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我是 PHP 和 Objective-C 的新手，我已经搜索过这个问题的答案，但一切似乎都很复杂，我无法理解。我正在尝试使用 PHP 将图像文件上传到我的 ftp 服务器。

我在我的应用程序中使用此代码上传图片：

UIImage *myImage = [UIImage imageNamed:@"black_strip.png"];
    NSData *imageData = UIImagePNGRepresentation(myImage);
    // setting up the URL to post to
    NSString *urlString = @"http://www.myurl.net/GagVidApp/uploadProfImage.php";

    // setting up the request object now
    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
    [request setURL:[NSURL URLWithString:urlString]];
    [request setHTTPMethod:@"POST"];

    NSString *boundary = @"---------------------------14737809831466499882746641449";
    NSString *contentType = [NSString stringWithFormat:@"multipart/form-data; boundary=%@",boundary];
    [request addValue:contentType forHTTPHeaderField: @"Content-Type"];

    /*
     now lets create the body of the post
     */
    NSMutableData *body = [NSMutableData data];
    [body appendData:[[NSString stringWithFormat:@"\r\n--%@\r\n",boundary] dataUsingEncoding:NSUTF8StringEncoding]];
    [body appendData:[@"Content-Disposition: form-data; name=\"userfile\"; filename=\"ipodfile.png\"\r\n" dataUsingEncoding:NSUTF8StringEncoding]];
    [body appendData:[@"Content-Type: application/octet-stream\r\n\r\n" dataUsingEncoding:NSUTF8StringEncoding]];
    [body appendData:[NSData dataWithData:imageData]];
    [body appendData:[[NSString stringWithFormat:@"\r\n--%@--\r\n",boundary] dataUsingEncoding:NSUTF8StringEncoding]];
    // setting the body of the post to the reqeust
    [request setHTTPBody:body];

    // now lets make the connection to the web
    NSData *returnData = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];
    NSString *returnString = [[NSString alloc] initWithData:returnData encoding:NSUTF8StringEncoding];

    NSLog(@"returnString: %@", returnString);

这是我的php代码：

<?php
$file = basename($_FILES['userfile']['name']);
$remote_file = basename($_FILES['userfile']['name']);
//$remote_file = 'readme.txt';

// set up basic connection
$ftp_server = "www.myurl.net";
$ftp_user_name = "myusername";
$ftp_user_pass = "mypassword";
$conn_id = ftp_connect($ftp_server); 

// login with username and password
$login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass); 

// check connection
if ((!$conn_id) || (!$login_result)) { 
    echo "FTP connection has failed!";
    echo "Attempted to connect to $ftp_server for user $ftp_user_name"; 
    exit; 
} else {
    echo "Connected to $ftp_server, for user $ftp_user_name";
}
// upload a file
if (ftp_put($conn_id, $remote_file, $file, FTP_ASCII)) {
 echo "successfully uploaded $file\n";
} else {
 echo "There was a problem while uploading $file\n";
}

// close the connection
ftp_close($conn_id);
?>

我在我的应用程序中得到的返回字符串是：

returnString: Connected to www.myurl.net, for user myuser<br />
<b>Warning</b>:  ftp_put(ipodfile.png) [<a href='function.ftp-put'>function.ftp-put</a>]: **failed to open stream: No such file or directory in** <b>/home/load2unet/domains/myurl.net/public_html/GagVidApp/uploadProfImage.php</b> on line <b>24</b><br />
There was a problem while uploading ipodfile.png

我一直在努力寻找答案，但没有任何运气。任何帮助将不胜感激！谢谢

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2 回答 2

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文件上传存储在临时文件夹中。因为您只提供文件的基本名称 ( basename($_FILES['userfile']['name'])，所以在当前路径中搜索文件，但它不存在。提供完整路径，您的脚本将起作用。

于 2013-11-02T15:50:35.323 回答
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正如彼得所说，我怀疑问题是未能引用临时文件的完整路径，即$_FILES["userfile"]["tmp_name"]. 这是我过去使用过的内容的再现，使用move_uploaded_file而不是 ftp，但您可能明白了：

<?php

header('Content-Type: application/json');

$allowedExts = array("jpg", "jpeg", "gif", "png");
$extension = end(explode(".", $_FILES["userfile"]["name"]));
if ((($_FILES["userfile"]["type"] == "image/gif")
     || ($_FILES["userfile"]["type"] == "image/jpeg")
     || ($_FILES["userfile"]["type"] == "image/png")
     || ($_FILES["userfile"]["type"] == "image/pjpeg"))
    && ($_FILES["userfile"]["size"] < 200000)
    && in_array($extension, $allowedExts))
{
    if ($_FILES["userfile"]["error"] > 0)
    {
        echo json_encode(array("error" => $_FILES["userfile"]["error"]));
    }
    else
    {
        if (file_exists("upload/" . $_FILES["userfile"]["name"]))
        {
            echo json_encode(array( "error" => $_FILES["userfile"]["name"] . " already exists"));
        }
        else
        {
            move_uploaded_file($_FILES["userfile"]["tmp_name"], "upload/" . $_FILES["userfile"]["name"]);
            echo json_encode(array("success"   => true,
                                   "upload"    => $_FILES["userfile"]["name"],
                                   "type"      => $_FILES["userfile"]["type"],
                                   "size"      => ($_FILES["userfile"]["size"] / 1024) . " kB",
                                   "stored in" => "upload/" . $_FILES["userfile"]["name"]));
        }
    }
}
else
{
    echo json_encode(array( "error" => "Invalid file"));
}
?>

也许您不需要文件大小检查（或者 200k 可能不是正确的阈值），这取决于您。但请注意，对于编程接口，我将响应格式化为 JSON，因此我的应用程序可以轻松解析响应。

于 2013-11-02T16:05:32.977 回答