jquery check value present in an object array

Question

I am trying to check if a value is present inside an array of object

function hasProperties(id){
    jQuery(JSON.parse(jQuery("#PropertiesField").html())).each(function () {
        if(id== jQuery(this)[0].properties.id) {
            console.log((id== jQuery(this)[0].properties.id));
            return "Present";
        }
    })
};
var something = hasProperties("someid");

the above snippet returns undefined for something, but also true is logged in console. why is it not returning present when condition satisfies, what is the mistake that I have done?

Imesha Sudasingha · Accepted Answer · 2016-01-05 08:22:05Z

1

The function provided in the each method is an anonymous inner function. Therefore, nothing is returned outside of each() context. To tackle this you can do something like,

function getProperty(id){
var result;
    $('your element').each(function(){
        //If your condition is true
        result=expectedValue
    });
    return result;
}

answered Jan 5, 2016 at 8:22

Imesha Sudasingha

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Comments

void · Accepted Answer · 2016-01-05 08:16:50Z

1

I don't think you actually want to parse #PropertyField html as JSON and then want to make its jQuery object. Do a check on it.
Instead of doing jQuery(this)[0].properties.id, just do this.id, that is not a right syntax.

answered Jan 5, 2016 at 8:16

void

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2 Comments

Vignesh Subramanian Over a year ago

thanks, now i am using this.properties.id, but the problem persists it is logging "true" in console, but the value is not getting assigned to the var

void Over a year ago

@vignesh just do this.id

Vignesh Subramanian · Accepted Answer · 2016-01-05 08:27:02Z

0

I found the issue, the return i had is for the .each(). I added a return outside the foreach function and it works now

function hasProperties(id){
var found =false;
    jQuery(JSON.parse(jQuery("#PropertiesField").html())).each(function () {
        if(id== jQuery(this)[0].properties.id) {
            console.log((id== jQuery(this)[0].properties.id));
            found= true;
            return;
        }
    })
return found;
};
var something = hasProperties("someid");

answered Jan 5, 2016 at 8:27

Vignesh Subramanian

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