Mik*_*ike 20 javascript base64 geolocation
我正在尝试在网站上实现一个简单的脚本,该脚本将从google的ajax API返回base64编码信息.这就是我到目前为止所玩的:
<html>
<head>
<script src="http://www.google.com/jsapi?key=ABQIAAAA0duujonFsEX871htGWZBHRS76H0qhS7Lb-D1Gd0Mnaiuid8Z7BQIyz2kMpojKizoyiCQA4yRkKAKug" type="text/javascript"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"></script>
<script type="text/javascript">
jQuery(document).ready(function() {
var location = 'Unable to determine your location.';
if (google.loader.ClientLocation) {
var loc = google.loader.ClientLocation;
location = 'Country: <strong>' + loc.address.country + '</strong>, Region: <strong>' + loc.address.region + '</strong>, City: <strong>' +
loc.address.city + '</strong>, Lat/Long: <strong>' + loc.latitude + ', ' + loc.longitude + '</strong>';
}
jQuery('.geolocation').html(location);
});
</script>
</head>
<body>
<span class="geolocation"></span>
</body>
</html>
它返回我正在尝试正确的信息,但我需要base64编码单独的部分,如国家,地区,城市,纬度和经度.在PHP中它会很简单,但我无法弄清楚如何在javascript中做到这一点.任何帮助,将不胜感激.
Tim*_*own 20
Mozilla中,WebKit和Opera都具有btoa()和atob()分别用于基部64的编码和解码的功能.尽可能使用它们,因为它们几乎肯定会比JavaScript实现快得多,并且会回到在进行Web搜索时出现的许多脚本之一.
编辑2013年9月10日:atob()并且btoa()不处理ASCII范围之外的Unicode字符.MDN有解决方法,但我无法保证.感谢@larspars指出这一点.
例如,如果您使用的是amphetamachine答案中的示例,则可以执行以下操作:
if (!window.btoa) {
window.btoa = function(str) {
return Base64.encode(str);
}
}
if (!window.atob) {
window.atob = function(str) {
return Base64.decode(str);
}
}
alert( btoa("Some text") );
amp*_*ine 15
这个答案似乎符合您的要求.
还有一个更优雅的:
/**
*
* Base64 encode / decode
* http://www.webtoolkit.info/
*
**/
var Base64 = {
// private property
_keyStr : "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=",
// public method for encoding
encode : function (input) {
var output = "";
var chr1, chr2, chr3, enc1, enc2, enc3, enc4;
var i = 0;
input = Base64._utf8_encode(input);
while (i < input.length) {
chr1 = input.charCodeAt(i++);
chr2 = input.charCodeAt(i++);
chr3 = input.charCodeAt(i++);
enc1 = chr1 >> 2;
enc2 = ((chr1 & 3) << 4) | (chr2 >> 4);
enc3 = ((chr2 & 15) << 2) | (chr3 >> 6);
enc4 = chr3 & 63;
if (isNaN(chr2)) {
enc3 = enc4 = 64;
} else if (isNaN(chr3)) {
enc4 = 64;
}
output = output +
this._keyStr.charAt(enc1) + this._keyStr.charAt(enc2) +
this._keyStr.charAt(enc3) + this._keyStr.charAt(enc4);
}
return output;
},
// public method for decoding
decode : function (input) {
var output = "";
var chr1, chr2, chr3;
var enc1, enc2, enc3, enc4;
var i = 0;
input = input.replace(/[^A-Za-z0-9\+\/\=]/g, "");
while (i < input.length) {
enc1 = this._keyStr.indexOf(input.charAt(i++));
enc2 = this._keyStr.indexOf(input.charAt(i++));
enc3 = this._keyStr.indexOf(input.charAt(i++));
enc4 = this._keyStr.indexOf(input.charAt(i++));
chr1 = (enc1 << 2) | (enc2 >> 4);
chr2 = ((enc2 & 15) << 4) | (enc3 >> 2);
chr3 = ((enc3 & 3) << 6) | enc4;
output = output + String.fromCharCode(chr1);
if (enc3 != 64) {
output = output + String.fromCharCode(chr2);
}
if (enc4 != 64) {
output = output + String.fromCharCode(chr3);
}
}
output = Base64._utf8_decode(output);
return output;
},
// private method for UTF-8 encoding
_utf8_encode : function (string) {
string = string.replace(/\r\n/g,"\n");
var utftext = "";
for (var n = 0; n < string.length; n++) {
var c = string.charCodeAt(n);
if (c < 128) {
utftext += String.fromCharCode(c);
}
else if((c > 127) && (c < 2048)) {
utftext += String.fromCharCode((c >> 6) | 192);
utftext += String.fromCharCode((c & 63) | 128);
}
else {
utftext += String.fromCharCode((c >> 12) | 224);
utftext += String.fromCharCode(((c >> 6) & 63) | 128);
utftext += String.fromCharCode((c & 63) | 128);
}
}
return utftext;
},
// private method for UTF-8 decoding
_utf8_decode : function (utftext) {
var string = "";
var i = 0;
var c = c1 = c2 = 0;
while ( i < utftext.length ) {
c = utftext.charCodeAt(i);
if (c < 128) {
string += String.fromCharCode(c);
i++;
}
else if((c > 191) && (c < 224)) {
c2 = utftext.charCodeAt(i+1);
string += String.fromCharCode(((c & 31) << 6) | (c2 & 63));
i += 2;
}
else {
c2 = utftext.charCodeAt(i+1);
c3 = utftext.charCodeAt(i+2);
string += String.fromCharCode(((c & 15) << 12) | ((c2 & 63) << 6) | (c3 & 63));
i += 3;
}
}
return string;
}
}
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