如何根据选择值显示表单输入字段？

Question

I know this is simple, and I need to search in Google. I tried my best and I could not find a better solution. I have a form field, which takes some input and a select field, which has some values. It also has "Other" value.

What I want is:

If the user selects the 'Other' option, a text field to specify that 'Other' should be displayed. When a user selects another option (than 'Other') I want to hide it again. How can I perform that using JQuery?

This is my JSP code

<label for="db">Choose type</label>
<select name="dbType" id=dbType">
   <option>Choose Database Type</option>
   <option value="oracle">Oracle</option>
   <option value="mssql">MS SQL</option>
   <option value="mysql">MySQL</option>
   <option value="other">Other</option>
</select>

<div id="otherType" style="display:none;">
  <label for="specify">Specify</label>
  <input type="text" name="specify" placeholder="Specify Databse Type"/>
</div>

Now I want to show the DIV tag**(id="otherType")** only when the user selects Other. I want to try JQuery. This is the code I tried

<script type="text/javascript"
    src="jquery-ui-1.10.0/tests/jquery-1.9.0.js"></script>
<script src="jquery-ui-1.10.0/ui/jquery-ui.js"></script>
<script>    
$('#dbType').change(function(){

   selection = $('this').value();
   switch(selection)
   {
       case 'other':
           $('#otherType').show();
           break;
       case 'default':
           $('#otherType').hide();
           break;
   }
});
</script>

But I am not able to get this. What should I do? Thanks

Answer 1

You have a few issues with your code:

1. you are missing an open quote on the id of the select element, so: <select name="dbType" id=dbType">

should be <select name="dbType" id="dbType">

2. $('this') should be $(this) : there is no need for the quotes inside the paranthesis.

3. use .val() instead of .value() when you want to retrieve the value of an option

4. when u initialize "selection" do it with a var in front of it, unless you already have done it at the beggining of the function

try this:

   $('#dbType').on('change',function(){
        if( $(this).val()==="other"){
        $("#otherType").show()
        }
        else{
        $("#otherType").hide()
        }
    });

http://jsfiddle.net/ks6cv/

UPDATE for use with switch:

$('#dbType').on('change',function(){
     var selection = $(this).val();
    switch(selection){
    case "other":
    $("#otherType").show()
   break;
    default:
    $("#otherType").hide()
    }
});

UPDATE with links for jQuery and jQuery-UI:

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js" ></script>
<script src="//ajax.googleapis.com/ajax/libs/jqueryui/1.10.2/jquery-ui.min.js"></script>‌​

Answer 2

Demo on JSFiddle

$(document).ready(function () {
    toggleFields(); // call this first so we start out with the correct visibility depending on the selected form values
    // this will call our toggleFields function every time the selection value of our other field changes
    $("#dbType").change(function () {
        toggleFields();
    });

});
// this toggles the visibility of other server
function toggleFields() {
    if ($("#dbType").val() === "other")
        $("#otherServer").show();
    else
        $("#otherServer").hide();
}

HTML:

    <p>Choose type</p>
    <p>Server:
        <select id="dbType" name="dbType">
          <option>Choose Database Type</option>
          <option value="oracle">Oracle</option>
          <option value="mssql">MS SQL</option>
          <option value="mysql">MySQL</option>
          <option value="other">Other</option>
        </select>
    </p>
    <div id="otherServer">
        <p>Server:
            <input type="text" name="server_name" />
        </p>
        <p>Port:
            <input type="text" name="port_no" />
        </p>
    </div>
    <p align="center">
        <input type="submit" value="Submit!" />
    </p>

Answer 3

You have to use val() instead of value() and you have missed starting quote id=dbType" should be id="dbType"

Live Demo

Change

selection = $('this').value();

To

selection = $(this).val();

or

selection = this.value;

Answer 4

I got its answer. Here is my code

<label for="db">Choose type</label>
<select name="dbType" id=dbType">
   <option>Choose Database Type</option>
   <option value="oracle">Oracle</option>
   <option value="mssql">MS SQL</option>
   <option value="mysql">MySQL</option>
   <option value="other">Other</option>
</select>

<div id="other" class="selectDBType" style="display:none;">
<label for="specify">Specify</label>
<input type="text" name="specify" placeholder="Specify Databse Type"/>
</div>

And my script is

$(function() {

        $('#dbType').change(function() {
            $('.selectDBType').slideUp("slow");
            $('#' + $(this).val()).slideDown("slow");
        });
    });

Answer 5

<!DOCTYPE html>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<script>
function myfun(){
$(document).ready(function(){

    $("#select").click(
    function(){
    var data=$("#select").val();
        $("#disp").val(data);
                     });
});
}
</script>
</head>
<body>

<p>id <input type="text" name="user" id="disp"></p>

<select id="select" onclick="myfun()">
<option name="1"value="1">first</option>
<option name="2"value="2">second</option>
</select>

</body>
</html>

Answer 6

$('#dbType').change(function(){

   var selection = $(this).val();
   if(selection == 'other')
   {
      $('#otherType').show();
   }
   else
   {
      $('#otherType').hide();
   } 

});