Question

<form method="post" enctype="multipart/form-data">
    <input type="file" name="image">
    <input type="submit" name="upload">
</form>

如何在不保存的情况下获取上传的文件，如何显示？

Answer 1

使用PHFetchOptions* fetchOptions = [[PHFetchOptions alloc] init]; fetchOptions.sortDescriptors = @[[NSSortDescriptor sortDescriptorWithKey:@"creationDate" ascending:NO]]; PHFetchResult *allPhotosResult = [PHAsset fetchAssetsWithMediaType:PHAssetMediaTypeImage options: fetchOptions]; [allPhotosResult enumerateObjectsUsingBlock:^(PHAsset *asset, NSUInteger idx, BOOL *stop) { PHImageRequestOptions* options = [[PHImageRequestOptions alloc] init]; options.synchronous = NO; options.networkAccessAllowed = YES; options.deliveryMode = PHImageRequestOptionsDeliveryModeHighQualityFormat; [[PHImageManager defaultManager] requestImageDataForAsset: asset options: options resultHandler:^(NSData *imageData, NSString *dataUTI, UIImageOrientation orientation, NSDictionary *info) { NSString* val = [asset valueForKey: @"uniformTypeIdentifier"]; if( !dataUTI ) { dataUTI = val; } }]; }];来检索图像。

$_FILES['image']

Answer 2

<?php
if(isset($_POST["submit"])) {
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["image"]["name"]);


    if (move_uploaded_file($_FILES["image"]["tmp_name"], $target_file)) {
        echo "The file ". basename( $_FILES["image"]["name"]). " has been uploaded.";
    } else {
        echo "Sorry, there was an error uploading your file.";
    }
}
?>

Answer 3

以下是jquery的使用方法。看看这个：

＆＃13;

$("#imgInp").change(function(){
        readURL(this);
    });

function readURL(input) {
        if (input.files && input.files[0]) {
            var reader = new FileReader();
            
            reader.onload = function (e) {
                $('#blah').attr('src', e.target.result);
            }
            
            reader.readAsDataURL(input.files[0]);
        }
    }

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<img id="blah" src="images/defaultUser.jpg" alt="your image"/>
<input type='file' id="imgInp" accept="image/*"/>

＆＃13;

如何在不保存的情况下在php上传图像

3 个答案: