CODEWITHSUNDEEP
javascriptarrays
ylcnky
ylcnky

Reputation: 775

JavaScript: search user input in number of arrays

I am working with D3js library. On the web page, there is a search box which user can input their parameter. I receive arrays of data via Ajax post to my database. Following is the code:

..........
var aa;
var bb;
$(function(){
$.ajax({
    type: "POST",
    url: "http://localhost:7474/db/data/transaction/commit",
    accepts: {json: "application/json"},
    dataType: "json",
    contentType: "application/json",
    data: JSON.stringify(query), //query is somewhere above the code
    //pass a callback to success to do something with the data
    success: function (data) {
        aa = data.results[0].data;
        aa.forEach(function (entry) {
             passVar(entry.row[0].name)
        });}}
    );
});

function passVar(smth){
    bb =[smth];
    console.log (bb);
    //Should search the user input..........
    }
//if the user input matches, filter function should run.........
function filterData() {
    var value = d3.select("#constraint")[0][0].value;

    inputValue = value;

    ............
}

As the result of console.log(bb)I receive the following on console:

["Direct Digital Control System"]
["Fire Protection"]
["HVAC Cooling- Waterside"]
["HVAC Heating- Waterside"]
["HVAC System"]
["HVAC-01"]
["HVAC-02"]

What I want to do: If the user input match with one of the results in var bb, then program should run function filterdata() {....for querying. If not, don't do anything.

How should I write the code to make the search and run the other function? Thanks for the any help/suggestion.

Upvotes: 2

Views: 113

Answers (1)

Hanushka Suren
Hanushka Suren

Reputation: 793

You can loop through the array and find whether the user input is equals to the current index value of the array. if equals you can call your function and break the loop since no need to loop further more. 

for(int i=0; i<bb.length; i++){

    if(bb[i] == userInput){

        filterdata();

        break;

    }

}

Upvotes: 1

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