js input file 获取其url

<input ref="fileInput" class="file-input" type="file" id="change" accept="image" @change="change"/>

methods: {

openCropper(srcUrl) {

this.imageUrl = srcUrl;

},

//创建url路径

getObjectURL(file) {

let url = null;

if (window.createObjectURL != undefined) {

// basic

url = window.createObjectURL(file);

} else if (window.URL != undefined) {

// mozilla(firefox)

url = window.URL.createObjectURL(file);

} else if (window.webkitURL != undefined) {

// webkit or chrome

url = window.webkitURL.createObjectURL(file);

}

return url;

},

change(e) {

console.log(e)

let files = e.target.files || e.dataTransfer.files;

if (!files.length) return;

let type = files[0].type; //文件的类型，判断是否是图片

let size = files[0].size; //文件的大小，判断图片的大小

// if (this.imgCropperData.accept.indexOf(type) == -1) {

// alert("请选择我们支持的图片格式！");

// return false;

// }

if (size > 5242880) {

alert("请选择5M以内的图片！");

return false;

}

}