This seems like a no brainer but surely there is either an internal js method or a jquery one to take a string like:
这似乎没什么好事但是肯定有一个内部的js方法或一个jquery来获取一个字符串:
intTime=1324443870&fltOriginalAmount=0.00&strOriginalCurrency=GBP
...then a lot more vals and turn it into a JSON object?
...然后更多的val并将其转换为JSON对象?
I had a dig around this site and google and surprisingly drew blanks... Anyone got an easy way to do this?
我在这个网站和谷歌进行了挖掘,并且令人惊讶地画了空白......任何人都有一个简单的方法来做到这一点?
2 个解决方案
#1
11
jQuery BBQ does exactly this. See $.deparam
, "The opposite of jQuery.param, pretty much."
jQuery BBQ正是这样做的。请参阅$ .deparam,“与jQuery.param完全相反。”
> var obj = $.deparam('intTime=1324443870&fltOriginalAmount=0.00&strOriginalCurrency=GBP')
> JSON.stringify(obj)
'{"intTime":"1324443870","fltOriginalAmount":"0.00","strOriginalCurrency":"GBP"}'
#2
2
i used this hack...
我用过这个黑客......
$.parseJSON('{"' + qs.replace(/&/g, '","').replace(/=/g, '":"') + '"}');
demo here http://jsbin.com/niqaw/
在这里演示http://jsbin.com/niqaw/
#1
11
jQuery BBQ does exactly this. See $.deparam
, "The opposite of jQuery.param, pretty much."
jQuery BBQ正是这样做的。请参阅$ .deparam,“与jQuery.param完全相反。”
> var obj = $.deparam('intTime=1324443870&fltOriginalAmount=0.00&strOriginalCurrency=GBP')
> JSON.stringify(obj)
'{"intTime":"1324443870","fltOriginalAmount":"0.00","strOriginalCurrency":"GBP"}'
#2
2
i used this hack...
我用过这个黑客......
$.parseJSON('{"' + qs.replace(/&/g, '","').replace(/=/g, '":"') + '"}');
demo here http://jsbin.com/niqaw/
在这里演示http://jsbin.com/niqaw/