在标签中使用冒号访问JSON数据

时间:2020-12-15 15:56:01

I have json data with a colon in the label (see responsedata) which I'm finding difficult to access in Angular with the following code:

我在标签中有一个冒号的json数据(参见responsedata),我发现很难在Angular中使用以下代码访问它:

<li ng-repeat="i in items.autnresponse.responsedata | searchFor:searchString"> <p>{{i.autn:numhits}}</p> </li>

I keep getting an error like this:

我一直收到这样的错误:

Error: [$parse:syntax] Syntax Error: Token ':' is an unexpected token at column 7 of the expression [i.autn:numhits] starting at [:numhits].

错误:[$ parse:syntax]语法错误:令牌':'是从[:numhits]开始的表达式[i.autn:numhits]第7列的意外标记。

JSON data excerpt:

JSON数据摘录:

"autnresponse": {
    "action": {
        "$": "QUERY"
    },
    "response": {
        "$": "SUCCESS"
    },
    "responsedata": {
        "autn:numhits": {
        "$": "92"
    },
    "autn:totalhits": {
        "$": "92"
    },
    "autn:totaldbdocs": {
        "$": "188"
    },
    "autn:totaldbsecs": {
        "$": "188"
    },

Does anybody know a way around this?

有人知道解决这个问题吗?

2 个解决方案

#1


4  

I'll assume I know the answer to my question from the comments and post what would be my response:

我会假设我从评论中知道了我的问题的答案,并发布了我的回答:

Assumption

假设

Your JSON parses fine but your code can't access something in the resulting data structure

您的JSON解析正常,但您的代码无法访问结果数据结构中的某些内容

Answer

回答

Use square bracket notation with a string:

使用带有字符串的方括号表示法:

var x = i['autn:numhits'];

The same can be used when you have a property name in a variable. Using your same example:

当您在变量中具有属性名称时,可以使用相同的名称。使用相同的示例:

var propertyName = 'autn:numhits';
var x = i[propertyName];

Addendum

附录

For Angular template, try

对于Angular模板,请尝试

{{i['autn:numhits']}}

#2


0  

Use brackets to access it like a dictionary rather than dot notation. Replace {{i.autn:numhits}} with {{i['autn:numhits']}}

使用括号可以像字典而不是点符号一样访问它。将{{i.autn:numhits}}替换为{{i ['autn:numhits']}}

As a heads up, if you want to wrap autn:numhits with double quotes you will need to html escape them.

作为一个提示,如果你想用双引号包装autn:numhits,你将需要html转义它们。

#1


4  

I'll assume I know the answer to my question from the comments and post what would be my response:

我会假设我从评论中知道了我的问题的答案,并发布了我的回答:

Assumption

假设

Your JSON parses fine but your code can't access something in the resulting data structure

您的JSON解析正常,但您的代码无法访问结果数据结构中的某些内容

Answer

回答

Use square bracket notation with a string:

使用带有字符串的方括号表示法:

var x = i['autn:numhits'];

The same can be used when you have a property name in a variable. Using your same example:

当您在变量中具有属性名称时,可以使用相同的名称。使用相同的示例:

var propertyName = 'autn:numhits';
var x = i[propertyName];

Addendum

附录

For Angular template, try

对于Angular模板,请尝试

{{i['autn:numhits']}}

#2


0  

Use brackets to access it like a dictionary rather than dot notation. Replace {{i.autn:numhits}} with {{i['autn:numhits']}}

使用括号可以像字典而不是点符号一样访问它。将{{i.autn:numhits}}替换为{{i ['autn:numhits']}}

As a heads up, if you want to wrap autn:numhits with double quotes you will need to html escape them.

作为一个提示,如果你想用双引号包装autn:numhits,你将需要html转义它们。