Question

我有一个非常嵌套的JSON文件。如何删除包含我的值的数组？例如：我想从JSON文件中删除{"customer":"Customer1","date":"2017-06-03"...}，我已经知道“Customer1”

{
            "info": [{
                "customer": "Customer1",
                "date": "2017-06-03",
                "beacons": [{
                    "data1": "1234",
                    "data2": "Test1",
                }, {
                    "data1": "0088",
                    "data2": "Test2",
                }]
            },{
                "customer": "Customer2",
                "date": "2017-06-03",
                "beacons": [{
                    "data1": "dcdd4",
                    "data2": "Test3",
                }, {
                    "data1": "0088",
                    "data2": "Test4",
                }]
            }]
        }

谢谢！

Answer 1

您的json数据存在一些问题。这不是有效的json数据;我解码了josn数据＆amp;然后检查＆＃34;客户＆＃34; value =＆＃39; Customer1＆＃39;，然后从主数组中删除数组。

应该是这样的：

$jsonData = '{"info ": [{
    "customer ": "customer1 ",
    "date ": "2017 - 06 - 03 ",
    "beacons ": [{
        "data1 ": "1234",
        "data2 ": "Test1"
    }]
}, {
    "customer": "customer2 ",
    "date": "2017 - 06 - 04 ",
    "beacons": [{
        "data1": "dcdd4",
        "data2": "Test3"
    }]
 }]
}';

$myData = json_decode($jsonData,true);


foreach($myData["info"] as $k=>$arr) {
    if($arr["customer"] == "customer1") {
       unset($myData["info"][$k]);
    }
}

Answer 2

使对象分配给一个obj示例：=＆gt;

var myobj={"info": [{ ...... }]}

。这个下面的函数将用于搜索对象数组中的确切值使用如下所示的参数调用该函数

searchObj (myobj, Customer1);

function searchObj (obj_name, searchingval) {
    for (var key in obj_name) {
        var value = obj_name[key];

        if (typeof value === 'object') {
            searchObj(value, searchingval);
        }

        if (value === searchingval) {
            console.log('property name=' + key + '    property value=' + value);
        }

    }
}

Answer 3

上一个答案中的一个小修正。

请拨打这样的功能。搜索值应该是astring格式的值。

searchObj (myobj, 'Customer1');

PHP：搜索并删除JSON对象

3 个答案: