Well, right now I am sending JSON to my php script which then uses json_decode() to make them into vars. But, I am wondering, is sending JSON the best way? Or should I just send the data as a normal query string, rather than JSON? Which is best?
On Jun 22, 5:13 pm, cs <chuck.schleut...@gmail.com> wrote: > Why not make each field that has a nested key-value pair an object > instead? In short, any array with a key but without a value is an > object since your value is just an array of elements. What language is > in charge of your back-end? And how is it handling the inputs your > pass it? This is what really counts. > > JSON is > > On Jun 22, 7:01 pm, Nic Hubbard <nnhubb...@gmail.com> wrote: > > > > > Yeah, I am not too hot on using a plugin for this. I had hoped there > > was just a quick function that I could pass in an array and return > > JSON. > > > On Jun 22, 8:28 am, diogobaeder <diogobae...@gmail.com> wrote: > > > > Hmmm... why would one use this plugin if we have Douglas Crockford's > > > json2.org, which is the original JSON library from the same creator of > > > the concept "JSON"? And why load more modules into the main jQuery > > > object, making it heavier, if we don't use DOM element wrappers in the > > > plugin, only conversions to/from JS objects? > > > > Sorry, I don't mean to be harsh, but it seems to me that this plugin > > > is a reinvented wheel, and also that it didn't have to be a "jQuery > > > plugin" to work. Does it have anything that we can't see in json2.org? > > > > Diogo > > > > On Jun 21, 11:39 pm, kranthi <kranthi...@gmail.com> wrote: > > > > > are u looking forhttp://www.google.co.in/search?q=jquery+json+plugin?
