option

[jemz]

Ok, well it sounds like we're almost there.
What we need to know is what toselectdropdown.php is returning when you have select * from employee in your code.
Can you therefore change this:

success:function(data)
    {
      var items = [];
      $.each(data, function(key, val) {
        items.push('<option value="' + key + '">' + val + '</option>');
      });
      $('<select/>', {
        'class': 'my-new-select',
        html: items.join('')
      }).appendTo('body');
    }

back to this:

success:function(data)
    {
      console.log(data);
    }

and post the output.

Cheers.

BTW, Array.join() is a method which joins the elements of an array into a string, and returns the string.

Hi pullo,

Thank you for the reply...do i need to do this?

$('<select/>', {
'class': 'my-new-select',
html: items.join('')
}).appendTo('body');

but i have already <select> element in my html...
my html code goes something like this

    
 <label>choose</label>

<select id="optselect">
  <option></option>
</select>

[Pullo]

but i have already <select> element in my html...
my html code goes something like this

<label>choose</label>
<select id="optselect">
  <option></option>
</select>

Hi,

Thanks for that.
If you have a select option with an id, things are even easire.
But first, I need to know what toselectdropdown.php is returning when you have select * from employee in your code.

Re-read my last post, alter the code accordingly and post the output here.

[jemz]

Hi,

Thanks for that.
If you have a select option with an id, things are even easire.
But first, I need to know what toselectdropdown.php is returning when you have select * from employee in your code.

Re-read my last post, alter the code accordingly and post the output here.

Hi pullo, this is the result of log

(an empty string)

[jemz]

Hi,

Thanks for that.
If you have a select option with an id, things are even easire.
But first, I need to know what toselectdropdown.php is returning when you have select * from employee in your code.

Re-read my last post, alter the code accordingly and post the output here.

hi pullo, i tried to use this

error:(jqXHR,textStatus){
  alert("error" + textStatus);
},
success: function(){
  some code....
});

when my code runs it will go to the error and it alert to "pareserror".

maybe in sql statement the way i select *.?

[jemz]

Hi,

Thanks for that.
If you have a select option with an id, things are even easire.
But first, I need to know what toselectdropdown.php is returning when you have select * from employee in your code.

Re-read my last post, alter the code accordingly and post the output here.

okay i fixed now the parse error: it's in my die statement

now i have new problem when it display to the dropdownlist my text value is in vertical align like this
sample

w
e
l
c
o
m
e
{
"

etc...

here is how i select.
please correct me if i am wrong.

if(isset($_POST['id'])){




          $con = mysql_query("select * from employee");

          if(!$con)
              die("problem".mysql_error());


      while($row = mysql_fetch_array($con,MYSQL_BOTH)){

               $empname= $row['emp_name'];
               $emaddr= $row['emp_addr'];

            $empdata = array('empname=>$empname,'empaddr'=>$emaddr);


          echo json_encode($empdata);
      }




    }

[Pullo]

Hi jemz,
Back to the original question, I am afraid.
What data is being returned to the success callback?
Can you log it to the console, as per Post#25
Thanx

[Pullo]

BTW, just looking at your PHP code.
I would have thought the correct way to do things, was to declare the $empdata array before your while loop, then append values to it within the while loop, then after the while loop and ouside of the if statement, echo the array using json_encode.

[jemz]

Hi jemz,
Back to the original question, I am afraid.
What data is being returned to the success callback?
Can you log it to the console, as per Post#25
Thanx

Hi,pullo

here is the log

{"empname":"jack","empaddr":"Australia"}

[Pullo]

Is this the output when you have "select * from employee" in your PHP code?
How many records are in the data base?

Try changing the PHP according to my suggestions in post 32

[jemz]

Is this the output when you have "select * from employee" in your PHP code?
How many records are in the data base?

