使用jQuery基于下拉选择选择单选按钮

Question

In the below code, if I select try1,it should search db and display corresponding value in input box as well as radio button. Here input box value is displayed correctly but radio button is not working. Need help

Script:

$("#test1").change(function()
    {
        $.post('test_file.php', {test_name: $(this).val() },
        function(opt){
        $("#name33").val(opt.test_code);
        $('#ok').val(opt.test_code2);
        }, 'json');

    }); 

Html:

<table><tr>
<td>
<select name="test1" id="test1" >
<option >try1</option>
<option >try2</option>
</select>
</td>
<td>
<input type="text" name="name33" id="name" />
</td>
<td>
<input type="radio" name="ok" id="ok" value="ok"/>ok
<input type="radio" name="ok" id="ok" value="not_ok"/>Not ok                
</td>

test_file.php

$db_sql = $db->queryUniqueObject("SELECT * FROM client WHERE test_name='".$_POST['test_name']."'" );
$test_code1=$db_sql->name;
$test_code2=$db_sql->try;

if($db_sql!=NULL)
{
$arr = array ("test_code1"=>"$test_code1","test_code2"=>"$test_code2");
echo json_encode($arr);
}
else
{
$arr1 = array ("no"=>"no");
echo json_encode($arr1);

}

Answer 1

Use

if(opt.test_code2=="ok")
{
  $('input:radio[value=ok]').prop("checked",true);
}
else if(opt.test_code2=="not_ok")
{
  $('input:radio[value=not_ok]').prop("checked",true);
}

Instead Of

$('#ok').val(opt.test_code2);

Because you set only value using .val() function and id of radio button is not unique so you have to take radio value as selector. If you want to checked your radio button according to its value then you have to use .prop() function.

Demo