jquery class prefix name selection with variable

Question

Ok i have a problem with selecting class with prefix name using variable. This is my code:

$('.product_type').on("click", function(event) {
  type_id = event.target.id;
  $("div[class*='product_']").fadeOut("fast");
  var dbg = $("div[class*='product_']") + type_id;
  dbg.fadeIn("fast");
  $('#echo').text(show_ids);
});

Problem is at var dbg = $("div[class*='product_']") + type_id; Selection is not working when I add variable + type_id...

Welcome to Stack Overflow! Please take the tour (you get a badge!), have a look around, and read through the help center, in particular How do I ask a good question? It's not clear what you're asking to do. Are you trying to add the contents of type_id onto the class in the selector? Or...? It's often helpful if you post a minimal reproducible example demonstrating the problem, ideally a runnable one using Stack Snippets (the [<>] toolbar button; here's how to do one). — T.J. Crowder
– T.J. Crowder, Commented Aug 25, 2018 at 12:48
Without your HTML, and without a clear description of the problem, we are mostly unable to help you, or any future visitors, without making guesses and assumptions. — David Thomas
– David Thomas, Commented Aug 25, 2018 at 12:49
In this case, though, you're adding the string variable type_id to a jQuery object. If you want to add it to the selector, add it to the string inside of $(). — Heretic Monkey
– Heretic Monkey, Commented Aug 25, 2018 at 12:53
I try that but not working... var dbg = $("div[class*='product_']" + type_id); — Balkan Design
– Balkan Design, Commented Aug 25, 2018 at 12:55

Mahdi Aslami Khavari · Accepted Answer · 2018-08-25 12:56:25Z

2

i think you want to use var dbg = $("div[class*='product_" + type_id + "']"); instead of var dbg = $("div[class*='product_']") + type_id;

i think your mistake is added your name after getting object

answered Aug 25, 2018 at 12:56

Mahdi Aslami Khavari

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Erubiel · Accepted Answer · 2018-08-25 12:56:54Z

0

Try with this code, you were missing the var declaration and you can just append inside the $()

$('.product_type').on("click", function(event) {
    var type_id = event.target.id;
    $("div.product_" + type_id).fadeOut("fast");
    var dbg = $("div.product_" + type_id);
    dbg.fadeIn("fast");
    $('#echo').text(show_ids); // i don't know what this line makes, just copied it from your code
});

answered Aug 25, 2018 at 12:56

Erubiel

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