Question

我有一个AJAX调用从Github API获取信息。它以base64编码返回，但是当我尝试解码时，我得到了上述错误。

之前有没有人遇到这个并知道发生了什么？服务器返回的信息通过基于Web的解码器运行良好。也许它在变量中有点受损？ IDK：/

(function () {
    'use strict';

    var objectsList = [];
    var isParsing = 0;
    var insertArea = $('body');

    function makeAJAXCall(hash, cb) {
        $.ajaxSetup({
            Accept: 'Application/vnd.github.raw+json',
            dataType: 'jsonp'
        });

        $.ajax({
            url: hash,
            success: function (json) {
                if (cb) {
                    cb(json);
                }
            },
            error: function (error) {
                console.error(error);
                throw error;
            }
        });
    }

    function parseBlob(hash, cb) {
        makeAJAXCall(hash, function (returnedJSON) {  // no loop as only one entry
            if (cb) {
                cb(returnedJSON.data.content);
            }
        });
    }

    function addSVGToPage(SVGToAdd) {
        var entry, decodedEntry;
        makeAJAXCall(SVGToAdd, function (returnedJSON) {
            console.info(window.atob(returnedJSON.data.content));
        });
    }

    function parseTree(hash) {
        var i, entry;
        var tree = 'https://api.github.com/repos/myusername/SVG-Shapes/git/trees/' + hash;

        makeAJAXCall(tree, function (returnedJSON) {
            for (i = 0;  i < returnedJSON.data.tree.length; i += 1) {
                entry = returnedJSON.data.tree[i];

                if (entry.type === 'blob') {
                    if (entry.path.slice(-4) === '.svg') {     // we only want the svg images not the ignore file and README etc
                        addSVGToPage(entry.url);
                    }
                } else if (entry.type === 'tree') {
                    parseTree(entry.sha);
                }
            }
        });
    }

    $(document).ready(function () {
        parseTree('master');
    });
}());

编辑：下面是从console.info返回的内容示例（returnedJSON.data.content）;

PD94bWwgdmVyc2lvbj0iMS4wIiBlbmNvZGluZz0idXRmLTgiPz4KPCFET0NU WVBFIHN2ZyBQVUJMSUMgIi0vL1czQy8vRFREIFNWRyAxLjEvL0VOIiAiaHR0 cDovL3d3dy53My5vcmcvR3JhcGhpY3MvU1ZHLzEuMS9EVEQvc3ZnMTEuZHRk Ij4KPHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHht bG5zOnhsaW5rPSJodHRwOi8vd3d3LnczLm9yZy8xOTk5L3hsaW5rIiB2ZXJz aW9uPSIxLjEiIGhlaWdodD0iMjQiIHdpZHRoPSI0MDAiPgkKCTxkZWZzPgoJ CTxwYXR0ZXJuIGlkPSJzdHJpcGVkYmFjayIgaGVpZ2h0PSIyMCIgd2lkdGg9 IjIwIiB5PSIyIiBwYXR0ZXJuVW5pdHM9InVzZXJTcGFjZU9uVXNlIiBvdmVy Zmxvdz0ic2Nyb2xsIj4KCQkJPGcgdHJhbnNmb3JtPSJza2V3WCgtMjApIj4K CQkJCTxyZWN0IHg9Ii0xMCIgeT0iMCIgd2lkdGg9IjEwIiBoZWlnaHQ9IjIw IiBzdHJva2U9Im5vbmUiIGZpbGw9IiNGMkYyRjIiLz4KCQkJCTxyZWN0IHg9 IjAiIHk9IjAiIHdpZHRoPSIxMCIgaGVpZ2h0PSIyMCIgc3Ryb2tlPSJub25l IiBmaWxsPSIjRTJFMkUyIi8+CgkJCQk8cmVjdCB4PSIxMCIgeT0iMCIgd2lk dGg9IjEwIiBoZWlnaHQ9IjIwIiBzdHJva2U9Im5vbmUiIGZpbGw9IiNGMkYy RjIiLz4KCQkJCTxyZWN0IHg9IjIwIiB5PSIwIiB3aWR0aD0iMTAiIGhlaWdo dD0iMjAiIHN0cm9rZT0ibm9uZSIgZmlsbD0iI0UyRTJFMiIvPgoJCQk8L2c+ CgkJPC9wYXR0ZXJuPgoKCQk8cmVjdCBpZD0iaW5uZXJjdXQiIHg9IjIiIHkg PSIxIiB3aWR0aD0iMzk2IiBoZWlnaHQ9IjIyIi8+CgoJCTxjbGlwUGF0aCBp ZD0ibG9hZGNsaXAiPgoJCQk8dXNlIHhsaW5rOmhyZWY9IiNpbm5lcmN1dCIv PgoJCTwvY2xpcFBhdGg+Cgk8L2RlZnM+CgoJPHJlY3Qgd2lkdGg9IjQwMCIg aGVpZ2h0PSIyNCIgeD0iMCIgeT0iMCIgc3R5bGU9InN0cm9rZS13aWR0aDoy cHg7c3Ryb2tlOmdyZXk7Ii8+ICA8IS0tIEJhY2tpbmcgcmVjdC9mcmFtZSAt LT4KCgk8ZyBjbGlwLXBhdGg9InVybCgjbG9hZGNsaXApIj4KCQk8cmVjdCB4 PSItMTgiIHkgPSIyIiB3aWR0aD0iNTE2IiBoZWlnaHQ9IjIwIiBzdHlsZT0i c3Ryb2tlLXdpZHRoOjBweDtzdHJva2U6Z3JleTsiIGZpbGw9InVybCgjc3Ry aXBlZGJhY2spIj4KCQkJPGFuaW1hdGVUcmFuc2Zvcm0KCQkgICAgICAgIGF0 dHJpYnV0ZVR5cGU9IlhNTCIKCQkgICAgICAgIGF0dHJpYnV0ZU5hbWU9InRy YW5zZm9ybSIKCQkgICAgICAgIHR5cGU9InRyYW5zbGF0ZSIKCQkJCWZyb209 IjIwIiB0bz0iMCIKCQkJCWJlZ2luPSIwcyIgZHVyPSIwLjZzIgoJCQkJcmVw ZWF0Q291bnQ9ImluZGVmaW5pdGUiCgkJCQlhZGRpdGl2ZT0ic3VtIi8+CgkJ PC9yZWN0PgoJPC9nPgo8L3N2Zz4K

