how can I use database data in form validation?

Question

Hi I have a form that using database to validation, Now validation just working for latest database data, What should i do to validate for each of data.

DB information

demo-form.php

<form name="login1" action="demo-login.php" onsubmit="return validateForm1()" method="post" id="form1">
<p>user name: <input type="text" name="luser" id="luser"/></p>
<p>password: <input type="password" name="lpass" id="lpass" /></p>
<input type="submit" value="login" name="submit"/>
</form>

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "users";

$conn2 = new mysqli($servername, $username, $password, $dbname);

if ($conn2->connect_error) {
     die("Connection failed: " . $conn2->connect_error);
} 


$sql2 = "SELECT * FROM users";
$result2 = $conn2->query($sql2);

if ($result2->num_rows > 0) {

     while($row = $result2->fetch_assoc()) {
         $luser = $row["user"];
         $lpass = $row["pass"];
     }
} else {
echo "err_2";
}

$conn2->close();
?> 

<script>
function validateForm1(){
var luser = document.getElementById("luser").value;
var lpass = document.getElementById("lpass").value;
        if (luser == "" || lpass == ""){
            document.getElementById("luser").placeholder="Fill, it's necessary";
            document.getElementById("lpass").placeholder="Fill, it's necessary";
            return false;
        }
        if (luser !== "<?php echo $luser; ?>" || lpass !== "<?php echo $lpass; ?>"){
            alert("user or pass is not true");
            return false;
        }
}
</script>

I think, it's better to use ajax for this.

502_Geek
– 502_Geek

2015-06-02 10:35:00 +00:00
Commented Jun 2, 2015 at 10:35 — 502_Geek
– 502_Geek, Commented Jun 2, 2015 at 10:35

DerStoffel · Accepted Answer · 2015-06-02 10:15:42Z

2

It looks like you are using the IDs "luser" and "lpass" multiple times. You cannot allocate markup IDs multiple times. Use classes instead and select with those.

See getElementsByClassName

answered Jun 2, 2015 at 10:15

DerStoffel

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1 Comment

DerStoffel Over a year ago

then show what you've tried. You need to iterate over the elements to validate, not just simplay replace getElementById with getElementsByClassName.

mzage · Accepted Answer · 2015-06-02 13:10:44Z

I Found my answer that using php ajax. i hope it helps to other people.

demo-form.php

<form name="login1" action="demo-login.php" onsubmit="return validateForm1()" method="post" id="form1">
<p>user name: <input type="text" name="luser" id="luser" onmouseleave="showUser(this.value)" onchange="showUser(this.value)"/></p>
<p>password: <input type="password" name="lpass" id="lpass" /></p>
<input type="submit" value="login" name="submit"/>
</form>
<div id="passh"></div>


<script>
function validateForm1(){
var luser = document.getElementById("luser").value;
var lpass = document.getElementById("lpass").value;
    if (luser == "" || lpass == ""){
        document.getElementById("luser").placeholder="Fill, it's necessary";
        document.getElementById("lpass").placeholder="Fill, it's necessary";
        return false;
    }
    if (lpass !== document.getElementById("passh").value){
        alert("user or pass is not true");
        return false;
    }
}

function showUser(str) {
    if (str == "") {
        document.getElementById("lpass").placeholder="";
        return;
    } else { 
        if (window.XMLHttpRequest) {
            xmlhttp = new XMLHttpRequest();
        } else {
            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
        }
        xmlhttp.onreadystatechange = function() {
            if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
    document.getElementById("passh").value = xmlhttp.responseText;
            }
        }
        xmlhttp.open("GET","loginpass.php?q="+str,true);
        xmlhttp.send();
    }
}
</script>

loginpass.php

<?php
$q = $_GET['q'];

$con = mysqli_connect('localhost','root','','users');
if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM users WHERE user = '". $q ."' ";
$result = mysqli_query($con,$sql);

while($row = mysqli_fetch_array($result)) {
    echo  $row['pass'] ;
}

mysqli_close($con);
?>

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