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图片上传后端处理的几种方式

图片上传后端处理的几种方式

作者: 大饼脸me | 来源:发表于2017-09-27 17:17 被阅读133次

图片上传前端可以采用两种方式:

  • formData
  • base64
    下面的例子都是采用formData来上传图片的
一、前端部分
<div>
    <input type="file" id="file" name="file" multiple>
    <input type="submit" class="submit" value="Upload">
</div>
<script type="text/javascript">   
    $(".submit").on("click", function(){
        var file = $("#file")[0].files[0];
        var formData = new FormData();
        formData.append('file', file);

        $.ajax({
            url: "/api/upload",
            method: "POST",
            data: formData,
            processData: false,
            contentType: false,
            success: function(){ 

            },
            error: function(){

            }
        })
    })
</script>

Q:为何不用form?
A:使用form提交,一般会这样写

<form action="/uploads" method="POST" enctype="multiple/form-data">
  <input type="file" name="file">
</form>

这种form表单提交,会跳转页面,就算去掉action,改成dosubmits="fn();",也还是会跳转页面,具体的有时间再去研究一下。

二、后端部分

1、方式一:简单粗暴的读取图片与写入图片,没考虑图片读取与写入过程中的出错情况。

const router = require('koa-router')()
const multer = require('koa-multer')
const fs = require('fs')
const path = require('path')

const upload = multer({ dest: __dirname + '/uploads/' });

router.post('/upload', upload.single('file'), async function(ctx, next){
    var file = ctx.req.file;
    if (file != null) {
        const reader = fs.createReadStream(file.path);
        const filePath = path.join(path.resolve(__dirname, '..'), '/uploads');
        const fileName = Math.random().toString(16).substr(2) + '.' + getSuffixName(file.originalname);
        const uploadFilePath = path.join(filePath, fileName);
    const stream = fs.createWriteStream(uploadFilePath);
    reader.pipe(stream);
    console.log('uploading %s -> %s', file.name, stream.path);
    } else {
        console.log('No upload file!');
    }
    ctx.body = 'upload success';
});

/**
 * 获取上传文件的后缀名
 * @param  {string} fileName 获取上传文件的后缀名
 * @return {string}          文件后缀名
 */
function getSuffixName( fileName ) {
  let nameList = fileName.split('.')
  return nameList[nameList.length - 1]
}

2、方式二:使用busboy

const router = require('koa-router')()
const Busboy = require('busboy')
const path = require('path')

router.post('/upload', async function(ctx, next){
    var result = { success: false };
    var filePath = path.join(path.resolve(__dirname, '..'), '/public/images/uploads');

    result = await uploadFile(ctx, {
        path: filePath
    });
    ctx.body = result;
});

function uploadFile(ctx, options){
    var busboy = new Busboy({ headers: ctx.req.headers });

    return new Promise((resolve, reject) => {
        console.log("文件上传中");
        var result = { 
            success: false,
            message: '',
            data: null   
        };

        // 解析请求文件事件
        busboy.on('file', function(fieldname, file, filename, encoding, mimetype){
            var filePath = options.path;
            var fileName = Math.random().toString(16).substr(2) + '.' + getSuffixName(filename);
            var uploadPath = path.join(filePath, fileName);

            //文件保存到指定目录
            file.pipe(fs.createWriteStream(uploadPath));

            // 文件写入事件结束
            file.on('end', function() {
                result.success = true;
                result.message = '文件上传成功';
                result.data = {
                  pictureUrl: `http://${ctx.host}/images/uploads/${fileName}`
                }
                console.log('文件上传成功!')
                resolve(result)
            })
        });

        busboy.on('finish', function() {
            console.log('文件上传结束');
            resolve(result);
        });  

        busboy.on('error', function(){
            console.log('文件上传出错');
            reject(result);
        });

        ctx.req.pipe(busboy);

    })
}

/**
 * 获取上传文件的后缀名
 * @param  {string} fileName 获取上传文件的后缀名
 * @return {string}          文件后缀名
 */
function getSuffixName( fileName ) {
  let nameList = fileName.split('.')
  return nameList[nameList.length - 1]
}

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