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jquery.parsejson转义怎么用
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How to use jquery.parsejson escaping

Jun 15, 2022 am 10:58 AM
jquery

在jquery中,“jquery.parsejson()”函数用于将格式完好的JSON字符串转为与之对应的JavaScript对象;该函数要求指定的字符串必须符合严格的JSON格式,若JSON格式不完好则会在转义过程中抛出异常,语法为“$.parseJSON(需要解析并转为JS对象的JSON格式字符串)”。

How to use jquery.parsejson escaping

本教程操作环境:windows10系统、jquery3.6.0版本、Dell G3电脑。

jquery.parsejson转义怎么用

jQuery.parseJSON()函数用于将格式完好的JSON字符串转为与之对应的JavaScript对象。

语法为:

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$.parseJSON( json )

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json String类型 需要解析并转为JS对象的JSON格式字符串

所谓”格式完好”,就是要求指定的字符串必须符合严格的JSON格式,例如:属性名称必须加双引号、字符串值也必须用双引号。

如果传入一个格式不”完好”的JSON字符串将抛出一个JS异常,例如:以下字符串均符合JSON格式,但它们不是格式完好的JSON字符串(因此会抛出异常):

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// 以下均是字符串值,省略了两侧的引号,以便于展现内容

{id: 1} // id是属性名称,必须加双引号

{'id': 1} // id是属性名称,必须用双引号(不能用单引号)

{"name": 'CodePlayer'} // name属性的值是字符串,必须用双引号(不能用单引号)

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JSON标准不允许字符串中出现”控制字符”,例如:一个Tab或换行符。

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$.parseJSON('{"name": "Code\tPlayer\n"}');

 // 多数情况下,它会抛出一个错误,因为JS解析器会将字符串中的\t或\n等转义直接视作字面值,起到Tab或换行的效果。

//正确写法应该如下(使用两个反斜杠,以免被JS解析器直接转义\t或\n):

$.parseJSON('{"name": "Code\\tPlayer\\n"}');

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如果浏览器提供了一个本地的JSON.parse()方法,则jQuery将使用它来解析JSON字符串。

示例如下:

以下是与jQuery.parseJSON()函数相关的jQuery示例代码,以演示jQuery.parseJSON()函数的具体用法:

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var result1 = $.parseJSON( "{}" ); // 一个空的Object对象

var result2 = $.parseJSON( '{ "name": "CodePlayer", "age": 1 }' ); // 一个Object对象

alert( result2.name ); // CodePlayer

var result3 = $.parseJSON( '[ 1, true, "CodePlayer" ]' ); // 一个Object对象

alert( result3[2] ); // CodePlayer

var result4 = $.parseJSON( 'true' ); // 布尔值:true

var result5 = $.parseJSON( 'null' ); // null

var result6 = $.parseJSON( '0' ); // 整数:0

var result7 = $.parseJSON( '"专注于编程开发技术分享"' ); // 字符串值:专注于编程开发技术分享

var result8 = $.parseJSON( "\"专注于编程开发技术分享\"" ); // 字符串值:专注于编程开发技术分享

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/* 以下均是无效的JSON字符串或非常规的参数值 */

var result9 = $.parseJSON( 100 ); // 1.9之前返回null;1.9+返回100

var result10 = $.parseJSON( 1.23 ); // 1.9之前返回null;1.9+返回1.23

var result11 = $.parseJSON( true ); // 1.9之前返回null;1.9+返回true

var result12 = $.parseJSON( null ); // 返回null

var result13 = $.parseJSON( "" ); // 1.9之前返回null;1.9+抛出错误

var result14 = $.parseJSON( undefined ); // 1.9之前返回null;1.9+抛出错误

var result15 = $.parseJSON( "undefined" ); // 抛出错误

var result16 = $.parseJSON( "12x;xx'无效的JSON字符串x4546" ); // 抛出错误

var result17 = $.parseJSON( '{ name: "CodePlayer", age: 1 }' ); // 抛出错误

var result18 = $.parseJSON( '{ "name": \'CodePlayer\', "age": 1 }' ); // 抛出错误

var result19 = $.parseJSON( '{ "name": "Code\tPlayer\n" }' ); // 抛出错误

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