Determine if the object's value is present

Question

Determine if the object's value is present

My current JavaScript setup looks like this:

var NAMES = []; 
function INFO(id,first,middle,last){ 
  var newMap = {};
  newMap[id] = [first, middle, last];
  return newMap ;   
}

Next, I have the following code block:

for (var j = 0; j < NUMBER.length; j++) {   //let's say there are three values
    var my_name = all_names[j]; // which contains "185, 185, 185"              
    if (NAMES[my_name] !== 185){ //This is where I need to perform a check              
        NAMES.push(INFO(my_name,"sean","sdfsd","sdfsfd"));                      
    }else{

    }               
}
alert(JSON.stringify(NAMES , null, 4));

Below is a screenshot of the alert displayed:

I used the number "185" for demonstration purposes. What I'm trying to achieve is checking if the id of 185 already exists, and if it does, proceed to the else block. I've attempted using typeof, undefined, etc., but haven't been successful. (In essence, I should only have one instance of "185").

Any suggestions or advice on how to accomplish this? Thank you!

javascript

Answer 1

Answer №1

From what I gather, it seems like you need to loop through NAMES and inspect each element. One approach could be utilizing the javascript function [].some:

if (!NAMES.some(function(item){return item[my_name]})) {
    ...
} else {

}

Answer 2

From what I gather, it seems like you need to loop through NAMES and inspect each element. One approach could be utilizing the javascript function [].some:

if (!NAMES.some(function(item){return item[my_name]})) {
    ...
} else {

}

Answer 3

Answer №2

To avoid duplicate entries, you can utilize the NAMES object instead of an array like so:

var all_names = [185, 185, 181], 
    NAMES = {};
for (var j = 0; j < all_names.length; j++) {   //assuming there are three values
    var my_name = all_names[j]; // contains "185, 185, 185"  
    NAMES[my_name] = ["sean","sdfsd","sdfsfd"];
}

alert(JSON.stringify(NAMES, null, 4));

Answer 4

To avoid duplicate entries, you can utilize the NAMES object instead of an array like so:

var all_names = [185, 185, 181], 
    NAMES = {};
for (var j = 0; j < all_names.length; j++) {   //assuming there are three values
    var my_name = all_names[j]; // contains "185, 185, 185"  
    NAMES[my_name] = ["sean","sdfsd","sdfsfd"];
}

alert(JSON.stringify(NAMES, null, 4));

Answer 5

Answer №3

To begin with, I suggest creating a JS Fiddle or CodePen to showcase the code in action.

The issue lies in the usage of NAMES[my_name]. Since NAMES is an array, calling NAMES[my_name] retrieves the entire object created in the INFO function, rather than checking if the object has an attribute matching the value from the my_names array.

While not the most elegant solution, here is a snippet that demonstrates what you actually want to achieve:

var NAMES = []; 
function INFO(id,first,middle,last){ 
  var newMap = {};
  newMap[id] = [first, middle, last];
  return newMap ;   
}

all_names = ["185", "186", "185"]

for (var j = 0; j < all_names.length; j++) {
    var my_name = all_names[j]; 
    if (NAMES.length == 0) {
        NAMES.push(INFO(my_name,"sean","sdfsd","sdfsfd"));                      
    } else {
        var match = false;
        for (var x = 0; x < NAMES.length; x++) {
            console.log(NAMES[x][my_name] + ' : ' + my_name);
            if(NAMES[x][my_name]) {
                match = true;
            } 
        }
        if (!match) {
            NAMES.push(INFO(my_name,"sean","sdfsd","sdfsfd"));
        }
    }
}
alert(JSON.stringify(NAMES , null, 4));

Note the condition checking NAMES[x][my_name] - it verifies if the item at array index 'x' possesses an attribute of 'my_name' (e.g. "185"). This likely aligns with your intended functionality. While there are more concise ways to handle this task in JavaScript, the provided code should clarify the fundamental issue at hand.

Answer 6

To begin with, I suggest creating a JS Fiddle or CodePen to showcase the code in action.

The issue lies in the usage of NAMES[my_name]. Since NAMES is an array, calling NAMES[my_name] retrieves the entire object created in the INFO function, rather than checking if the object has an attribute matching the value from the my_names array.

While not the most elegant solution, here is a snippet that demonstrates what you actually want to achieve:

var NAMES = []; 
function INFO(id,first,middle,last){ 
  var newMap = {};
  newMap[id] = [first, middle, last];
  return newMap ;   
}

all_names = ["185", "186", "185"]

for (var j = 0; j < all_names.length; j++) {
    var my_name = all_names[j]; 
    if (NAMES.length == 0) {
        NAMES.push(INFO(my_name,"sean","sdfsd","sdfsfd"));                      
    } else {
        var match = false;
        for (var x = 0; x < NAMES.length; x++) {
            console.log(NAMES[x][my_name] + ' : ' + my_name);
            if(NAMES[x][my_name]) {
                match = true;
            } 
        }
        if (!match) {
            NAMES.push(INFO(my_name,"sean","sdfsd","sdfsfd"));
        }
    }
}
alert(JSON.stringify(NAMES , null, 4));

Note the condition checking NAMES[x][my_name] - it verifies if the item at array index 'x' possesses an attribute of 'my_name' (e.g. "185"). This likely aligns with your intended functionality. While there are more concise ways to handle this task in JavaScript, the provided code should clarify the fundamental issue at hand.

Answer 7

Answer №4

You can try implementing this code snippet using the hasOwnProperty method :

for (var j = 0; j < NUMBER.length; j++) {
    var my_name = all_names[j];   
    if (!NAMES[my_name].hasOwnProperty("185")){              
        NAMES.push(INFO(my_name,"sean","sdfsd","sdfsfd"));                      
    }else{

    }               
}

Answer 8

You can try implementing this code snippet using the hasOwnProperty method :

for (var j = 0; j < NUMBER.length; j++) {
    var my_name = all_names[j];   
    if (!NAMES[my_name].hasOwnProperty("185")){              
        NAMES.push(INFO(my_name,"sean","sdfsd","sdfsfd"));                      
    }else{

    }               
}

Determine if the object's value is present

Answer №1

Answer №2

Answer №3

Answer №4

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