php javascript beginner update mysql

PHP_CHILD · February 7, 2013

i have a form. if user enter the data, the db table display. but i want the table to get displayed when the user logins for the first time.

simple one but i can't get it how...

aa.html

<html>
<body>


<script language="javascript" type="text/javascript">
<!-- 
//Browser Support Code

function ajaxFunction(){
var ajaxRequest;  // The variable that makes Ajax possible!

try{
 // Opera 8.0+, Firefox, Safari
 ajaxRequest = new XMLHttpRequest();
} catch (e){
 // Internet Explorer Browsers
 try{
  ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
 } catch (e) {
  try{
   ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
  } catch (e){
   // Something went wrong
   alert("Your browser broke!");
   return false;
  }
 }
}


// Create a function that will receive data sent from the server
ajaxRequest.onreadystatechange = function(){
 if(ajaxRequest.readyState == 4){
  var ajaxDisplay = document.getElementById('ajaxDiv');
  ajaxDisplay.innerHTML = ajaxRequest.responseText;
 }
}
var age = document.getElementById('age').value;
var wpm = document.getElementById('wpm').value;
var sex = document.getElementById('sex').value;
var name = document.getElementById('name').value;
var queryString = "?age=" + age + "&wpm=" + wpm + "&sex=" + sex + "&name="  + name;
$a=ajaxRequest.open("GET", "ss1.php" + queryString, true);




ajaxRequest.send(null); 

}


//-->
</script>




<form name='myForm'>
Name:<input type='text' id='name' /> <br />
Age: <input type='text' id='age' /> <br />
WPM: <input type='text' id='wpm' />
<br />
Sex: <select id='sex'>
<option value='m'>m</option>
<option value='f'>f</option>
</select>
<input type='button' onclick='ajaxFunction()' value='Insert New row' />
</form>
<div id='ajaxDiv'>Your result will display here</div>
</body>
</html>

ss1.php

$age = $_GET['age'];
$sex = $_GET['sex'];
$wpm = $_GET['wpm'];
$name= $_GET['name'];
// Escape User Input to help prevent SQL Injection
$a=isset($_GET['age']) && $_GET['age'];
$b=isset($_GET['sex']) && $_GET['sex'];
$c=isset($_GET['wpm']) && $_GET['wpm'];
$d=isset($_GET['name']) && $_GET['name'];




  $age = mysql_real_escape_string($age);
  $sex = mysql_real_escape_string($sex);
  $wpm = mysql_real_escape_string($wpm);
  $name = mysql_real_escape_string($name);
   //build query
  $query = "INSERT INTO `ajax_example`(`ae_name`, `ae_age`, `ae_sex`, `ae_wpm`)
   VALUES ('$name','$age','$sex','$wpm')";
   $qry = mysql_query($query) or die(mysql_error());







$a="SELECT `ae_name`, `ae_age`, `ae_sex`, `ae_wpm` FROM `ajax_example`";
$qry_result=mysql_query($a) or die(mysql_error());





//Build Result String
$display_string = "<table>";
$display_string .= "<tr>";
$display_string .= "<th>Name</th>";
$display_string .= "<th>Age</th>";
$display_string .= "<th>Sex</th>";
$display_string .= "<th>WPM</th>";
$display_string .= "</tr>";


// Insert a new row in the table for each person returned
while($row = mysql_fetch_array($qry_result)){
$display_string .= "<tr>";
$display_string .= "<td>$row[ae_name]</td>";
$display_string .= "<td>$row[ae_age]</td>";
$display_string .= "<td>$row[ae_sex]</td>";
$display_string .= "<td>$row[ae_wpm]</td>";
$display_string .= "</tr>";

}


$display_string .= "</table>";
echo $display_string;
?>

if u could tell me show good websites to practice javascript. woulg b gr8.. thank u in advances.

Christian F. · February 7, 2013

You have to add a field to the table, where you save a boolean that tells whether or not this is the first time the user has logged on. Set it to 0 (false) by default.

Then, when the user logs in, retrieve the flag with the rest of the details. If it's set to 0 then show the form, and update the flag to read 1 (true) at the same time as you save the other details from the form.

Edited February 7, 2013 by Christian F.

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php javascript beginner update mysql

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