Answer 1

atob由于响应中的换行符而中断。删除它们以使您的代码正常工作：

function addSVGToPage(SVGToAdd) {
    var entry, decodedEntry;         // <-- What is this doing here? It's unused.
    makeAJAXCall(SVGToAdd, function (returnedJSON) {
        console.info(window.atob(returnedJSON.data.content.replace(/\s/g, '')));
        //                                                ^^^^^^^^^^^^^^^^^^^
    });
}

Answer 2

根据MDN docs，您可能需要escape然后decodeURIComponent来处理unicode：

function utf8_to_b64( str ) {
    return window.btoa(unescape(encodeURIComponent( str )));
}

function b64_to_utf8( str ) {
    return decodeURIComponent(escape(window.atob( str )));
}

// Usage:
utf8_to_b64('✓ à la mode'); // "4pyTIMOgIGxhIG1vZGU="
b64_to_utf8('4pyTIMOgIGxhIG1vZGU='); // "✓ à la mode"

Answer 3

我在这里有一个更好的解决方案：

实时演示：https://codepen.io/arliang/pen/mNQayE?editors=1111

// https://best33.com/311.moe

function base64encode(str) {
  let encode = encodeURIComponent(str).replace(/%([a-f0-9]{2})/gi, (m, $1) => String.fromCharCode(parseInt($1, 16)))
  return btoa(encode)
}
function base64decode(str) {
  let decode = atob(str).replace(/[\x80-\uffff]/g, (m) => `%${m.charCodeAt(0).toString(16).padStart(2, '0')}`)
  return decodeURIComponent(decode)
}

function tests() {
  Array.from([
    /*鼠、牛、虎、兔、龙、蛇、马、羊、猴、鸡、狗、猪*/
    '????????????',
    '测试'
  ])
  .forEach(text => {
    try{
      console.assert(btoa(atob(text)) === text, `btoa(atob('${text}')) === '${text}' fail`)
    } catch(e) {
      // console.error(`btoa(atob('${text}')) === '${text}' fail`, e)
    }
    console.assert(base64decode(base64encode(text)) === text, 'pass')
  })
  console.log(`Open the console and click this link?: data:text/html;base64,${base64encode('????????????')}`)
}
tests()

javascript atob返回'字符串包含无效字符'

3 个答案: