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javascriptjquery
Soft Technoes
Soft Technoes

Reputation: 1105

How can I increment the id in jquery with for loop?

I have 'for loop' and I want to send the values through jquery but I can't increment the value of id, 'my_id' is the id of my input fields.

for(i=0;i<6;i++){
    $("#my_id_+i").val('test');
}

Upvotes: 0

Views: 2899

Answers (8)

Soft Technoes
Soft Technoes

Reputation: 1105

Thanks I found the solution. This part is working fine..

for(i=0;i<6;i++){
    $("#my_id_"+i).val('test');
}

Upvotes: 0

Djaouad
Djaouad

Reputation: 22766

You can use a template string to reference a variable from within a string literal:

for(i = 0; i < 6; i++){
    $(`#my_id_${i}`).val('test');
}

You can also use + but with the + i out of the string literal:

for(i = 0; i < 6; i++){
    $("#my_id_" + i).val('test');
}

If you don't have more 6 elements whose id starts with my_id_, you can avoid a loop and use:

$('[id^="my_id_"]').val('test');

Upvotes: 1

bik
bik

Reputation: 339

$("#my_id_+i").val('test'); and maybe it would be better to declare the counter with let declaration 

for(let i = 0; i < 6; i++){
    $("#my_id_" + i).val('test');
}

Upvotes: 1

Mamun
Mamun

Reputation: 68933

i has dynamic value but your code treat that as a string.

Try

$("#my_id_"+i).val('test');

Upvotes: 2

Clarity
Clarity

Reputation: 10873

With ES6 you can use let to properly scope variable to the loop and string interpolation to make the code cleaner:

for (let i = 0; i < 6; i++) {
  $(`#my_id_${i}`).val('test');
}

Upvotes: 3

behnam shateri
behnam shateri

Reputation: 1403

for this purpose you can combine string with i that you get from for each loop:

for(i=0;i<6;i++){
    $("#my_id_"+i).val('test');
}

Upvotes: 1

Bastian Springer
Bastian Springer

Reputation: 281

Just put the +i outside of the String

for(i=0;i<6;i++){
    $("#my_id_"+i).val('test');
}

Upvotes: 1

Bhushan Kawadkar
Bhushan Kawadkar

Reputation: 28513

you are putting i inside double quote hence it is part of string and not getting evaluated as variable. keep it outside and it will work

for(i=0;i<6;i++){
    $("#my_id_" +i).val('test');
}

Upvotes: 1

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