Try changing the PHP according to my suggestions in post 32

Hi, you mean like this

if(isset($_POST['id'])){




          $con = mysql_query("select * from employee");

          if(!$con)
              die("problem".mysql_error());

      $empdata = array();
      while($row = mysql_fetch_array($con,MYSQL_BOTH)){

               $empname= $row['emp_name'];
               $emaddr= $row['emp_addr'];

            $empdata = array('empname=>$empname,'empaddr'=>$emaddr);


         
      }




    }
 echo json_encode($empdata);

i have 20 records..

[Pullo]

Hi jemz,

Let's put it another way.
What do you want the function mydropdown(id) to do.

Should it fetch a list of employees from your database table and automatically select the current one?

[jemz]

Hi jemz,

Let's put it another way.
What do you want the function mydropdown(id) to do.

Should it fetch a list of employees from your database table and automatically select the current one?

Hi pullo, thank you for the reply...yes i want to fetch all list of employees and put it in the dropdown list and then select the current one

[Pullo]

I want to fetch all list of employees and put it in the dropdown list and then select the current one

Ok, that should be no problem.
A select looks like this:

<select>
  <option value = "value">Some text</option>
  <option value = "value">Some text</option>
</select>

What should the value in the option be?
The employee's id?

What should "Some text" be?
The employees name?

Edit:
I've just coded up something along these lines:

If you alter your "toselectdropdown.php" file like this:

<?php $con = mysql_connect("localhost","root","*****");
  if (!$con){
    die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("test", $con);
  $result = mysql_query("SELECT * FROM employee");
    
  while($row = mysql_fetch_array($result)){
    $empdata[$row['id']] = $row['emp_name'];
  }

  mysql_close($con);
  echo json_encode($empdata);
?>

Then use the JavaScript code from post 22 to create the drop down, you should see something approaching what you are trying to achieve.
It's not the finished product, but let me know if this works for you.

[jemz]

Ok, that should be no problem.
A select looks like this:

<select>
  <option value = "value">Some text</option>
  <option value = "value">Some text</option>
</select>

What should the value in the option be?
The employee's id?

What should "Some text" be?
The employees name?

Edit:
I've just coded up something along these lines:

If you alter your "toselectdropdown.php" file like this:

<?php $con = mysql_connect("localhost","root","*****");
  if (!$con){
    die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("test", $con);
  $result = mysql_query("SELECT * FROM employee");
    
  while($row = mysql_fetch_array($result)){
    $empdata[$row['id']] = $row['emp_name'];
  }

  mysql_close($con);
  echo json_encode($empdata);
?>

Then use the JavaScript code from post 22 to create the drop down, you should see something approaching what you are trying to achieve.
It's not the finished product, but let me know if this works for you.

Hi pullo, but what if i have made already in my html file the <select></select>
do i need to do this

$('<select/>', {
'class': 'my-new-select',
html: items.join('')
}).appendTo('body');

[jemz]

Ok, that should be no problem.
A select looks like this:

<select>
  <option value = "value">Some text</option>
  <option value = "value">Some text</option>
</select>

What should the value in the option be?
The employee's id?

What should "Some text" be?
The employees name?

Edit:
I've just coded up something along these lines:

If you alter your "toselectdropdown.php" file like this:

<?php $con = mysql_connect("localhost","root","*****");
  if (!$con){
    die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("test", $con);
  $result = mysql_query("SELECT * FROM employee");
    
  while($row = mysql_fetch_array($result)){
    $empdata[$row['id']] = $row['emp_name'];
  }

  mysql_close($con);
  echo json_encode($empdata);
?>

Then use the JavaScript code from post 22 to create the drop down, you should see something approaching what you are trying to achieve.
It's not the finished product, but let me know if this works for you.

Hi pullo, there is no output...but i tried some google i found this solution and it works for me.

success: function(data){

$.each(data, function(key, val) {
$('#my-new-select').append($('<option></option>').val(key).html(val));

});

but the problem is i don't know how to make selected option.

when all the key and value will be loaded in the dropdown,i want that it will select automatic which id is being search.
example:

data: 'id='+0005;

i want that employee having 0005 will be selected automatically in the dropdownlist..how can i do this?...


