If else in Jquery

i am a noob to jquery and i want to know how to make use of if else for the following:

on the server side there is a if for number of rows is equal to 0 and else some JSON part.

$age= mysql_query("SELECT title FROM parent WHERE id ='$name'");
$age_num_rows = mysql_num_rows($age);   
  if ($age_num_rows==0) 
  {
   echo "true";  
  }
  else
  {
$sql ="SELECT * FROM parentid WHERE id = '$name'"; //$name is value from html      
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))
     {
$abc_output = array('title' => $row['title'],'age' => $row['age']);
     }
echo json_encode($abc_output);
  }

Now coming to Jquery part :

If the above PHP code go to if part then i want to 开发者_StackOverflow中文版display an alert box or if it goes to else part it needs to insert some values into the forms. Here is something i tried but it did not work.

   $(document).ready(function(){  
      $("#button1").click(function(){
        $.getJSON('script_1.php',function(data){
            if (data=='true') {
                 alert ('hello')
            }
            else {
                $.post('script_1.php', 
                    { id: $('input[name="id"]', '#myForm').val() }, 
                    function(json) { 
                        $("input[name='title']").val(json.title);
                        $("input[name='age']").val(json.age); 
                    }, 
                    "json");  
            }
      });
  });  

Edited:

  $(document).ready(function(){  
$("#button1").click(function(){
    $.post(
        'script.php',
        { id: $('input[name="id"]', '#myForm').val() },
        function(json) { 
            var data = JSON.parse(json);
            if (data.length === 0){
             alert('no data');   
            }
            else{
            $("input[name='title']").val(json.title);
                    $("input[name='age']").val(json.age); 

            }},
        "json"
    );
});
});

PHP side

$name = mysql_real_escape_string($_POST['id']); 
$sql ="SELECT * FROM parentid WHERE id = '$name'";       
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
if ($row) {
$row=  array('title' => $row['title'],'age' => $row['age']);
echo json_encode($row);
} else {
echo json_encode(array());
}

You need to parse the JSON before you can use it like that. Modern browsers will have built in support for JSON.parse(yourJSON), but to account for those that don't, you should use Douglas Crockford's JavaScript JSON library. Including it will provide JSON.parse() if the browser doesn't have it already.

For the if-else stuff you're doing in the PHP, the common practice is to echo out an empty JSON object or array, so you don't have to test for things like no rows on the server side. You could do something as simple as this, later accounting for your database column names:

$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
if ($row) {
  echo json_encode($row);
} else {
  echo json_encode(array());
}

Back in the JavaScript, you could then do something like this:

var data = JSON.parse(json);
if (data.length === 0) {
  alert('no data');
} else {
  $("input[name='title']").val(data.title);
  $("input[name='age']").val(data.age);
}

jeroen is right though, you only need to use one AJAX call.

There seem to be a few problems:

1. You are treating the data returned from getJSON as if it is plain text
2. The php that you are calling from javascript does not always return json
3. You are doing 2 ajax requests; getJSON and post where you only need one: The first call to getJSON without any data will never reach the else condition

By the way, where does $name come from in your php script? For your second ajax call to work, it needs to be something like mysql_real_escape_string($_POST['id']) or (int) $_POST['id'] if it is an integer.

Edit: I think it would be easiest to get rid of the .post and just use the first ajax call. So you will need to change:

$.getJSON('script_1.php',function(data){

to something like:

$.getJSON('script_1.php?id=' + $('input[name="id"]').val(), function(data) {

and in your php you need to use something like:

$name = mysql_real_escape_string($_GET['id']);