Php 如何从数据库中插入、检索多个图像
这是提供图像路径的php代码, 我从另一个ios应用程序获取的多个图像,如何将这些图像插入到具有不同名称的数据库中并显示Php 如何从数据库中插入、检索多个图像,php,Php,这是提供图像路径的php代码, 我从另一个ios应用程序获取的多个图像,如何将这些图像插入到具有不同名称的数据库中并显示 <?php $myparam = $_POST['userfile']; //getting multiple image Here //$mytextLabel= $_POST['filenames'] //getting textLabe Here //echo $myparam; //echo $mytextLabel; $target_path
<?php
$myparam = $_POST['userfile']; //getting multiple image Here
//$mytextLabel= $_POST['filenames'] //getting textLabe Here
//echo $myparam;
//echo $mytextLabel;
$target_path = "uploads/";
$target_path = $target_path . basename( $_FILES['userfile']['name']);
if(move_uploaded_file($_FILES['userfile']['tmp_name'], $target_path)) {
echo "Success ! The file has been uploaded";
//insert query will run here right?
//but how to get those multiple images and save it m not getting
} else{
echo "There was an error uploading the file, please try again!";
}
?>
您可以使用。此类为您管理文件上载。简而言之,它管理上传的文件,并允许您对该文件执行任何操作,特别是如果它是一个图像,并且可以执行任意次数 它是在您的站点中快速集成文件上传的理想类。如果文件是图像,可以通过多种方式进行转换、调整大小和裁剪。您还可以应用过滤器、添加边框、文本、水印等。。。例如,这就是你所需要的画廊脚本。支持的格式有PNG、JPG、GIF和BMP 这是一个示例类:
class UploadFile
{
protected $uploadFileModel;
public function __construct()
{
$this->uploadFileModel = new UploadFileModel();
}
public function insert($params)
{
$handle = new upload($params['file'], 'fa_IR');
$handle->file_max_size = '20000000';
$handle->allowed = array(
'image/png',
'image/jpeg',
'image/gif');
$handle->mime_check = TRUE;
if ($handle->uploaded) {
$handle->process($params['file_path']);
if ($handle->processed) {
$params['file_name'] = $handle->file_dst_name;
// insertToDB is a function that insert data to database
$insToDB = $this->insertToDB(array(
'title' => $params['title'],
'file_name' => $params['file_name'],
'path' => $params['path'],
'created_at' => $params['created_at']
));
if ($insToDB !== FALSE) {
return intval($insToDB);
} else {
unlink($params['file_path'] . $params['file_name']);
return FALSE;
}
}
}
return $handle->error;
}
public function insertToDB($params)
{
return $this->uploadFileModel->insert($params);
}
public function showForm()
{
$str = '<input type="file" id="fileToUpload" name="fileToUpload">';
return $str;
}
}
因为,Ashish提供的数据/代码不多。我从我这边假设 ImageUpload.php
<?
extract($_POST);
// This will count total images selected.
$TotalImage = count($_FILES['userfile']['name']);
for($i = 0;$i<$TotalImage;$i++)
{
$image_name = $_FILES['userfile']['name'][$i];
$target_path = "uploads/".$imagename;
if(move_uploaded_file($_FILES['userfile']['tmp_name'][$i], $target_path))
{
//Write Query Here To Insert Into Database
}
}
?>
首先,您需要使用enctype=“multipart/form data”
上传图像。
其次,您需要将name
声明为数组类型。(因为您正在上载多个图像)
SubmitImageUpload.php
<?
extract($_POST);
// This will count total images selected.
$TotalImage = count($_FILES['userfile']['name']);
for($i = 0;$i<$TotalImage;$i++)
{
$image_name = $_FILES['userfile']['name'][$i];
$target_path = "uploads/".$imagename;
if(move_uploaded_file($_FILES['userfile']['tmp_name'][$i], $target_path))
{
//Write Query Here To Insert Into Database
}
}
?>
而且,由于您没有向我们提供您的数据库结构,因此,我不能假设您的数据库结构并编写代码进行检索。我感到困惑。那你对哪一部分有意见?代码似乎与您询问的内容不匹配。如何获取多个图像以及如何插入它,我在这里结结巴巴地问您的代码Ashish在哪里,您尝试过多个图像上载?我想你被困在丝绸板交易中了嘿,穆罕默德,你怎么知道他的项目中有MVC结构?哦,是的。。。我很抱歉,但这是一个答案,你可以在没有MVC的情况下编写它的函数。
<?
extract($_POST);
// This will count total images selected.
$TotalImage = count($_FILES['userfile']['name']);
for($i = 0;$i<$TotalImage;$i++)
{
$image_name = $_FILES['userfile']['name'][$i];
$target_path = "uploads/".$imagename;
if(move_uploaded_file($_FILES['userfile']['tmp_name'][$i], $target_path))
{
//Write Query Here To Insert Into Database
}
}
?>