可能重复:
Javascript: Determine whether an array contains a value
var thelist = new Array();
function addlist(){
thelist.push(documentgetElementById('data').innerHTML);
}
如何检查我推送的数据是否已存在于数组thelist
中?
答案 0 :(得分:4)
var thelist = []; // Use the array literal, not the constructor.
function addlist(){
// get the data we want to make sure is unique
var data = documentgetElementById('data').innerHTML;
// make a flag to keep track of whether or not it exists.
var exists = false;
// Loop through the array
for (var i = 0; i < thelist.length; i++) {
// if we found the data in there already, flip the flag
if (thelist[i] === data) {
exists = true;
// stop looping, once we have found something, no reason to loop more.
break;
}
}
// If the data doesn't exist yet, push it on there.
if (!exists) {
thelist.push(data);
}
}
答案 1 :(得分:1)
如果你不关心IE&lt; 9你也可以使用Array方法“some”。 看看这个例子:
var thelist = [1, 2, 3];
function addlist(data) {
alreadyExists = thelist.some(function (item) {
return item === data
});
if (!alreadyExists) {
thelist.push(data);
}
}
addlist(1);
addlist(2);
addlist(5);
console.log(thelist);
有些人确定是否存在至少一个具有给定约束的元素(回调返回值=== true)是否存在。
https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/Array/some
答案 2 :(得分:0)
如果您不关心IE 8或更低版本,可以使用Array.filter
:
var thelist = new Array();
function addlist(){
var val = documentgetElementById('data').innerHTML;
var isInArray = theList.filter(function(item){
return item != val
}).length > 0;
if (!isInArray)
thelist.push(val);
}
或者,您可以使用Array.indexOf
:
var thelist = new Array();
function addlist(){
var val = documentgetElementById('data').innerHTML;
var isInArray = theList.indexOf(val) >= 0;
if (!isInArray)
thelist.push(val);
}
答案 3 :(得分:0)
查看underscore.js
:underscore.js
然后你可以检查数组
_.contains(thelist, 'value you want to check');
// The full example
var thelist = new Array();
function addlist(){
var data = documentgetElementById('data').innerHTML;
if(!_.contains(thelist, data)) theList.push(data);
}
或者您可以将值添加到数组而不考虑重复值,并且在添加过程完成后,您可以通过
删除重复的元素theList = _.uniq(theList);
第二种方法当然效率低下。