我有一个json信息，我想用它来创建像这样的对象：
var legends = {
"Altimetry" : {
0 : {
legenrowname : "0m - 250m ",
fillcolor : "005500",
bordercolor : ""
},
1 : {
legenrowname : "250m - 1000m ",
fillcolor : "ffff7f",
bordercolor : ""
},
2 : {
legenrowname : "1000m - 5000m ",
fillcolor : "B47811",
bordercolor : ""
},
3 : {
legenrowname : "+ 5000m ",
fillcolor : "482400",
bordercolor : ""
}
}
};
alert(legends); //the result is this:
Object []
它被认为是一个对象（这就是我下一步所需要的）
现在，我正在尝试使用legends2
动态创建它var dataL = ${dataJsonLegends};
var legends2 ="";
var lengthL = dataL.length;
$.each(dataL, function(i, item){
legendName = item.legendname;
legends2= '"' + legendName + '":{';
dataR = item.rows;
var lengthR = dataR.length;
$.each(dataR, function(j, item2){
id=item2.uid;
rowname=item2.legendrowname;
fillcolor=item2.fillcolor;
rowobject = id + ' : { legenrowname:"' + rowname + '", fillcolor:"' + fillcolor + '"}';
if(j!=lengthR-1) rowobject = rowobject + ",";
legends2= legends2+ rowobject;
});
legends2= legends2+ '}';
if(i!=lengthL-1) legends2= legends2+ ",";
});
alert(legends2); //the result is this:
"Altimetry" : { 0 : { legenrowname : "0m - 250m ",
fillcolor : "005500",
bordercolor : ""
},1 : {
legenrowname : "250m - 1000m ",
fillcolor : "ffff7f",
bordercolor : ""
},2 : {
legenrowname : "1000m - 5000m ",
fillcolor : "B47811",
bordercolor : ""
},3 : {
legenrowname : "+ 5000m ",
fillcolor : "482400",
bordercolor : ""
}}
legends2值是我的预期，但我希望它被识别为一个对象，而不是一个具有对象结构的字符串。
我尝试了很多东西来解决它。 JSON.parse，eval等等......但没人工作。
¿有人可以帮忙或给我一个提示吗？
编辑：传入json
我发布了我的json对象（dataL），用于创建我需要的结构。
[{"uid":1,"legendname":"Altimetry","legend_description":null,"rows":[{"uid":2,"legendrowname":"250m - 1000m","fillcolor":"ffff7f","bordercolor":null,"rast_value":null,"fillcolorrgb":null,"bordercolorrgb":null,"legendrow_description":null,"legend":null},{"uid":4,"legendrowname":"+ 5000m","fillcolor":"482400","bordercolor":null,"rast_value":null,"fillcolorrgb":null,"bordercolorrgb":null,"legendrow_description":null,"legend":null},{"uid":1,"legendrowname":"0m - 250m","fillcolor":"005500","bordercolor":null,"rast_value":null,"fillcolorrgb":null,"bordercolorrgb":null,"legendrow_description":null,"legend":null},{"uid":3,"legendrowname":"1000m - 5000m","fillcolor":"B47811","bordercolor":null,"rast_value":null,"fillcolorrgb":null,"bordercolorrgb":null,"legendrow_description":null,"legend":null}],"layer":{"id":1,"description":"Altimetry","geoserver":"/geoserver/Sopcawind/wms","name":"Sopcawind:srtm","transparent":true,"format":"image/png","viewparams":"","isbaselayer":false,"opacity":0.5,"color":"#2b383b","visibility":false,"groupid":1,"infocontrol":true,"geometry":"","featuretype":"","srs":"","extradata":"","legends":null,"project":null}}];
Edited2： 我试图使用代码，但我真的不需要一个json对象。我使用json对象创建具有此结构的另一个基本对象：
//loop legends
legends={"legendname": {
//loop rows
row.id : {
legenrowname : row.legendrowname,
fillcolor : row.fillcolor,
bordercolor : row.bordercolor
}
}}
答案 0 :(得分：0)
你不应该这么复杂，你帖子顶部的Json是无效的。 也许这是你想要的东西？
{
"Altimetry": [
{
"legenrowname": "0m - 250m ",
"fillcolor": "005500",
"bordercolor": ""
},
{
"legenrowname": "250m - 1000m ",
"fillcolor": "ffff7f",
"bordercolor": ""
},
{
"legenrowname": "1000m - 5000m ",
"fillcolor": "B47811",
"bordercolor": ""
},
{
"legenrowname": "+ 5000m ",
"fillcolor": "482400",
"bordercolor": ""
}
]
}
检查传入的JSON是否为http://jsonlint.com/，如果它有效。 你自己生成了JSON吗？ 执行console.log（jsonString）并将其粘贴到jsonlint。
使用有效的JSON，您可以使用JSON.parse(jsonString) - 不要添加任何额外的括号
比如var dataL = ${dataJsonLegends};
现在看起来你正在循环遍历响应字符串中的每个字符
修改
您更新的JSON是有效的，这就是我处理它的方式：
var json = [{"uid":1,"legendname":"Altimetry","legend_description":null,"rows":[{"uid":2,"legendrowname":"250m - 1000m","fillcolor":"ffff7f","bordercolor":null,"rast_value":null,"fillcolorrgb":null,"bordercolorrgb":null,"legendrow_description":null,"legend":null},{"uid":4,"legendrowname":"+ 5000m","fillcolor":"482400","bordercolor":null,"rast_value":null,"fillcolorrgb":null,"bordercolorrgb":null,"legendrow_description":null,"legend":null},{"uid":1,"legendrowname":"0m - 250m","fillcolor":"005500","bordercolor":null,"rast_value":null,"fillcolorrgb":null,"bordercolorrgb":null,"legendrow_description":null,"legend":null},{"uid":3,"legendrowname":"1000m - 5000m","fillcolor":"B47811","bordercolor":null,"rast_value":null,"fillcolorrgb":null,"bordercolorrgb":null,"legendrow_description":null,"legend":null}],"layer":{"id":1,"description":"Altimetry","geoserver":"/geoserver/Sopcawind/wms","name":"Sopcawind:srtm","transparent":true,"format":"image/png","viewparams":"","isbaselayer":false,"opacity":0.5,"color":"#2b383b","visibility":false,"groupid":1,"infocontrol":true,"geometry":"","featuretype":"","srs":"","extradata":"","legends":null,"project":null}}];
for(i = 0; i<json.length; i++) {
console.log("new array item");
console.log(json[i].uid);
console.log(json[i].legendname);
console.log(json[i].legenddescription);
for(j = 0; j<json[i].rows.length; j++) {
console.log("new row");
console.log('\t' + json[i].rows[j].uid);
console.log('\t' + json[i].rows[j].legendrowname);
console.log('\t' + json[i].rows[j].fillcolor);
console.log('\t' + json[i].rows[j].bordercolor);
// and so on...
}
}
编辑2
为什么不只是创建为javascript对象，如果只需要字符串化后需要json
这样的事情（可能不准确）：
var objArr = new [];
$.each(dataL, function(i, item){
var leg = new Object();
leg.uid = item.uid;
leg.legendname = item.legendname;
leg.rows = [];
$.each(dataR, function(j, item2){
var rw = new Object();
rw.uid = item2.uid;
rw.legendrowname = item2.legendrowname
// and so on
leg.rows.push(rw);
});
objArr.push(leg);
});
alert(JSON.stringify(objArr));
答案 1 :(得分：0)
输出缺乏围绕它的支撑。添加它们，它应该工作。