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Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

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Php 无法将映像上载到服务器ios_Php_Ios_Objective C_File Upload_Image Uploading - Fatal编程技术网

Php 无法将映像上载到服务器ios

Php 无法将映像上载到服务器ios,php,ios,objective-c,file-upload,image-uploading,Php,Ios,Objective C,File Upload,Image Uploading,您好,我正在尝试将图像从IOS设备上载到服务器 这是我上传图片的代码 - (IBAction)btnUpload:(id)sender { if (self.imageViewGallery.image == nil) { UIAlertView *ErrorAlert = [[UIAlertView alloc] initWithTitle:@"Wait"

您好,我正在尝试将图像从IOS设备上载到服务器

这是我上传图片的代码

- (IBAction)btnUpload:(id)sender {

    if (self.imageViewGallery.image == nil) {
        UIAlertView *ErrorAlert = [[UIAlertView alloc] initWithTitle:@"Wait"
                                                             message:@"Please Select An Image To Upload." delegate:nil
                                                   cancelButtonTitle:@"OK"
                                                   otherButtonTitles:nil, nil];
        [ErrorAlert show];
        NSLog(@"error");
    }
    else{
    NSData *imageData = UIImageJPEGRepresentation(self.imageViewGallery.image, 90);
    NSString *urlString = @"http://localhost/ColorPicker/api.upload.php";


    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"];
        NSString *imgName = LIbraryImage;
        NSLog(@"image name : %@",imgName);

    NSMutableData *body = [NSMutableData data];
    [body appendData:[[NSString stringWithFormat:@"\r\n--%@\r\n",boundary] dataUsingEncoding:NSUTF8StringEncoding]];

        [body appendData:[[NSString stringWithString:[NSString stringWithFormat:@"Content-Disposition: form-data; name=\"files\"; filename=\"%@\"\r\n", imgName]] 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]];

    [request setHTTPBody:body];
        NSLog(@"setRequest : %@", request);


    NSData *returnData = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];
    NSString *returnString = [[NSString alloc] initWithData:returnData encoding:NSUTF8StringEncoding];

    NSLog(@"returnstring is : %@",returnString);
    }
这是我的服务器端代码。有点长

  <?php

$Image = $_FILES['files']['name'];
    print_r($_FILES);
foreach ($Image as $f => $name) {

    $allowedExts = array("gif", "jpeg", "jpg", "png");
    $temp = explode(".", $name);
    $extension = end($temp);

    if ((($_FILES['files']['type'][$f] == "image/gif") || ($_FILES['files']['type'][$f] == "image/jpeg") || ($_FILES['files']['type'][$f] == "image/jpg") || ($_FILES['files']['type'][$f] == "image/png")) && ($_FILES['files']['size'][$f] < 2000000) && in_array($extension, $allowedExts)) {
        if ($_FILES['files']['error'][$f] > 0) {
            echo "Return Code: " . $_FILES['files']['error'][$f] . "<br>";
        } else {

            if (file_exists("upload/" . $name)) {

            } else {
                move_uploaded_file($_FILES['files']['tmp_name'][$f], 'upload/' . uniqid() . "_" . $name);
            }
        }
    } else {
        $error = "Invalid file";
    }

}
?>

现在,每次我尝试上传时都会发生什么?我得到了这个警告

<b>Warning</b>:  Invalid argument supplied for foreach() in <b>/Applications/XAMPP/xamppfiles/htdocs/ColorPicker/uploadAction.php</b> on line <b>5</b><br />
警告:为第5行/Applications/XAMPP/xamppfiles/htdocs/ColorPicker/uploadAction.php中的foreach()提供的参数无效

当您以多部分方式上载图像数据时,它最终会以。自PHP4.3.0以来,
$\u REQUEST
不再包含任何关于以这种方式上载的文件的信息


代码的其余部分基于这样一种思想:图像不是作为多部分上传的,而是以标准形式作为base-64字符串上传的。您不需要将base 64数据写入文件,只需从
$\u FILES
中获取有关文件的信息,并将其移动到目标目录。

从注释中,即
print\u r
返回:

Array ( 
    [files] => Array ( 
        [name] => iphone.jpg 
        [type] => application/octet-stream 
        [tmp_name] => /Applications/XAMPP/xamppfiles/temp/php4VsKOv 
        [error] => 0 
        [size] => 13484 
    )
) 
同时,您的代码正在执行以下操作:

$Image = $_FILES['files']['name'];
print_r($_FILES);
foreach ($Image as $f => $name) {
您将$\u文件['FILES']['name']视为一个数组,而实际上它不是。这就是为什么PHP说“为foreach()提供的参数无效”

请注意,如果使用相同的表单元素名称上载多个文件,则它可以成为一个数组。这是迄今为止关于
$\u文件
最令人恼火的事情


上载代码不会尝试使用一次上载多个文件所需的PHP语法,因此PHP代码不需要担心内部数组的内容是数组本身。您可以消除
foreach
循环和对
[$f]
索引的所有引用。

PHP中运行的是哪个else块?“未发布”或“错误”一个?始终未发布…打印($\u文件)的输出是什么?请发布打印($\u文件)的输出。ye sir它返回以下数组([FILES]=>array([name]=>iphone.jpg[type]=>application/octet流[tmp_name]=>/Applications/XAMPP/xamppfiles/temp/php4VsKOv[error]=>0[size]=>13484])是的。请验证该文件是用
print_r($\u FILES)
上载的,然后重写PHP。先生,我做了一些更改,请您帮助我。