Thank you in advance.

[Pullo]

Hi jemz,

Hi pullo, there is no output...but i tried some google i found this solution and it works for me.

Did you use the PHP code I provided?
It sounds like you have managed to make the PHP file return something sensible and have constructed a <select> menu from this data.
Is that the case?

Is 'id' in scope?
If so, you can mark the current employee as selected thus:

$("#my-new-select option").each(function(){
  if($(this).val()==id){
    $(this).attr("selected","selected");    
  }
});

Let me know if that works.

Happy Christmas!

[jemz]

Hi jemz,



Did you use the PHP code I provided?
It sounds like you have managed to make the PHP file return something sensible and have constructed a <select> menu from this data.
Is that the case?

Is 'id' in scope?
If so, you can mark the current employee as selected thus:

$("#my-new-select option").each(function(){
  if($(this).val()==id){
    $(this).attr("selected","selected");    
  }
});

Let me know if that works.

Happy Christmas!

Hi pullo,

 success: function(data){
                  $.each(data, function(key, val) {
                      $('#my-new-select').append($('<option></option>').val(key).html(val));

                          if($(this).val()==id){
                              $(this).attr("selected","selected");
                          }

                  });
              }

It will not select automatically...

Did you use the PHP code I provided?

yeah like this.

 $result = mysql_query("SELECT * FROM employee");
    
  while($row = mysql_fetch_array($result)){
    $empdata[$row['id']] = $row['emp_name'];
  }

  mysql_close($con);
  echo json_encode($empdata);

Merry Christmas!

[jemz]

Hi jemz,



Did you use the PHP code I provided?
It sounds like you have managed to make the PHP file return something sensible and have constructed a <select> menu from this data.
Is that the case?

Is 'id' in scope?
If so, you can mark the current employee as selected thus:

$("#my-new-select option").each(function(){
  if($(this).val()==id){
    $(this).attr("selected","selected");    
  }
});

Let me know if that works.

Happy Christmas!

I got it now to select automatically

success: function(data){
                  $.each(data, function(key, val) {

                     if(key==id)
                       $('#my-new-select').append($('<option></option>').val(key).html(val).attr('selected','selected'));
                     else
                       $('#my-new-select').append($('<option></option>').val(key).html(val));

                  });
              }

[Pullo]

I got it now to select automatically

Yay

Just for my personal interest, could you please post your final code (similar to what you posted in the first post, but obviously the working version).
Thanks!

[jemz]

Yay

Just for my personal interest, could you please post your final code (similar to what you posted in the first post, but obviously the working version).
Thanks!

Hi pullo,

Here is now the working code,but please correct me if i am wrong.

 $(function(){ 
  $.ajax
	({
	   type: "POST",
	  data: "id="+myid,
	  url: "members.php",
	 	
	  success: function(r)
	  {
              var id = r.empid;
	     $('#emp').val(r.empname);
             mydropdown(id);

	 }
  });
});


 function  mydropdown(id){
               
	$.ajax({
	     type: 'post',
	     data: 'id='+id,    
	     url: 'toselectdropdown.php',
	      success: function(data){
                  $.each(data, function(key, val) {

                     if(key==clscode)
                       $('#my-new-select').append($('<option></option>').val(key).html(val).attr('selected','selected'));
                     else
                       $('#my-new-select').append($('<option></option>').val(key).html(val));

                  });
              }
			
			});
		    
		 
		 }
 }

toselectdropdown.php

if(isset($_POST['id'])){




          $con = mysql_query("select * from employee");

          if(!$con)
              die("Unable to load".mysql_error());

            $data=array();
      while($row = mysql_fetch_array($con)){

               $empdata[$row['emp_id']]=$row['emp_address'];


      }




    }


   header('Content-type: application/json');


    echo json_encode($empdata);
    }

[Pullo]

Thanx jemz,

It was just with all of the back and forth, I'd lost sight of what the working solution ended up being.
The code looks ok.
There are a few small optimizations you could make, for example creating your <options> as a document fragment, then appending them to the <select>, but if this code works for you, then I would leave it at that.

