php PHP图片上传功能，保存在一个目录然后返回保存图片的url
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PHP image upload function, save in a dir and then return save image url
提问by user007
I am trying to upload an image to server using PHP and the save inside a dir, and then returning the image url.
我正在尝试使用 PHP 将图像上传到服务器并保存在目录中，然后返回图像 url。
HTML:
HTML：
<input name="photo" type="file" />
PHP
PHP
save_string_to_database( upload_img($_POST['photo']));
I have not much idea of PHP, I got a code from SO, but it did not do anything. Kindly help me to fix this code, or give a simple code to perform an upload:
我对 PHP 不太了解，我从 SO 那里得到了一个代码，但它没有做任何事情。请帮助我修复此代码，或提供一个简单的代码来执行上传：
function upload_img($img){
if ((($_FILES[$img]["type"] == "image/gif")
|| ($_FILES[$img]["type"] == "image/jpeg")
|| ($_FILES[$img]["type"] == "image/pjpeg")
|| ($_FILES[$img]["type"] == "image/jpg")
|| ($_FILES[$img]["type"] == "image/png"))
&& ($_FILES[$img]["size"] < 20000)
&& (strlen($_FILES[$img]["name"]) < 51)){
if ($_FILES[$img]["error"] > 0){
echo "Return Code: " . $_FILES[$img]["error"];
}
else{
// echo "Upload: " . $_FILES["image"]["name"] . "<br />";
// echo "Type: " . $_FILES["image"]["type"] . "<br />";
// echo "Size: " . ($_FILES["image"]["size"] / 1024) . " Kb<br />";
// echo "Temp file: " . $_FILES["image"]["tmp_name"] . "<br />";
if (file_exists(THEME_DIR."/images/" . $_FILES[$img]["name"])){
echo $_FILES[$img]["name"] . " already exists. ";
}
else{
move_uploaded_file($_FILES[$img]["tmp_name"],THEME_DIR."/images/" . $_FILES[$img]["name"]);
return THEME_DIR."/images/" . $_FILES[$img]["name"];
}
}
}
}
回答by Shankar Damodaran
Here's a simple one.
这是一个简单的。
HTML form to upload image
用于上传图片的 HTML 表单
<form enctype="multipart/form-data" action="upload.php" method="POST">
<input type="hidden" name="MAX_FILE_SIZE" value="512000" />
Send this file: <input name="userfile" type="file" />
<input type="submit" value="Send File" />
</form>
Your PHP file that does the Upload
执行上传的 PHP 文件
<?php
$uploaddir = '/var/www/uploads/';
$uploadfile = $uploaddir . basename($_FILES['userfile']['name']);
echo "<p>";
if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
echo "File is valid, and was successfully uploaded.\n";
} else {
echo "Upload failed";
}
echo "</p>";
echo '<pre>';
echo 'Here is some more debugging info:';
print_r($_FILES);
print "</pre>";
?>
回答by Michael Mitch
First you need a multipart/form-dataform for uploading. This is a must :)
首先你需要一个multipart/form-data表单来上传。这是必须的:)
<form action="upload_file.php" method="post"
enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"><br>
<input type="submit" name="submit" value="Submit">
</form>
The PHP part is fairly simple: This would result your file stored in "upload/{filename}"The main part you want to consider is how to get the filename and back to your write_string_to_database procedure, you could do a simple script after the upload page like
PHP部分相当简单：这将导致您的文件存储在“upload/{filename}”中您要考虑的主要部分是如何获取文件名并返回到您的write_string_to_database过程，您可以在上传后执行一个简单的脚本页面喜欢
save_string_to_database("upload/" . $_FILES["file"]["name"]);
would do the trick.
会做的伎俩。
<?php
if ($_FILES["file"]["error"] > 0)
{
echo "Error: " . $_FILES["file"]["error"] . "<br>";
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"upload/" . $_FILES["file"]["name"]);
}
}
回答by chirag ode
for file upload try this
对于文件上传试试这个
<?php if(isset($_POST['submit']))
{
$ImageName = $_FILES['photo']['name'];
$fileElementName = 'photo';
$path = 'images/';
$location = $path . $_FILES['photo']['name'];
move_uploaded_file($_FILES['photo']['tmp_name'], $location);
} ?>
<form name="form1" id="form1" method="post" action="" enctype="multipart/form-data">
<input type="file" name="photo">
<input type="submit" name="submit">
</form>
回答by user11956710
<form method='post' action='' enctype='multipart/form-data'>
Name : <input type="text" name="name" required=""/><br><br>
Code : <input type="text" name="code" required=""/><br><br>
Price : <input type="text" name="price" required=""/><br><br>
Image : <input type="file" name="image" required=""/><br><br>
<button type='submit' class='buy' name="submit">Add Now</button>
</form>
<!--insert data -->
<?php
session_start();
include('db.php');
if(isset($_POST["submit"]));
{
/*echo "<pre>";
print_r($_POST);
print_r($_FILES);*/
$name = $_POST["name"];
$code = $_POST["code"];
$price = $_POST["price"];
$image = $_FILES["image"]["name"];
/* folder image save */
// $target_dir = "/var/www/html/shivam/new/upload/";
// $target_file = $target_dir.basename($_FILES["image"]["name"]);
// /*echo "1121".$target_file;*/
// $name = basename($_FILES["image"]["name"]);
// mysqli_query($con,$qry);
// /* move file */
// move_uploaded_file($_FILES['image']['tmp_name'],$target_dir.$name);
/* move_uploaded_file($tmp_name, "$target_dir/$name");*/
/* end */
$uploaddir = '/var/www/html/uploads/';
$uploadfile = $uploaddir . basename($_FILES['image']['name']);
echo '44'.$uploadfile;
echo "<p>";
if (move_uploaded_file($_FILES['image']['tmp_name'], $uploadfile)) {
echo "File is valid, and was successfully uploaded.\n";
} else {
echo "Upload failed";
}
echo "</p>";
echo '<pre>';
echo 'Here is some more debugging info:';enter code here
print_r($_FILES);
print "</pre>";
}
?>
回答by trung nguyen
This my function, variable $ten_anh is name of file image in html.
这是我的函数，变量 $ten_anh 是 html 中文件图像的名称。
function upload_anh($ten_anh){ //$ten_anh la ten tren html vi du "avatar"
if(isset($_FILES[$ten_anh])){
$errors= array();
$file_name = $_FILES[$ten_anh]['name'];
$file_size =$_FILES[$ten_anh]['size'];
$file_tmp =$_FILES[$ten_anh]['tmp_name'];
$file_type=$_FILES[$ten_anh]['type'];
$file_ext=strtolower(end(explode('.',$_FILES[$ten_anh]['name'])));
$expensions= array("jpeg","jpg","png");
if(in_array($file_ext,$expensions)=== false){
$errors[]="Kh?ng ch?p nh?n ??nh d?ng ?nh có ?u?i này, m?i b?n ch?n JPEG ho?c PNG.";
}
if($file_size > 2097152){
$errors[]='Kích c? file nên là 2 MB';
}
if(empty($errors)==true){
move_uploaded_file($file_tmp,"../images/".$file_name);
echo "Thành c?ng!!!";
}
else{
print_r($errors);
}
}
}
}
Example: - html code:
示例： - html 代码：
<input type="file" id="avatar" name="avatar"accept="image/png, image/jpeg" required/>
- call function php:
upload_anh('avatar');
- 调用函数php：
upload_anh('avatar');
