Replacing values doesn't work ok when array field is empty

Question

I have this code:

$('#myTextArea').val($('#myTextArea').val().replace(linesText[4] + '\n', ""));

and it works fine. The problem is in this case:

$('#myTextArea').val() = "\n\n33333333333\n\n\n"

and linesText is this array:

0: ""
1: ""
2: "33333333333"
3: ""
4: ""
5: ""

What I want to happen: $('#myTextArea').val() to become "\n\n33333333333\n\n".

What happens:

$('#myTextArea').val()

becomes

"\n33333333333\n\n".

This happens because I actually replace "" + "\n" with "" and it takes the first "\n". I want to take the fourth. How to fix this? This works when linesText's fields aren't empty.

Alberto Fecchi · Accepted Answer · 2013-04-20 15:20:27Z

i've made a small function that probably needs improvement, but seems to work:

var ok = "\n\n33333333333\n\n\n";

function replaceSymbol(dataStr, toFind, elemPos) {
    var spacing = toFind.length;
    var indexToReplace = 0 - spacing;
    var curString;
    for (var i = 0; i < elemPos; i++) {
        curString = dataStr.substr(indexToReplace + spacing);
        if (curString.indexOf(toFind) == -1) 
            return false;
        indexToReplace = indexToReplace + curString.indexOf(toFind) + spacing;
    }
    return dataStr.substr(0, indexToReplace) + dataStr.substr(indexToReplace + spacing);
}

replaceSymbol(ok, '\n', 4);

this function ask for 3 parameters, the string (ok), the symbol to replace ('\n') and the position (in this case the 4th occurence of the symbol)

if the function can't find the symbol before/in the position, the function return false, while if all is ok the function will return the string without the element in Nth position

Jacob T. Nielsen · Accepted Answer · 2013-04-20 14:31:29Z

0

I believe this could be what you want to do

var parts = $('#myTextArea').val().split('\n');
parts[4].replace(linesText[4] + '\n', "");
$('#myTextArea').val(parts.join(''));

answered Apr 20, 2013 at 14:31

Jacob T. Nielsen

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