我正在尝试使用PHP将图像上传到服务器并在目录中保存,然后返回图像网址。
HTML:
<input name="photo" type="file" />
PHP
save_string_to_database( upload_img($_POST['photo']));
我对PHP不太了解,我从SO获得了代码,但它没有做任何事情。请帮我修复此代码,或者提供一个简单的代码来执行上传:
function upload_img($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"];
}
}
}
}
答案 0 :(得分:16)
这是一个简单的。
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>
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>";
?>
<强> Source 强>
答案 1 :(得分:6)
首先,您需要 multipart / form-data 表单进行上传。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>
PHP部分相当简单: 这会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write_string_to_database过程,您可以在上传页面后执行一个简单的脚本,如
save_string_to_database("upload/" . $_FILES["file"]["name"]);
会做到这一点。
<?php
if ($_FILES["file"]["error"] > 0)
{
echo "Error: " . $_FILES["file"]["error"] . "<br>";
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"upload/" . $_FILES["file"]["name"]);
}
}
答案 2 :(得分:1)
用于文件上传试试这个
<?php if(isset($_POST['submit']))
{
$ImageName = $_FILES['photo']['name'];
$fileElementName = 'photo';
$path = 'images/';
$location = $path . $_FILES['photo']['name'];
move_uploaded_file($_FILES['photo']['tmp_name'], $location);
} ?>
<form name="form1" id="form1" method="post" action="" enctype="multipart/form-data">
<input type="file" name="photo">
<input type="submit" name="submit">
</form>
答案 3 :(得分:0)
这是我的函数,变量$ 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){
$errors[]="Không chấp nhận định dạng ảnh có đuôi này, mời bạn chọn JPEG hoặc PNG.";
}
if($file_size > 2097152){
$errors[]='Kích cỡ file nên là 2 MB';
}
if(empty($errors)==true){
move_uploaded_file($file_tmp,"../images/".$file_name);
echo "Thành công!!!";
}
else{
print_r($errors);
}
}
}
示例: -html代码:
<input type="file" id="avatar" name="avatar"accept="image/png, image/jpeg" required/>
upload_anh('avatar');
答案 4 :(得分:0)
<form method='post' action='' enctype='multipart/form-data'>
Name : <input type="text" name="name" required=""/><br><br>
Code : <input type="text" name="code" required=""/><br><br>
Price : <input type="text" name="price" required=""/><br><br>
Image : <input type="file" name="image" required=""/><br><br>
<button type='submit' class='buy' name="submit">Add Now</button>
</form>
<!--insert data -->
<?php
session_start();
include('db.php');
if(isset($_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>";
}
?>