PHP JAVASCRIPT XML MYSQL AJAX

[ajaxl]

Hi I am a total nube with ajax (in fact just broke the ice with it today) I have created a link to mysql database and can view these records as well. I am trying to create a form that will edit the details that are displayed as well.

 

This will update when i hit the Update button but only the first line is this a common problem. Any help would be great.

 

Here is the html and function from my ajax app i can post more if it is useful.

 

HTML/PHP

 

while($row = mysql_fetch_array($result))

{

echo "<form name='meals".$row['ID']."'>";

echo "<input type='text' id='itemID' value='".$row['ID']."'/> ";

echo "<input type='text' id='mealName' value='". $row['mealName'] ."' /><input type='button' onClick='updateRequest();' value='Update' /> <a href='javascript:deleteRequest('".$row['ID']."');'>Delete</a>";

echo "<br/>";

echo "</form>";

}

 

mysql_close($con);

 

 

 

XML/JAVASCRIPT.

 

 

function updateRequest() {

var url="update.php";

//xmlHttp.open('get', 'update.php?' + mealName + '&ID=' + ID);

var mealName = document.getElementById('mealName').value;

var ID = document.getElementById('itemID').value;

 

url=url+"?mealName="+mealName;

url=url+"&ID="+ID;

 

xmlHttp.open("GET",url,true);

 

xmlHttp.onreadystatechange = handleResponse;

xmlHttp.send(null);

}

 

thanks

[priti]

Hi,

 

You are creating multiple forms??? I think to update any row your don't need a form .My advice would be create your <from> tab above the while loop and inside the loop try to make the form fields.It will ease your task.

 

Let me know if you feel some problem after this.

 

regards