您好我正在尝试将Image从我的IOS设备上传到服务器。
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);
}
现在这是我的服务器端代码。这有点长。
<?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";
}
}
?>
现在每次尝试上传时都会发生这种警告
<b>Warning</b>: Invalid argument supplied for foreach() in <b>/Applications/XAMPP/xamppfiles/htdocs/ColorPicker/uploadAction.php</b> on line <b>5</b><br />
答案 0 :(得分:2)
从评论中,print_r
正在返回:
Array (
[files] => Array (
[name] => iphone.jpg
[type] => application/octet-stream
[tmp_name] => /Applications/XAMPP/xamppfiles/temp/php4VsKOv
[error] => 0
[size] => 13484
)
)
与此同时,您的代码正在执行此操作:
$Image = $_FILES['files']['name'];
print_r($_FILES);
foreach ($Image as $f => $name) {
您将$_FILES['files']['name']
视为数组,而不是数组。这就是为什么PHP说“为foreach()提供的无效参数”
请注意,如果使用相同的表单元素名称上载多个文件,可以成为一个数组。到目前为止,这是 关于$_FILES
最令人愤怒的事情。
您的上传代码不尝试使用上传多个文件所需的PHP语法,因此您的PHP代码不需要担心内容内部数组本身就是数组。您可以消除foreach
循环以及对[$f]
索引的所有引用。
答案 1 :(得分:1)
当您将图片数据上传为多部分时,它会以$_FILES
结尾。自PHP 4.3.0起,$_REQUEST
不再包含有关以这种方式上传的文件的任何信息。
您的其余代码基于图像未被上传为多部分的想法,而是以标准格式上传为base-64字符串。您只需从$_FILES
获取有关该文件的信息,然后使用move_uploaded_file
将其移至目标目录,而不是将基本64位数据写入文件。