Question

我创建了一个名为＆＃34; employee.php＆＃34;的表单页面。用于获取用户数据。另外我还有一个名为SQLConnectionProcess.php的文件，其中包含将employee.php中的表单链接到sql表的代码。数据库的名称是＆＃34;员工信息＆＃34;表的名称是＆＃34;员工信息＆＃34;。我使用phpmyadmin和XAMPP进行本地服务器测试。

employee.php代码：

<html>
<body>

<form name="EmployeeDatabase" action="SQLConnectionProcess.php" method="post">

<link rel="stylesheet" href="css.css">

<h1>EMPLOYEE DATABASE</h1>

Employe Card NO: <input type="text" name="cardNO" ><br><br>
Employee NO: <input type="text" name="employeeNO" ><br><br>
Employee Name: <input type="text" name="employeename"><br><br>
Nationality: <input type="text" name="nationality"><br><br>
Profession: <input type="text" name="profession"><br><br>
DOB: <input type="text" name="DOB"><br><br>
DOJ: <input type="text" name="DOJ"><br><br>
DOA(VisitVisa): <input type="text" name="DOA"><br><br>
Company Code: <input type="text" name="companycode"><br><br>
Sponsor Code: <input type="text" name="sponsorcode"><br><br>
Visa Type: <input type="text" name="visatype"><br><br>
Status: <input type="text" name="status"><br><br>

<input type="submit" name="formSubmit" value="Submit">

</form>

</body>
</html>

SQLConnectionProcess.php代码：

if(isset($_POST['formSubmit'])){
  $cardNO= $_POST['cardNO'];
  $employeeNO= $_POST['employeeNO'];
  $employeename= $_POST['employeename'];
  $nationality= $_POST['nationality'];
  $profession= $_POST['profession'];
  $DOB= $_POST['DOB'];
  $DOJ= $_POST['DOJ'];
  $DOA= $_POST['DOA'];
  $companycode = $_POST['companycode'];
  $sponsorcode= $_POST['sponsorcode'];
  $visatype= $_POST['visatype'];
  $status= $_POST['status'];
  mysqli_connect('localhost','root','password','employee information');
  $sql = sprintf("INSERT INTO table_employee info(Employee Card NO,Employee NO,Employee Name,Nationality,Profession,DOB,DOJ,DOA(VisitVisa),Company Code,Sponsor Code,Visa Type,Status) VALUES ('','%s','%s','%s','%s','%s','%s','%s','%s','%s','%s','%s','%s')",$cardNO,$employeeNO,$employeename,$nationality,$profession,$DOB,$DOJ,$DOA,$companycode,$sponsorcode,$visatype,$status);
  mysqli_query($sql);

但是当我从employee.php提交表单时，我被重定向到一个新页面，显示SQLConnectionProcess.php中的所有代码。我无法弄清楚编码错误。请帮助我......

Answer 1

使用以下代码。

<强> SQLConnectionProcess.php

<?php
if(isset($_POST['formSubmit'])){
  $cardNO= isset($_POST['cardNO']) ? $_POST['cardNO'] : 0;
  $employeeNO= isset($_POST['employeeNO']) ? $_POST['employeeNO'] : 0;
  $employeename= isset($_POST['employeename']) ? $_POST['employeename'] : "";
  $nationality= isset($_POST['nationality']) ? $_POST['nationality'] : "";
  $profession= isset($_POST['profession']) ? $_POST['profession'] : "";
  $DOB= isset($_POST['DOB']) ? $_POST['DOB'] : "";
  $DOJ= isset($_POST['DOJ']) ? $_POST['DOJ'] : "";
  $DOA= isset($_POST['DOA']) ? $_POST['DOA'] : "";
  $companycode = isset($_POST['companycode']) ? $_POST['companycode'] : 0;
  $sponsorcode= isset($_POST['sponsorcode']) ? $_POST['sponsorcode'] : 0;
  $visatype= isset($_POST['visatype']) ? $_POST['visatype'] : "";
  $status= isset($_POST['status']) ? $_POST['status'] : "";
  $con = mysqli_connect('localhost','root','','employee information');
  $sql = sprintf("INSERT INTO table_employee info(Employee Card NO,Employee NO,Employee Name,Nationality,Profession,DOB,DOJ,DOA(VisitVisa),Company Code,Sponsor Code,Visa Type,Status) VALUES ('','%s','%s','%s','%s','%s','%s','%s','%s','%s','%s','%s','%s')",$cardNO,$employeeNO,$employeename,$nationality,$profession,$DOB,$DOJ,$DOA,$companycode,$sponsorcode,$visatype,$status);
  mysqli_query($con,$sql);
}
?>

Answer 2

您的服务器不解析PHP文件，而是服务文件“raw”。这是你需要解决的问题。如果你找到这个问题，你可以在这里找到解决方案的建议：

XAMPP not parsing PHP

顺便说一句 - 这可能很明显，但你有开放和关闭<?php标签吗？或者是您在整个文件中发布的文本

编辑：你需要像这样传递连接：

$connection = mysqli_connect('localhost','root','','employee information');
$sql = "...";
mysqli_query($connection, $sql);

Answer 3

SELECT t.id,MIN(t.date) as start_date,MAX(t.date) as end_date,t1.name
FROM t,t1
where t.id=t1.id
GROUP BY id

检查发布后是否正常工作

我无法将我的html表单链接到mysql表？

3 个答案: