我有一個字符串形式爲「IT/Internet/Web Development/Ajax」。我試圖創建一個PHP函數,分析它,並創建一個JSON對象像創建一個JSON對象
[{
"name": "IT",
"subcategories":[
{
"name": "Internet",
"subcategories" : [
{
"name": "Web Development",
"subcategories" : [
{
"name":"Ajax"
}]}]}]
我有寫一個函數問題,做到這一點。這是我迄今爲止。
$category = "IT /Internet /Web Development";
$categoryarray = split("\/", $category);
$categoryLength = count($categoryarray);
$subcategory_collection = array();
$lastCategory = array("name"=>$categoryarray[$categoryLength-1]);
array_push($subcategory_collection, $lastCategory);
for($i=$categoryLength-2; $i>=0; $i--) {
$subcategory = array("name" => $categoryarray[$i], "subcategories" => $subcategory_collection);
array_push($subcategory_collection, $subcategory);
}
這不會產生所需的輸出。我希望該函數能夠解析任何以「父/子/孫/孫子孫」形式出現的字符串,並將其轉換爲JSON對象。如果有人能指導我在正確的方向,那將不勝感激
可能重複[創建JSON對象的正確方法(http://stackoverflow.com/questions/3281354/create-json-object-the-correct-way) –