[jemz]

Thanx jemz,

It was just with all of the back and forth, I'd lost sight of what the working solution ended up being.
The code looks ok.
There are a few small optimizations you could make, for example creating your <options> as a document fragment, then appending them to the <select>, but if this code works for you, then I would leave it at that.

your welcome .....yup this code works for me.

[jemz]

Thanx jemz,

It was just with all of the back and forth, I'd lost sight of what the working solution ended up being.
The code looks ok.
There are a few small optimizations you could make, for example creating your <options> as a document fragment, then appending them to the <select>, but if this code works for you, then I would leave it at that.

Hi pullo,I'm back...I just want to ask in this php code that you gave

$result = mysql_query("SELECT * FROM employee");

while($row = mysql_fetch_array($result)){
$empdata[$row['id']] = $row['emp_name'];
}

mysql_close($con);
echo json_encode($empdata);

how can i make if i want 3 columns in my employee table to be populate,the 2 columns will be on the dropdown,the one is for the textfield,i tried something like this but it doesn't work.

$result = mysql_query("SELECT * FROM employee");

while($row = mysql_fetch_array($result)){
$empdata[$row['id']] = $row['emp_name']=$row['emp_age'];
}

mysql_close($con);
echo json_encode($empdata);

Thank you in advance.

[Pullo]

Hi jemz,

One thing you could do is to create an array of arrays in your PHP script.
For example:
[ [id=>'', name=>'', age=>''], [id=>'', name=>'', age=>''], [id=>'', name=>'', age=>''] ]

You could then return it with json_encode and access all of the values you need on a per employee basis.

<?php 
  $arr = Array();

  $con = mysql_connect("localhost","...","...");
  if (!$con){
    die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("test", $con);
  $result = mysql_query("SELECT * FROM employee");
  
  while($row = mysql_fetch_array($result)){
   $temp['id'] =  $row['id'];
   $temp['name'] =  $row['emp_name'];
   $temp['age'] =  $row['emp_age'];
   array_push($arr, $temp);
  }

  mysql_close($con);
  echo json_encode($arr);
?>

Then in your JavaScript:

function  mydropdown(id){
  $.ajax
  ({
    type: "POST",
    url: "toselectdropdown.php",
    data: { "id": id },
    dataType: "json",
    success:function(data)
    {
      for (var i=0;i<data.length;i++){
        console.log(data[i]['id']);
        console.log(data[i]['name']);
        console.log(data[i]['age'] + '\n');
      }
    }
  });
}

Hope this helps.

[jemz]

Hi jemz,

One thing you could do is to create an array of arrays in your PHP script.
For example:
[ [id=>'', name=>'', age=>''], [id=>'', name=>'', age=>''], [id=>'', name=>'', age=>''] ]

You could then return it with json_encode and access all of the values you need on a per employee basis.

<?php 
  $arr = Array();

  $con = mysql_connect("localhost","...","...");
  if (!$con){
    die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("test", $con);
  $result = mysql_query("SELECT * FROM employee");
  
  while($row = mysql_fetch_array($result)){
   $temp['id'] =  $row['id'];
   $temp['name'] =  $row['emp_name'];
   $temp['age'] =  $row['emp_age'];
   array_push($arr, $temp);
  }

  mysql_close($con);
  echo json_encode($arr);
?>

Then in your JavaScript:

function  mydropdown(id){
  $.ajax
  ({
    type: "POST",
    url: "toselectdropdown.php",
    data: { "id": id },
    dataType: "json",
    success:function(data)
    {
      for (var i=0;i<data.length;i++){
        console.log(data[i]['id']);
        console.log(data[i]['name']);
        console.log(data[i]['age'] + '\n');
      }
    }
  });
}

Hope this helps.

Hi pullo, thank you for the reply...Okay i will write back to you if i have some doubt.