Skipping non matching strings when looping through an array in javascript

Question

In the code below I have a form input. When the user searches for a string that happens to be in the array I want it to output the query. When a user happens to search for a string not in the array I want to output an error message. The problem is when a user searches for a string that is other than item [0] in the array (in this case ipsum) they get an error message and then they get their query returned. I want to know if this can be remedied using the code below or if a different methodology for doing this should be pursued ( I know that's an opinion ).

<form>
<input type="text" id="formInput"></input>
<input type = "button" id="search"></input>

</form>


<script>


var search = document.getElementById("search");


var data = ["lorim", "ipsum"];

search.onclick = function(){
var formInput = document.getElementById("formInput").value;
for (i=0; i<data.length; i++){

    if (data[i] === formInput) {

      alert(data[i]);
    }   
      else{ alert("not working yet");   }
}
};

</script>

jbabey · Accepted Answer · 2012-10-05 19:22:42Z

1

you don't need a loop, just use indexOf:

search.onclick = function(){
    var formInput = document.getElementById("formInput").value;

    if (data.indexOf(formInput) === -1) {
        // they entered a bad search term
        return;
    }

    // do the rest of your search logic
};

answered Oct 5, 2012 at 19:22

jbabey

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Comments

Mutahhir · Accepted Answer · 2012-10-05 19:22:17Z

1

:) Keep at it.

The thing to remember is that you can only say 'nope didn't find it' after searching everything. So... keep a flag :)

var didntFind = true;
for (var i = 0; i < data.length; i++) {
    if (data[i] === formInput) {
        alert(data[i]);
        didntFind = false;
        break;
    }
}

if (didntFind) alert('error!');

You can also check if i === data.length-1 after the loop, but the above code should be less confusing for you. Hope this helps

answered Oct 5, 2012 at 19:22

Mutahhir

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1 Comment

William Over a year ago

Thanks for this. The other answer was easier and more to the point but I appreciate this regardless.

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