我正在使用ngx-image-cropper在我的angular应用程序中裁剪图像。我使用返回'File‘的输出属性(imageCroppedFile)来捕获裁剪后的图像。我需要裁剪后的图像有一个文件名,这样我就可以使用name属性遍历后端的文件,而它在默认情况下是未定义的。我怎么给它起个名字呢?我尝试了以下方法:在FormData上,
var formData:any = new FormData();
console.log('The number of files is '+files.length);//Logs the number of files is 1
for(var i=0; i<files.length;i++) {
formData.append("uploads[]", files[i].name, 'image'+i);
console.log('File name '+ i + ' ' +files[i].name);//Logs File name 0 undefined
}以及由裁剪触发的方法
imageCroppedFile(image: File) {
this.filesToUpload = [];
console.log('imageCroppedFile method '+image.name+ ' size is '+image.size);// Logs imageCroppedFile method undefined size is 380284
this.filesToUpload[0]=image;
console.log('The filesToUpload is '+this.filesToUpload[0].name);// Logs The filesToUpload is undefined
}上传器在没有裁剪器的情况下工作。
发布于 2018-10-03 17:52:29
将你的文件转换为blob,你可以自己命名你的blob
const blobImage = file as Blob;参考:https://github.com/Mawi137/ngx-image-cropper/issues/91#issuecomment-422252629
发布于 2021-04-09 22:26:57
这对我来说很有效:
// grab the cropped area to file for upload
var b64File = this.dataURLtoFile(this.croppedImage, 'hello.jpg'); // you can name it anything.
dataURLtoFile(dataurl, filename): File {
var arr = dataurl.split(','), mime = arr[0].match(/:(.*?);/)[1],
bstr = atob(arr[1]), n = bstr.length, u8arr = new Uint8Array(n);
while(n--){
u8arr[n] = bstr.charCodeAt(n);
}
return new File([u8arr], filename, {type:mime});
}https://stackoverflow.com/questions/52624169
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