Question

我有一个我想要过滤的对象。这就是我使用的：

query = {
   "teststring-123": "true",
   "teststring-12344566": "false",
   test: "true"
}

我想过滤查询，以便在过滤后我只有：

query = {
   "teststring-123": "true",
   "teststring-12344566": "false"
}

$(query).each(function(index, value) {
    $.each(value, function(i, v) {
        if(i.indexOf('teststring-')==-1) {
            // remove part of query object where index is this one       
            console.log(index)
        }
    });     
});

我该如何处理？

Answer 1

您是否尝试删除所有没有以＆＃34; teststring开头的键的键值对 - ＆＃34;？如果是的话......

for(var key in query){
    if(query.hasOwnProperty(key) && key.indexOf('teststring-') === -1){
        delete query[key];
    }
}

Answer 2

使用delete运算符：

$.each(query, function(key, value) {
    if(key.indexOf('teststring-') == -1) {
        delete query[key];
    }
});

http://jsfiddle.net/NvZyA/（在此演示中，Object.keys()用于显示所有密钥）。

Answer 3

您可能正在寻找delete运营商。

Answer 4

使用delete运算符：

var query = {
   "teststring-123": "true",
   "teststring-12344566": "false",
   test: "true"
}
$.each(query, function(sKey) {
    if (sKey.indexOf("teststring-") < 0) { // or check if it is not on first position: != 0
        delete query[sKey];
    }  
});

Answer 5

正如其他人所说，使用delete运算符。但是不需要迭代对象的属性：

var query = {
    "teststring-123" : true,
    "teststring-12344566" :  false,
    test: true
};

delete query['test'];

Answer 6

喜欢这个吗？

var query = {
    "teststring-123": "true",
    "teststring-12344566": "false",
    "test": "true"
}
var newobj = {};

$.each(query, function(i, v) {
    if(i.indexOf('teststring-') != -1) {
        // remove part of query object where index is this one        
        console.log(i);
        newarr[i] = v;
    }
});
console.log(newobj);

按键名过滤对象

6 个答案: