请帮我的家伙有代码插入姓名等。。对于插入图像,我不能组合代码
这是phpmyadmin中的输出,我只插入一个值。。。
谢谢你们的帮助
<---!THIS CODE BELOW IT INSERT ID,NAME AND ETC --->
<?php
require 'db.php';
$message = '';
$Error = '';
if (isset ($_POST['Attendee_id']) &&
isset($_POST['RFID_number']) &&
isset($_POST['Attendee_Name']) &&
isset($_POST['CourseOrDepartment']) &&
isset ($_POST['Status']) ) {
$Attendee_id = $_POST['Attendee_id'];
$RFID_number = $_POST['RFID_number'];
$Attendee_Name = $_POST['Attendee_Name'];
$CourseOrDepartment = $_POST['CourseOrDepartment'];
$Status = $_POST['Status'];
$sql = 'INSERT INTO tbl_listofregister(Attendee_id,
RFID_number,Attendee_Name,CourseOrDepartment,Status)
VALUES(:Attendee_id,
:RFID_number,:Attendee_Name,:CourseOrDepartment,:Status)';
$statement = $connection->prepare($sql);
if ($statement->execute([':Attendee_id' => $Attendee_id, ':RFID_number' =>
$RFID_number,':Attendee_Name' => $Attendee_Name,':CourseOrDepartment' =>
$CourseOrDepartment,':Status' => $Status])) {
$message = 'DATA INSERTED SUCCESSFULLY';
}
else
{
$Error = "ID SHOULD BE UNIQUE";
}
}
?>
<---! HERE IS FOR IMAGE --->
<?php
$msg = '';
if($_SERVER['REQUEST_METHOD']=='POST'){
$image = $_FILES['Image']['tmp_name'];
$img = file_get_contents($image);
$con = mysqli_connect('localhost','root','','dbattendancelibrary') or
die('Unable To connect');
$sql = "insert into tbl_listofregister (image) values(?)";
$stmt = mysqli_prepare($con,$sql);
mysqli_stmt_bind_param($stmt, "s",$img);
mysqli_stmt_execute($stmt);
$check = mysqli_stmt_affected_rows($stmt);
if($check==1){
$msg = 'Image Successfullly UPloaded';
}else{
$msg = 'Error uploading image';
}
mysqli_close($con);
}
?>
2条答案
按热度按时间mspsb9vt1#
在这些行之后,您可以为img添加这一行
移动上传的文件($文件['file']['tmp文件名],“filename/”。$文件['file']['name']);
然后,将其作为($\u files['file']['name'])添加到查询中。
在html文件中写下这一行()。
我希望它工作得很好
clj7thdc2#
这将把两个插入合并成一个动作。但我建议不要把图像存储在数据库里。而是存储相对于图像站点的路径。