。
  3. 在插入之前对用户的输入进行清理和验证。
  4. 按正确（逻辑）顺序（护士、患者、疾病、优先级）插入您的记录。
  5. 现在你所有的 id 都是通过 POST 来的。您可以考虑在 mysql 中使用 id 自动更新。
  6. 你有一个缺失的变量$priority_id。我已经把它放在修改后的代码中，假设你通过POST.
  7. 做正确的错误处理不仅仅是die().
  8. 最好考虑切换到 PDO 或 mysqli_* 并使用准备好的语句。

修改后的代码（更新）：

假设是为每个表的列auto_increment启用。id

$con = mysql_connect("localhost","root","") or die('Could not connect: ' . mysql_error());
mysql_select_db("a&e", $con) or or die('Could not select database.');

//get NURSE values from form
//We don't need to post an id for a Nurse since mysql will assign it for us
//$nurse_id = $_POST['nurse_id'];
$nurse_name = $_POST['nurse_name']; 
$nurse_lastname = $_POST['nurse_lastname'];

//get Disease values from form
// We don't need to post an id for a Disease since mysql will assign it for us
//$disease_id = $_POST['disease_id'];
$symptoms = $_POST['symptoms'];
$diagnosis = $_POST['diagnosis'];
$treatment = $_POST['treatment'];

//get Patient values from form
//We don't need to post an id for a Patient since mysql will assign it for us
//$patient_id = $_POST['patient_id'];
$patient_name = $_POST['patient_name'];
$patient_lastname = $_POST['patient_lastname'];
$gender = $_POST['gender'];
$dateOfBirth = $_POST['dateOfBirth'];
$monthOfBirth = $_POST['monthOfBirth'];
$yearOfBirth = $_POST['yearOfBirth'];
$address = $_POST['address'];
$history = $_POST['history'];
$illness = $_POST['illness'];
$priority = $_POST['priority'];

//We don't need to post an id for a Priority entity since mysql will assign it for us
//missing variable
//$priority_id = $_POST['priority_id'];

//Sanitize and validate your input here 
// ...skipped
// Validate
$date = $dateOfBirth.'-'.$monthOfBirth.'-'.$yearOfBirth;

//We don't provide an id for a Nurse since mysql will assign it for us
$sql ="INSERT INTO Nurse(Forename, Surname)
       VALUES('$nurse_name', '$nurse_lastname')";
mysql_query($sql,$con) or die('Error: ' . mysql_error());
$nurse_id = mysql_insert_id(); //get the assigned id for a nurse

//We don't provide an id for a Patient since mysql will assign it for us
$sql="INSERT INTO Patient(Forename, Surname, Gender, Date_Of_Birth, Address, Patient_History, Illness, Priority)
      VALUES('$patient_name', '$patient_lastname', '$gender', '$date', '$address', '$history', '$illness', '$priority')";
mysql_query($sql,$con) or die('Error: ' . mysql_error());
$patient_id = mysql_insert_id(); //get the assigned id for a patient

//We don't provide an id for a Disease since mysql will assign it for us
$sql ="INSERT INTO Disease(Illness, Symptoms, Diagnosis, Treatment, PatientID)
       VALUES('$illness', '$symptoms', '$diagnosis', '$treatment', '$patient_id')";
mysql_query($sql,$con) or die('Error: ' . mysql_error());

//We don't provide an id for a Priority since mysql will assign it for us
//But we use $nurse_id and $patient_id that we get earlier
$sql ="INSERT INTO Priority(NurseID, PatientID, Priority)
       VALUES('$nurse_id', '$patient_id', '$priority')";
mysql_query($sql,$con) or die('Error: ' . mysql_error());

echo "1 record added";
// close connection 
mysql_close($con);
于 2013-02-02T18:09:23.827 回答