PHP Undefined Index

omarsierra27 · Jan 5, 2017

Pa tulong naman, lumalabas kasi "Undefined index: user in C:\wamp64\www\----\testform.php "

<?php

$lhost = "localhost";
$user = "root";
$pss = "";
$dbn = "testarea";

$conn = mysqli_connect ($lhost,$user,$pss);

if (!$conn)
{
die("Connection Failed" .mysqli_connect_error());
}

$sdb = mysqli_select_db ($conn,$dbn);

if (!$sdb)
{
die("No Database" .mysqli_error());
}

$sqlii = "INSERT INTO person (name,contact,email) VALUES ('".$_POST['user']."','".$_POST['number']."','".$_POST['email']."')";

?>

<html>
<title>Test Area</title>
<body>
<form action="testform.php" method="post">
<input type="text" name="user"></br>
<input type="text" name="number"></br>
<input type="text" name="email">
<input type="submit" name="save" value="Save">
</form>
</body>
</html>

Image Attached.
View attachment 298625

filmore · Jan 5, 2017

hinahanap na kasi yong value ng

('".$_POST['user']."','".$_POST['number']."','".$_POST['email']."')";

during loading, kaya nag error dahil walang value or hindi na declare

so dalawa choice mo,

mag lagay ka ng

error_reporting(0);

para di na lumabas ang mga error

or enclose mo sa if yong sa php

<?php

if(!empty($_POST['user'])){
$lhost = "localhost";
$user = "root";
$pss = "";
$dbn = "testarea";

$conn = mysqli_connect ($lhost,$user,$pss);

if (!$conn)
{
die("Connection Failed" .mysqli_connect_error());
}

$sdb = mysqli_select_db ($conn,$dbn);

if (!$sdb)
{
die("No Database" .mysqli_error());
}

$sqlii = "INSERT INTO person (name,contact,email) VALUES ('".$_POST['user']."','".$_POST['number']."','".$_POST['email']."')";
}
?>

masterZEUS · Jan 8, 2017

nilagay mo lang sa variable ($sqlii) yung sql query mo na INSERT... i-execute mo para maprocess at mailagay sa database mo..

jhefreyzz · Jan 8, 2017

maganda siguro if i seperate mo ung connection at operation

//Connection.php

Code:

<?php

$lhost = "localhost";
$user  = "root";
$pss   = "";
$dbn   = "testarea";

$conn = new mysqli($lhost, $user, $pss, $dbn);

if ($conn->connect_errno)
{
    exit('Connection failed' . $conn->connect_error);
}

?>

//Operation Process

Code:

<?php

require_once 'connection.php';

if (isset($_POST['save']))
{
    $user   = $_POST['user'];
    $number = $_POST['number'];
    $email  = $_POST['email'];
    
    $sql = "INSERT INTO person (name,contact,email) VALUES ('$user', '$number', '$email' )";
    $conn->query($sql);
}

?>

eugcar · Jan 10, 2017

jhefreyzz said:

maganda siguro if i seperate mo ung connection at operation

//Connection.php

Code:

<?php

$lhost = "localhost";
$user  = "root";
$pss   = "";
$dbn   = "testarea";

$conn = new mysqli($lhost, $user, $pss, $dbn);

if ($conn->connect_errno)
{
    exit('Connection failed' . $conn->connect_error);
}

?>

//Operation Process

Code:

<?php

require_once 'connection.php';

if (isset($_POST['save']))
{
    $user   = $_POST['user'];
    $number = $_POST['number'];
    $email  = $_POST['email'];
    
    $sql = "INSERT INTO person (name,contact,email) VALUES ('$user', '$number', '$email' )";
    $conn->query($sql);
}

?>

dagdagan mo n lng nito.
$query = $conn->query($sql);
if($query){
echo "<script>alert('success!...');</script>";
}else{
echo "<script>alert('failed!...');</script>";
}

yet ow · Jan 10, 2017

omarsierra27 said:
Pa tulong naman, lumalabas kasi "Undefined index: user in C:\wamp64\www\----\testform.php "

<?php

$lhost = "localhost";
$user = "root";
$pss = "";
$dbn = "testarea";

$conn = mysqli_connect ($lhost,$user,$pss);

if (!$conn)
{
die("Connection Failed" .mysqli_connect_error());
}

$sdb = mysqli_select_db ($conn,$dbn);

if (!$sdb)
{
die("No Database" .mysqli_error());
}

$sqlii = "INSERT INTO person (name,contact,email) VALUES ('".$_POST['user']."','".$_POST['number']."','".$_POST['email']."')";

$query = mysqli_query($conn,$sqlii);

if($query){
echo "Record inserted!";
}else{
echo 'Mysql Error: '.mysqli_error();
}

mysqli_close($conn);
?>

<html>
<title>Test Area</title>
<body>
<form action="testform.php" method="post">
<input type="text" name="user"></br>
<input type="text" name="number"></br>
<input type="text" name="email">
<input type="submit" name="save" value="Save">
</form>
</body>
</html>

Image Attached.
View attachment 1174111

try mo yan ts, yung nka red..

- - - Updated - - -

Tsaka TS, pag irun mo yan ng wala kang variable na ipinasa sa POST mo, error talaga ang lumalabas.

bentesn1per · Jan 10, 2017

mag if ka GAMIT KA NG empty na function::

/// ang ibig sabihn nian kaga
if(!emprty($_POST['user']){
dito mo lagay ung query mo.
}

novhex · Jan 13, 2017

omarsierra27 said:
Pa tulong naman, lumalabas kasi "Undefined index: user in C:\wamp64\www\----\testform.php "

<?php

$lhost = "localhost";
$user = "root";
$pss = "";
$dbn = "testarea";

$conn = mysqli_connect ($lhost,$user,$pss);

if (!$conn)
{
die("Connection Failed" .mysqli_connect_error());
}

$sdb = mysqli_select_db ($conn,$dbn);

if (!$sdb)
{
die("No Database" .mysqli_error());
}

$sqlii = "INSERT INTO person (name,contact,email) VALUES ('".$_POST['user']."','".$_POST['number']."','".$_POST['email']."')";

?>

<html>
<title>Test Area</title>
<body>
<form action="testform.php" method="post">
<input type="text" name="user"></br>
<input type="text" name="number"></br>
<input type="text" name="email">
<input type="submit" name="save" value="Save">
</form>
</body>
</html>

Image Attached.
View attachment 1174111

From

Code:

$sqlii = "INSERT INTO person (name,contact,email) VALUES ('".$_POST['user']."','".$_POST['number']."','".$_POST['email']."')";

To

Code:

if(isset($_POST['user']) && isset($_POST['number']) && isset($_POST['email']){

$sqlii = "INSERT INTO person (name,contact,email) VALUES ('".$_POST['user']."','".$_POST['number']."','".$_POST['email']."')";

}

PHP Undefined Index

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