Question

我有一个带有+/- 300项的巨大JSON字符串。一个缩短的例子：

[
{"DateGps":"2016-03-25T19:28:19+01:00","DateReceived":"2016-03-25T19:28:20.163+01:00","Longitude":5.85294,"Latitude":51.84475,"Speed":55,"VehicleNumber":"678","TravelNumber":"4321"},
{"DateGps":"2016-03-25T19:28:13+01:00","DateReceived":"2016-03-25T19:28:14.065+01:00","Longitude":4.8139,"Latitude":52.43844,"Speed":23,"VehicleNumber":"2335","TravelNumber":"1234"}
]

用

$array = json_decode($json,true); // i will convert this into a array.

但是我怎样才能得到VehicleNumber与VehicleNumber 2335相同的行？

Answer 1

您可以像{/ p>一样使用array_filter

$array = json_decode($json,true);
$resulted_array = array_filter($array,function($v){ return ($v['VehicleNumber'] == 2335);});
print_r($resulted_array);

Answer 2

与Uchiha's answer几乎相同，但使用anonymous class作为FilterIterator。
在这种情况下，它只增加（不必要的）复杂性（特别是因为json_decode（，true）返回一个数组，因此无论如何整个数据都在内存中）。但是如果你有大量的数据和，你可以一步一步地接收它，迭代器和/或generators可能会派上用场; - ）

<?php
$data = json_decode(data(), true);
$it = new class(new ArrayIterator($data)) extends FilterIterator {
    public function accept() { return '2335'==$this->current()['VehicleNumber']; }
};

foreach($it as $v) {
    var_export($v);
}

function data() {
    return <<< eoj
[
{"DateGps":"2016-03-25T19:28:19+01:00","DateReceived":"2016-03-25T19:28:20.163+01:00","Longitude":5.85294,"Latitude":51.84475,"Speed":55,"VehicleNumber":"678","TravelNumber":"4321"},
{"DateGps":"2016-03-25T19:28:13+01:00","DateReceived":"2016-03-25T19:28:14.065+01:00","Longitude":4.8139,"Latitude":52.43844,"Speed":23,"VehicleNumber":"2335","TravelNumber":"1234"}
]
eoj;
}

PHP数组：如何过滤参数

2 个答案: