我有一个带有一些元素的json,在分支上我有一个名为cars的数组
for(var i=0;i<itemsArray;i++)
{
$('.red').append('<p class="col-xs-2 no-padding text-format">' + item[i].cars + '</p>');
}
数组如何显示:
"cars": ["dawdawd","wadwad","wadwad","wad"];
我需要的是将这个数组中的每个元素放在一个段落上,比如我正在使用append;
例如:
<p class="col-xs-2 no-padding text-format">dawdawd</p>
<p class="col-xs-2 no-padding text-format">wadwad</p>
修改
完整数组:
"description": "asdasd",
"name": "geaaar",
"shop": "http://google.com",
"date": "2016-12-31T00:00:00.000Z",
"cars": [
"dawdawd",
"wadwad",
"wadwad",
"wad"
],
var itemsArray = data.data.my_Array.length; //length of the array
答案 0 :(得分:2)
您应该使用length
属性来iterate
数组项。
for(var i=0;i<itemsArray.length;i++)
^^^^^^^^
{
$('.red').append('<p class="col-xs-2 no-padding text-format">' + itemsArray[i].cars + '</p>');
}
此外,您可以使用forEach
方法接受作为参数的回调函数。
itemsArray.forEach(function(item){
$('.red').append('<p class="col-xs-2 no-padding text-format">' + item.cars + '</p>');
});
答案 1 :(得分:0)
尝试
var paras ='';
for(var i=0;i<itemsArray.length;i++) {
paras += '<p>'+itemsArray[i]+'</p>';
}
循环后,
$('.red').html(paras);
答案 2 :(得分:0)
var obj = {
"description": "asdasd",
"name": "geaaar",
"shop": "http://google.com",
"date": "2016-12-31T00:00:00.000Z",
"cars": [
"dawdawd",
"wadwad",
"wadwad",
"wad"
]
};
var $container = $(".red");
obj.cars.forEach(function(car) { // for each car in obj.cars array
$container.append("<p>" + car + "</p>"); // add the car as a paragraph inside the div .red
});
&#13;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="red"></div>
&#13;
答案 3 :(得分:0)
在阵列上使用join
。
var obj = {
"description": "asdasd",
"name": "geaaar",
"shop": "http://google.com",
"date": "2016-12-31T00:00:00.000Z",
"cars": [
"dawdawd",
"wadwad",
"wadwad",
"wad"
]
};
$('.red').append(obj.cars.length ? '<p>' + obj.cars.join('</p><p>') + '</p>' : '');
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="red"></div>