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get value customize of output json_encode() by jQuery.?

开发者 https://www.devze.com 2023-04-01 21:08 出处:网络
I insert data with json_encode() in database, want get just values name_units on row units in database. how is it?

I insert data with json_encode() in database, want get just values name_units on row units in database. how is it?

This is output database in code php(by json_encode()):

my_table=>units=>name_units

[{"name":"jack","units":[{"name_units":"salam","price_units":"74开发者_运维问答,554","checkbox_units":["minibar","mobleman"]},{"name_units":"mokhles","price_units":"4,851,269","checkbox_units":["mobleman","tv"]},{"name_units":"fadat","price_units":"85,642","checkbox_units":["minibar","mobleman","tv"]}]}]

var dataObj = $(this).closest('form').serialize();

$.ajax({
    type: "POST",
    dataType: 'json',
    url: 'url',
    data: dataObj,
    cache: false,
    success: function(data) {
/////////////////////*HERE/////////////////////
        $.each(data, function(a, b) {
            //alert(b.units[name_units]);
            $('<p id="' + b.units[name_units] + '">' + b.units[name_units] + '</p>').appendTo('.class');
        });
/////////////////////HERE*/////////////////////
    };
})


Try this:

$.each(data, function(a, b) {
    $.each(b.units, function(c, d){
        $('<p id="' + d.name_units  + '">' + d.name_units + '</p>').appendTo('.class');
    });
});

Also, its a good idea to use jQuery templates for this scenario


units is an array, therefore you need to reference an item in the array or loop through all items in that array. b.units[0].name_units

Edit: like i said, you would have to loop through them.

$.each(b.units,function(i,unit){
  alert(unit.name_units);
});

you could also use a for loop if you would prefer.

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