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Detailed explanation of json array operation and each traversal example code under jquery

伊谢尔伦
伊谢尔伦Original
2017-07-17 14:37:591130browse

Traversal is used more when processing JSON arrays in jquery, but adding and removing them does not seem to be too much.

1. Creation of array

var arrayObj = new Array(); //创建一个数组 
var arrayObj = new Array([size]); //创建一个数组并指定长度,注意不是上限,是长度 
var arrayObj = new Array([element0[, element1[, ...[, elementN]]]]); //创建一个数组并赋值

It should be noted that although the second method creates an array with a specified length, in fact the array is variable length in all cases, that is to say Even if a length of 5 is specified, elements can still be stored outside the specified length. Note: the length will change accordingly.
2. Access to array elements

var testGetArrValue=arrayObj[1]; //获取数组的元素值 
arrayObj[1]= "这是新值"; //给数组元素赋予新的值

3. Addition of array elements

arrayObj. push([item1 [item2 [. . . [itemN ]]]]);// 将一个或多个新元素添加到数组结尾,并返回数组新长度 
arrayObj.unshift([item1 [item2 [. . . [itemN ]]]]);// 将一个或多个新元素添加到数组开始,数组中的元素自动后移,返回数组新长度 
arrayObj.splice(insertPos,0,[item1[, item2[, . . . [,itemN]]]]);//将一个或多个新元素插入到数组的指定位置,插入位置的元素自动后移,返回""。

4. Deletion of array elements

arrayObj.pop(); //移除最后一个元素并返回该元素值 
arrayObj.shift(); //移除最前一个元素并返回该元素值,数组中元素自动前移 
arrayObj.splice(deletePos,deleteCount); //删除从指定位置deletePos开始的指定数量deleteCount的元素,数组形式返回所移除的元素

5 , interception and merging of arrays

arrayObj.slice(start, [end]); //以数组的形式返回数组的一部分,注意不包括 end 对应的元素,如果省略 end 将复制 start 之后的所有元素 
arrayObj.concat([item1[, item2[, . . . [,itemN]]]]); //将多个数组(也可以是字符串,或者是数组和字符串的混合)连接为一个数组,返回连接好的新的数组

6, copying of arrays

arrayObj.slice(0); //返回数组的拷贝数组,注意是一个新的数组,不是指向 
arrayObj.concat(); //返回数组的拷贝数组,注意是一个新的数组,不是指向

7, sorting of array elements

arrayObj.reverse(); //反转元素(最前的排到最后、最后的排到最前),返回数组地址 
arrayObj.sort(); //对数组元素排序,返回数组地址

8, stringification of array elements

arrayObj.join(separator); //返回字符串,这个字符串将数组的每一个元素值连接在一起,中间用 separator 隔开。 
toLocaleString 、toString 、valueOf:可以看作是join的特殊用法,不常用

A simple jQuery example to iterate over a JavaScript array object.

var json = [
{"id":"1","tagName":"apple"},
{"id":"2","tagName":"orange"},
{"id":"3","tagName":"banana"},
{"id":"4","tagName":"watermelon"},
{"id":"5","tagName":"pineapple"}
];
 
$.each(json, function(idx, obj) {
alert(obj.tagName);
});

The above code snippet works fine, prompting "apple", "orange"... etc., as expected.
Issue: JSON String

The following example declares a JSON string (enclosed in single or double quotes) directly.

var json = '[{"id":"1","tagName":"apple"},{"id":"2","tagName":"orange"},
{"id":"3","tagName":"banana"},{"id":"4","tagName":"watermelon"},
{"id":"5","tagName":"pineapple"}]';
 
$.each(json, function(idx, obj) {
alert(obj.tagName);
});

In Chrome, it shows the following error in the console:

Uncaught TypeError: Cannot use 'in' operator to search for '156'
in [{"id" :"1","tagName":"apple"}...

Solution: Convert JSON string to JavaScript object.
To fix it, convert it to a JavaScript object via standard JSON.parse() or jQuery's $.parseJSON.

{"id":"3","tagName":"banana"},{"id":"4","tagName":"watermelon"},
{"id":"5","tagName":"pineapple"}]';
 
$.each(JSON.parse(json), function(idx, obj) {
alert(obj.tagName);
}); 
//or
$.each($.parseJSON(json), function(idx, obj) {
alert(obj.tagName);
});


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