Querying JSON array in JSON object value

Question

Using jsonSelect Plugin I am trying to do a query select as below. I am presenting this example.

How can I retrieve the All Project and Contractors who has the Projects in "North" ProjectArea? I already tried this to get the contractor's name but it didn't work

var contractors = [{
    "contractor1": [{
                     "project": "Project 2",
                     "projectArea": "South"
                    },
                    {
                     "project": "Project 101",
                     "projectArea": "East"
                    }],
     "contractor2": [{
                     "project": "Project 500",
                     "projectArea": "North"
                     },  
                     {
                     "project": "Projetc 108",
                     "projectArea": "West"
                    }]
}]

var stringSelect = 'has(:root > .projectArea:val("North"))';
JSONSelect.forEach(stringSelect, contractors, function (resultObj) {
    $('body').append('<p>' + resultObj + '</p>');
});

76484 · Accepted Answer · 2014-04-03 07:32:32Z

2

I think the only way you are going to be able to achieve what you want is by changing your JSON data structure. Specifically, you have your contractors' names as keys ('contractor1', contractor2'), but they really should be values. The library you are using returns a set of objects, but I don't see any way to step back a level and obtain each object's corresponding key. I suggest structuring your data like this:

var contractors = [
    {
        "name": "contractor1",
        "projects": [
            { "project": "Project 2", "projectArea": "South" },
            { "project": "Project 101", "projectArea": "North" }
        ]
    },
    {
        "name": "contractor2", 
        "projects": [
            { "project": "Project 500", "projectArea": "North" },
            { "project": "Project 108", "projectArea": "West" }
         ]
    }
];

Also, I think you will have to apply two levels of filters, first to get the contractors, and secondly to get each contractor's projects:

JSONSelect.forEach('object:has(.projects:has(.projectArea:val("North")))', contractors, function (contractor) {
    $('body').append('<p>' + contractor.name + '</p>');
    JSONSelect.forEach('object:has(.projectArea:val("North"))', contractor.projects, function (project) {
        $('body').append('<p>' + project.project + '</p>'); 
    });
});

I do not know of any other plugins that do this. If I were to come across this problem, my first inclination would be to solve it using jQuery's $.map and $.grep methods:

var filtered = $.map(contractors, function (contractor) {
    var filteredProjects = $.grep(contractor.projects, function (project) {
        return project['projectArea'] === 'North';
    });
    if (filteredProjects.length) {
        return { 
            'name': contractor.name, 
            'projects': filteredProjects
        };
    }
});

I am using $.grep to filter each contractor's projects to only those with ['projectArea'] === 'North'. If the contractor has any projects that match the filter, a new object with the contractor's name and the filtered projects gets added to my filtered var.

edited Apr 3, 2014 at 7:32

answered Apr 3, 2014 at 2:21

76484

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6 Comments

Mona Coder Over a year ago

Great , thanks User 76484, it was perfect explanation. just wondering if you know any better plugin to do this kind of thing easier? once again thanks.

76484 Over a year ago

Unfortunately, I do not know of another plugin. I've appended to my answer the way I would go about solving this issue without using a plugin.

Mona Coder Over a year ago

Thanks but in some complex querying like finding Max and Min values or returning a range between two x number values, can we still do that things by jquery $.map() and $.grep(). For Example if I have a pair of "projectCost" : number ; for each contractor is there any way to do querying on those number too?

76484 Over a year ago

If you want to filter projects by 'projectCost' instead of by 'projectArea', just alter the return line in the $.grep callback, for example, return project['projectCost'] > 100;. Max and Min values, since they require single objects instead of arrays, would not be suitable for $.grep - in such cases I would probably do a simple for each ($.each).

76484 Over a year ago

I think it is best to have a separate function for rendering the contrators to the page. We could call it renderContractors and pass to it an array of contractor objects. This keeps the presentation code (HTML) in one place and separate from the business logic. This will allow us to pass whatever filter result we want to be displayed. I have created a fiddle, jsfiddle.net/76484/5dtUd, where I first pass the unfiltered contractors to the render function, and then the filtered result.

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