在test2中,您可以看到,它不是从数组中删除节点,而是用'null'替换节点。
我做错了什么,如何完全删除?
编辑:使用Splice而不是删除。更新了小提琴:http://jsfiddle.net/J2KuY/1/
答案 0 :(得分:7)
答案 1 :(得分:3)
//say you have these arrays
var test1 = [{
"Country": "Spain",
"info info1": 0.329235716,
"info info2": 0.447683684,
"info info3": 0.447683747},
{
"Country": "Chile",
"info info1": 1.302673893,
"info info2": 1.357820775,
"info info3": 1.35626442},
{
"Country": "USA",
"info info1": 7.78805016,
"info info2": 26.59681951,
"info info3": 9.200900779}];
var test2 = [{
"Country": "Germany",
"info info1": 0.329235716,
"info info2": 0.447683684,
"info info3": 0.447683747},
{
"Country": "China",
"info info1": 1.302673893,
"info info2": 1.357820775,
"info info3": 1.35626442},
{
"Country": "France",
"info info1": 7.78805016,
"info info2": 26.59681951,
"info info3": 9.200900779}];
//similar to jQuery, wrap them in an object
function $W(param) {
var obj = {};
//set data
obj.data = param;
//augment the object with a remove function
obj.remove = function(key, val) {
var i = 0;
//loop through data
while (this.data[i]) {
if (this.data[i][key] === val) {
//if we have that data, splice it
//splice changes the array length so we don't increment
this.data.splice(i, 1);
} else {
//else move on to the next item
i++;
}
}
//be sure to return the object so that the chain continues
return this;
}
//return object for operation
return obj
}
//the following will look strangely similar to jQuery
//remove chile, then USA
var result1 = $W(test1).remove('Country', 'Chile').remove('Country', 'USA');
//remove germany and china
var result2 = $W(test2).remove('Country', 'China').remove('Country', 'Germany');
//should only have spain and france
$('#test2').val(JSON.stringify(result1)+JSON.stringify(result2));