jQuery - 使用值或文本以及嵌套在optgroups中的选项设置选择框的选定选项

Question

So I am writing an app that requires an address input and I have a select element for the user to select the state/province. It needs to support the US and Canada so it has nested optgroups to separate those out and a single, first level option as it's default value. Here is a basic example:

<select name="state" id="state">
  <option class="co" value="" data-placeholder="true" disabled selected>Choose your state...</option>
  <optgroup label="United States">
    <option class="co" value="AL">Alabama</option>
    <option class="co" value="AK">Alaska</option>
    <option class="co" value="AZ">Arizona</option>
  </optgroup>
  <optgroup label="Canada">
    <option class="co" value="AB">Alberta</option>
    <option class="co" value="BC">British Columbia</option>
    <option class="co" value="MB">Manitoba</option>
  </optgroup>

Now I need to programmatically select the option that matches input from an external source and I want to check for a match based on both the value of the option element or its text. Whichever option is a match would then be set as the selected option. I know you can set the selected option by value using

$("#state").val(myValue)

and I know you can set an option based on text in this way

var myText = "The state I want.";
$("#state").children().filter(function() {
  return $(this).text() == myText;
}).prop('selected', true);

Is there a clean way to do this without having to run through each child and checking if it's an optgroup and then running through all its children to check for a match? Is there an easy way through jQuery to combine the value and text methods of setting the selected option?

One other complication, I am going to be doing this within an external jQuery plugin. Within the function I need to modify I have the select element as a variable

$element

so I need a way to do it kind of like this if possible:

$element.descendents(":option").filter(function() {
  //do the selecting here
}).prop('selected', true);

Answer 1

If you want to select by the option value, use the value selector:

var myText = "AZ";
$('#state option[value="' + myText + '"]').prop('selected', true);

If you want to search by the option's label, use a filter:

var myText = "Arizona";
$('#state option').filter(function () { return $(this).html() == myText; }).prop('selected', true)

Answer 2

Solved. Since I already had my element passed to a function as a jQuery variable, $element, I couldn't just use the standard selector in the form of:

$("#state option").filter(
  // filter function
).prop('selected', true);

After a lot of trying, I got this and it works:

function functionIHadToChange($element, value) {
  // other code
  $element.find("option").filter(function(){
      return ( ($(this).val() == value) || ($(this).text() == value) )
    }).prop('selected', true);
}

Answer 3

I am not sure I understood completely your question but I am attempting to answer it in this fiddle

The trick being that you can select it by setting the value of the select box directly

$("#state").val( a_value );

Answer 4

You can set it by $("#select_id").prop("selectedIndex", 3); // Select index starts from zero. $("#select_id").prop("selectedIndex", 3); // Select index starts from zero.

Read here for example this .

Answer 5

$element = $('select#state');
$options = $element.find('option');
$wanted_element = $options.filter(function () {
  return $(this).val() == "Alabama" || $(this).text() == "Alabama"
});
$wanted_element.prop('selected', true);

Would be one way to do it.

But i would guess, without knowing the exact internas of the .find() method, in the end jQuery will use at least two loops itself to perform this...

Answer 6

I'm late here but for future visitor, easiest way to do that is :

html

<select name="dept">
<option value="">This doctor belongs to which department?</option>
<option value="1">Orthopaedics</option>
<option value="2">Pathology</option>
<option value="3">ENT</option>
</select>

jQuery

$('select[name="dept"]').val('3');

Output: This will active ENT .