我有一个返回此json的php文件(我已经验证了它):
{"images":{"2":"building.jpg","3":"campus.jpg","4":"grads.jpg","5":"trio.jpg"},"videos":[]}
在jquery中,如何返回"图像"的子级数量?当我使用:
$.post('php/file.php', {variable: variable}, function(returnedData) {
console.log(returnedData);
var obj = jQuery.parseJSON( returnedData );
console.log(obj.images.length);
});
我得到"未定义"。
这是我在 php 中构建 json 的方式?
$variable= $_POST['variable'];
$imgdir = '../' . $variable. '/img';
$weeds = array('.', '..');
$images = array_diff(scandir($imgdir), $weeds);
$viddir = '../' . $school . "/vid";
$vids = array_diff(scandir($viddir), $weeds);
$data = array();
$data['images'] = $images;
$data['videos'] = $vids;
echo json_encode($data);
发生这种情况是因为obj.images
不是Array
,你可以这样length
var returnedData = {"images":{"2":"building.jpg","3":"campus.jpg","4":"grads.jpg","5":"trio.jpg"},"videos":[]};
console.log(Object.keys(returnedData.images).length);
Object.keys() - 返回给定对象自己的数组 可枚举属性
或者你可以使用for..in
,就像这样
var returnedData = {"images":{"2":"building.jpg","3":"campus.jpg","4":"grads.jpg","5":"trio.jpg"},"videos":[]};
var count = 0;
for (var i in returnedData.images) {
if (returnedData.images.hasOwnProperty(i)) {
count += 1;
}
}
console.log(count);
为什么不使用 $.getJSON():
$.getJSON('php/file.php', {variable: variable}, function(returnedData) {
console.log(returnedData);
console.log(Object.keys(returnedData.images).length);
});