在 JSON Object 寻找房产

Question

I'm creating a JSON object like

tags = {"jon":["beef","pork"],"jane":["chicken","lamb"]};

which was generated using php from an array like

$arr = array(
        'jon' => array('beef', 'pork'),
        'jane' => array('chicken', 'lamb')
       );
$tags = json_encode($arr);

And I want to check if something is in one or the other. None of these seem to work, but something like

if('lamb' in tags.jane)) {
    console.log('YES');
} else {
    console.log('NO');
}

writes NO to the console

if('foo' in tags.jane)) {
    console.log('YES');
} else {
    console.log('NO');
}

also writes NO to the console

so looking at

typeof(tags.jane);

it shows it's an "object" but

console.log(tags);

shows the following:

Object
jane: Array[2]
    0: "chicken"
    1: "lamb"
    length: 2
    __proto__: Array[0]
jon: Array[2]
    0: "beef"
    1: "pork"
    length: 2
    __proto__: Array[0]
__proto__: Object

so i thought maybe tags.jane may actually be an array and tried

if($.inArray('lamb', tags.jane)) {
    console.log('YES');
} else {
    console.log('NO');
}

which writes YES to the console but

if($.inArray('foo', tags.jane)) {
    console.log('YES');
} else {
    console.log('NO');
}

also writes YES to the console.

Am I incorrectly building the JSON Object? Not targeting the value(s) properly? Any advice is greatly appreciated. If this would be easier as an array instead of an object, I have full control to change it. I'm just a bit stumped at how I should treat this.

Answer 1

jQuery.inArray returns -1 when element is not found. That's true value from the POV of Javascript. Try this:

if($.inArray('foo', tags.jane) != -1) {

Answer 2

Your second set of answers are the way you should go. However, $.inArray returns an index, not a boolean. Any non-zero integer is true, which means when foo is not found, it returns -1 which evaluates to true and prints YES .

Similarly, $.inArray('chicken', tags.jane) would return 0 and cast to false, which is also not the answer you want.

Instead, use $.inArray('foo', tags.jane) !== -1 as your condition.

Answer 3

tags.name will give you the array for that person. So $.inArray("chicken",tags.jane) would see if "chicken" is in jane's tags array. If it's not, you'd get -1, otherwise you'd it's position in the array (using your example, this would return zero, the first array element).

Answer 4

You're using the keyword in for the wrong reason. The statement ( prop 'in' obj ) checks to see if the object(associated array) has a property with the value of prop. Since you're using the 'in' keyword on an array, then false is going to be returned because tags.jane is an array with indexes and not an associated array with properties.

If you want to know was values are in the array then loop through and compare. If you want to use the 'in' keyword then convert your array to an object like so.

    tags = {};
    // old code
    tags.jane = ['lamb', 'food']; 
console.log(('lamb' in tags.jane) === false )
    // new code
    tags.jane = {
       'lamb':1,
        'food':1
    }
console.log(('lamb' in tags.jane) === true )

https://developer.mozilla.org/en/JavaScript/Reference/Statements/for...in

Answer 5

you can not use

if('foo' in tags.jane))

it should be used as

if (1 in tags.jane)

if you want to check 'foo' is in tags.jane, try this

var inIt = (function() {
    var inIt = false;
    tags.jane.forEach(function(item) {
        inIt = inIt || 'foo' == item;
    });
    return inIt;
})();