1
如果我的json_encode輸出是這樣的。我想輸出json_encode數組到javascript數組
[{"id":"nameid","src":"http#"},{"id":"nameid","src":"http#"}]
我怎麼會變成這樣的:
[["name1","address1"],["name2","address2"]]
如果我的json_encode輸出是這樣的。我想輸出json_encode數組到javascript數組
[{"id":"nameid","src":"http#"},{"id":"nameid","src":"http#"}]
我怎麼會變成這樣的:
[["name1","address1"],["name2","address2"]]
假設你的意思是name1
是從第一個對象的值3210,等...
其轉換爲陣列格式你想在傳遞給json_encode()
之前。像這樣的東西可能會爲你工作:
// Assuming your objects are an array $objects
$output_array = array();
foreach ($objects as $o) {
// Put the two properties from the object into an array
// and append it to $output_array
$output_array[] = array($o->id, $o->src);
}
// Encode the array as json
$json = json_encode($output_array);
var obj, result, source, _i, _len;
source = [
{
"id": "nameid",
"src": "http#"
}, {
"id": "nameid",
"src": "http#"
}
];
result = [];
for (_i = 0, _len = source.length; _i < _len; _i++) {
obj = source[_i];
result.push([obj.id, obj.src]);
}
(這是由CoffeeScript中產生的CoffeeScript源FYI要小得多。)
source = [{"id":"nameid","src":"http#"},{"id":"nameid","src":"http#"}]
result = []
result.push([obj.id, obj.src]) for obj in source
除了從一個陣列有兩個成員,這兩個數據結構似乎沒有任何共同之處,因此沒有合理的方式來執行轉換。 – Quentin