在CodeIgniter中,在PHP中为下拉列表创建类似嵌套的元素<select>
我正在使用此函数:
function list_categories(&$result, $cats, $sub = ''){ // <- THIS
foreach($cats as $cat){
//$cat['category']->id = $sub.$cat['category']->title;
$result[$cat['category']->id] = $sub.$cat['category']->title; // <- THIS
if( sizeof($cat['children']) > 0 ){
$sub2 = str_replace('—→ ', '–', $sub);
$sub2.= '–→ ';
list_categories($result, $cat['children'], $sub2); // <- THIS
}
}
}
// Make an array available outside the function
$categoryData = array(0 => '-- '.lang('FORM_SELECT_CATTOP').' --');
// Launch the function to fill out the array with nested sets
list_categories($categoryData, $categories);
生成HTML <select>
列表:
我想在jQuery中从检索到的JSON中做同样的事情但是找不到比使用这个jQuery函数的第一级JSON对象更深入的答案:
$.getJSON('../raw_get_categories/'+com_id_selected+'/0', function(data){
var items = [];
$.each(data, function(key, value, sub){
items.push('<option id="'+ key[0] + '">' + value[0] + '</option>');
});
console.log(items);
});
这是我想要管理的JSON,并从PHP代码中检索相同的值:
{
"28": {
"category": {
"id": "28", // <<< I need to retrieve THIS
"pid": "0",
"com_id": "2",
"route_id": "59",
"excerpt": "",
"ordering": "1",
"title": "Oblecenie a vybava",
"slug": "oblecenie-a-vybava1",
"description": "",
"image": "clothes.png",
"seo_title": "",
"meta": ""
},
"compname": {
"id": "2",
"name": "E-commerce"
},
"children": {
"61": {
"category": {
"id": "61", // <<< I need to retrieve THIS
"pid": "28",
"com_id": "2",
"route_id": "52",
"excerpt": "",
"ordering": "1",
"title": "Ubor na hlavu",
"slug": "ubor-na-hlavu1",
"description": "<p>Capice, Klobuky, Panamy, Baretky, Prilby, Kukly<br></p>",
"image": "clothes-head.png",
"seo_title": "",
"meta": ""
},
"compname": {
"id": "2",
"name": "E-commerce"
},
"children": [ ]
},
"30": {
"category": {
"id": "30",
"pid": "28",
"com_id": "2",
"route_id": "53",
"excerpt": "",
"ordering": "2",
"title": "Bundy a vetrovky",
"slug": "bundy-a-vetrovky",
"description": "",
"image": "",
"seo_title": "",
"meta": ""
},
"compname": {
"id": "2",
"name": "E-commerce"
},
"children": {
"34": {
"category": {
"id": "34",
"pid": "30",
"com_id": "2",
"route_id": "0",
"excerpt": "",
"ordering": "0",
"title": "Letne",
"slug": "letne",
"description": "",
"image": "",
"seo_title": "",
"meta": ""
},
"compname": {
"id": "2",
"name": "E-commerce"
},
"children": [ ]
},
"35": {
"category": {
"id": "35",
"pid": "30",
"com_id": "2",
"route_id": "0",
"excerpt": "",
"ordering": "0",
"title": "Vsesezonne",
"slug": "vsesezonne",
"description": "",
"image": "",
"seo_title": "",
"meta": ""
},
"compname": {
"id": "2",
"name": "E-commerce"
},
"children": [ ]
},
"33": {
"category": {
"id": "33",
"pid": "30",
"com_id": "2",
"route_id": "0",
"excerpt": "",
"ordering": "0",
"title": "Zimne",
"slug": "zimne",
"description": "",
"image": "",
"seo_title": "",
"meta": ""
},
"compname": {
"id": "2",
"name": "E-commerce"
},
"children": [ ]
}
}
},
"31": {
"category": {
"id": "31",
"pid": "28",
"com_id": "2",
"route_id": "54",
"excerpt": "",
"ordering": "3",
"title": "Nohavice a teplaky",
"slug": "nohavice-a-teplaky",
"description": "",
"image": "",
"seo_title": "",
"meta": ""
},
"compname": {
"id": "2",
"name": "E-commerce"
},
"children": [ ]
},
"32": {
"category": {
"id": "32",
"pid": "28",
"com_id": "2",
"route_id": "55",
"excerpt": "",
"ordering": "4",
"title": "Obuv",
"slug": "obuv",
"description": "",
"image": "",
"seo_title": "",
"meta": ""
},
"compname": {
"id": "2",
"name": "E-commerce"
},
"children": [ ]
}
}
},
"38": {
...
}
有人可以建议我在jQuery中使用相同的foreach
语句吗?
答案 0 :(得分:3)
Juste使用积分来获得更多孩子
$.getJSON('filename.php', function(data) {
$('#List li').remove();
$.each(data, function(index, item) {
$('#list').append('<li>'+item.category.id+'</li>');
});
});
答案 1 :(得分:0)
您可以这样做:
$.each(data.actions, function(entryIndex, entry) {
var html = '<li class="top-level">' + this.action + '</li>';
});