NodeJS + JSON + jQuery:将JSON-object表单Node-backend发送到jQuery-frontend

f1n*_*1nn 0 jquery post json ejs node.js

我正在尝试将JSON-object形式的MongoDB发送到带有ejs模板引擎的页面.这是我的DB(一个JSON对象)的响应:

"suppliers": [
    {
        "lastname": "lastname1",
        "firstname": "firstname1",
        "middlename": "middlename1",
        "mobtel": "mobtel1",
        "worktel": "worktel1",
        "email": "email1"
    },
    {
        "lastname": "lastname2",
        "firstname": "firstname2",
        "middlename": "middlename2",
        "mobtel": "mobtel2",
        "worktel": "worktel2",
        "email": "email2"
    }
]
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当我从数据库中获取对象时,我必须将其字符串化为str以作为EJS模板引擎的var发送:

var myJSONText = JSON.stringify(doc.suppliers);
profile["suppliers"] =myJSONText;
...
res.render('profile.ejs',{profile:profile});
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在前端,我得到一个带有json数据的普通字符串.我通过jQuery解析它到一个对象:

var jsonObject = jQuery.parseJSON('<%= profile.suppliers %>');
for (var i=0; i<jsonObject.length; i++){
    //trying to show field 'name' for every supplier
    alert(jsonObject[i]['lastname']);
}
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......它不起作用,但应该提醒我两次.我试图检查一个纯文本的JSON字符串,来到前端,并找到了这样的符号:

[{&quot;lastname&quot;: &quot; ... 
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似乎jQuery.parseJSON无法从此类字符串创建有效对象.我做错了什么?谢谢!

更新:嗯,我已将我的代码转换为:

NodeJS代码:

profile["suppliers"] =doc.suppliers;
res.render('profile.ejs', { profile:profile });
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jQuery代码:

for (var i=0; i<'<%= profile.suppliers %>'.length; i++){
//trying to show field 'name' for every supplier
alert('<%= profile.suppliers %>'[i]['lastname']);
}
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它给了我永无止境的未定义警报循环.

更新2:我已经更新了我的源代码,所以:

的NodeJS:

var myJSONText = JSON.stringify(doc.suppliers);
profile["suppliers"] =myJSONText;
res.render('profile.ejs',{profile:profile});
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客户:

var jsonObject = <%- profile.suppliers %>;
$.each(jsonObject, function(item) {
    alert(item.lastname);
});
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生成的源代码:

var jsonObject = [{"lastname":"lastname1","firstname":"firstname1","middlename":"middlename1","mobtel":"mobtel1","worktel":"worktel1","email":"email1"},{"lastname":"lastname2","firstname":"firstname2","middlename":"middlename2","mobtel":"mobtel2","worktel":"worktel2","email":"email2"}];
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似乎还可以,但仍然给我未定义的警报.但是,它只发出两次警报,这意味着它正确解析JSON并且无法访问lastname属性?

Eri*_*ric 9

正确诊断后,&quot;s会被插入到JSON中.这会使其无效,从而阻止其被解码.

这是因为在ejs中,<%= ... %>html输出是否会逃逸,因此&quot;.您需要使用<%- profile.suppliers %>以避免这种情况.

没有必要JSON.parse在客户端上使用:JSON已经是有效的javascript代码,因此可以直接插入到代码中,而不是存储为字符串文字.

var jsonObject = <%- profile.suppliers %>;
$.each(jsonObject, function() {
    alert(this.lastname);
});
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