CODEWITHSUNDEEP
javascriptphphtmlajax
user6577548
user6577548

Reputation:

pass the value to database

I want to pass the value from the Ajax code to the database as this program is to show the details of the user, but there is an error in the code in passing the value. What can I do now?

function showUser() {
  httpRequest = new XMLHttpRequest();

  if (!httpRequest) {
    alert('Giving up :( Cannot create an XMLHTTP instance');
    return false;
  }
  var id = document.getElementById("id").value;
  httpRequest.onreadystatechange = alertContents;

  httpRequest.open("GET", "http://localhost/cart/guser.php?id=" + id + "&rand=" + , true);
  httpRequest.send();
}

function alertContents() {
  if (httpRequest.readyState === XMLHttpRequest.DONE) {
    if (httpRequest.status === 200) {
      document.getElementById("txtHint").innerHTML = httpRequest.responseText;
    }
  }
  var id = document.getElementById('id').value;

}
<form>
  enter digit :
  <input type="text" id="id" name="id" />
  <br />
  <input type='button' onclick='showUser(this.value)' value='select' />

</form>
<br>
<div id="txtHint"><b>Person info will be listed here...</b>
</div>

below code is for guser.php

<!DOCTYPE html>
<html>
<head>
<style>
table {
    width: 100%;
    border-collapse: collapse;
}

table, td, th {
    border: 1px solid black;
    padding: 5px;
}

th {text-align: left;}
</style>
</head>
<body>

<?php
$q = intval($_GET['q']);

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "cart";
 
$conn = mysqli_connect($servername, $username, $password, $dbname);

if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }


mysqli_select_db('cart',$con);

$sql="SELECT * FROM user_details WHERE id = '".$q."'";
$result = mysqli_query($con,$sql);

echo "<table>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>email</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
    echo "<tr>";
    echo "<td>" . $row['FirstName'] . "</td>";
    echo "<td>" . $row['LastName'] . "</td>";
    echo "<td>" . $row['email'] . "</td>";
    echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</body>
</html>

Upvotes: 2

Views: 711

Answers (3)

Viks
Viks

Reputation: 227

Try this as your code have lots of problems code of new html file

<!DOCTYPE html>
<html>
<body>

<form>
  enter digit :
  <input type="text" id="id" name="id" onkeyup='showUser(this.value)'/>
  <br />


</form>
<br>
<div id="txtHint"><b>Person info will be listed here...</b>
</div> 

<script>
function showUser(id) {
  httpRequest = new XMLHttpRequest();

  if (!httpRequest) {
    alert('Giving up :( Cannot create an XMLHTTP instance');
    return false;
  }
else
{

        httpRequest.onreadystatechange = function() {
            if (httpRequest.readyState == 4 && httpRequest.status == 200) {
                document.getElementById("txtHint").innerHTML = httpRequest.responseText;
            }
        };
        httpRequest.open("GET", "localhost/cart/guser.php?id=" + id, true);
        httpRequest.send();
}

}
</script>

</body>
</html>

and code of new guser.php file

<!DOCTYPE html>
<html>
<head>
<style>
table {
    width: 100%;
    border-collapse: collapse;
}

table, td, th {
    border: 1px solid black;
    padding: 5px;
}

th {text-align: left;}
</style>
</head>
<body>

<?php
$id = intval($_GET['id']);

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "cart";

$conn = mysqli_connect($servername, $username, $password, $dbname);

if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }


mysqli_select_db('cart',$con);

$sql="SELECT * FROM user_details WHERE id = '".$id."'";
$result = mysqli_query($con,$sql);

echo "<table>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>email</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
    echo "<tr>";
    echo "<td>" . $row['FirstName'] . "</td>";
    echo "<td>" . $row['LastName'] . "</td>";
    echo "<td>" . $row['email'] . "</td>";
    echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</body>
</html>

Hope this helps!!..Comment for further queries

Upvotes: 1

Harshad Hirapara
Harshad Hirapara

Reputation: 462

Try This:

Add in head:<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.0.0/jquery.min.js"></script>

you have to type person id in inputbox and user info in texthint div:

you will get id in guser.php of user from post and run query.and echo user info into guser.php file

<script type="text/javascript">
    function showUser($id){

    $("#email").keyup(function(){
        $.ajax({
        type: "POST",
        url: "http://localhost/cart/guser.php",
        data:'keyword='$id,
        success: function(data){
            $("#txtHint").html(data);

         }
        });
    });
}
   </script>

Upvotes: 0

Arif
Arif

Reputation: 1643

Either you should remove the , from httpRequest.open("GET", "http://localhost/cart/guser.php?id="+id+"&rand="+,true); jsut before true to make rand=true

or remove the + from the url just after &rand="

Upvotes: 0

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