Trouble with text replace in array Javascript (Angular)

Question

I need to remove 'hello' substring from each object field in objects array. And i have an error "Cannot read property 'indexOf' of null". It happens, because i try to change object field in loop. But What to do? :) I use AngularJS.

var array = [
    {text: 'hello user1'},
    {text: 'hello user2'},
    {text: 'user3'},
    {text: 'hello user4'},
];
for (i = 0; i < array.length; i++) {
    if (array[i].text.indexOf('hello') + 1) {
        array[i].text = array[i].text.replace('hello','');
    }
}

// For demo
document.write(JSON.stringify(array));

@trigger I edited your question, care to hit "Run code snippet" and tell us what you see? — Siguza
– Siguza, Commented May 21, 2015 at 7:40

Satpal · Accepted Answer · 2015-05-21 07:36:18Z

1

Your condition condition is not correct.

The indexOf() method returns the index within the calling String object of the first occurrence of the specified value, starting the search at fromIndex. Returns -1 if the value is not found

change your condition to

if(array[i].text.indexOf('hello') > -1){
    //Rest of your code
}

edited May 21, 2015 at 7:36

answered May 21, 2015 at 7:33

Satpal

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2 Comments

Siguza Over a year ago

While I agree that this is better practise, the +1 method should actually work too.

briosheje Over a year ago

indeed, indexOf('hello') returning -1 +1 means 0, which should be false, else it should evaluate to true, so this shouldn't be the problem, but it's surely a better practice :P

SharpEdge · Accepted Answer · 2015-05-21 07:34:13Z

0

Change + to > -1

indexOf() returns the index of the string searched, in case it's not found returns -1.

if indexOf > -1 it means that there is some hello word in the string

var array = [
	{text: 'hello user1'},
	{text: 'hello user2'},
	{text: 'user3'},
	{text: 'hello user4'},
];
for (i = 0; i < array.length; i++) {
  if (array[i].text.indexOf('hello') > -1) {
    array[i].text = array[i].text.replace('hello','');
  }
}
    
console.log(array);

answered May 21, 2015 at 7:34

SharpEdge

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2 Comments

SharpEdge Over a year ago

Your issue is due to another piece of code, this one works perfectly.

SharpEdge Over a year ago

Depends on what you mean by "invalid"! If it's corrupted you can't!

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