i have this code for the user-edit.php
what i need with this code is to fetch user data from database and show it to the textbox and also the user able to edit the textbox value and updating the database
<?php
include("config/session.php");
include("config/connection.php");
$user_id = $_SESSION['LOGGED_USER_ID'];
?>
<?php
$sql_query = "SELECT * FROM table_users WHERE `SNo` = '$user_id'";
$query = mysql_query($sql_query);
//$i = 1;
$fetch = mysql_fetch_assoc($query);
//$user_id = $_GET['id'];
?>
<form name="form1" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" >
<fieldset>
<p>
<label for="simple-input" >User Name</label>
<input type="text" id="UserName" class="round default-width-input" autofocus name="UserName" value="<?php echo $fetch['UserName'];?>" readonly="readonly" />
</p>
<p>
<label for="simple-input" >Password</label>
<input type="text" id="pass_word" class="round default-width-input" autofocus name="pass_word" value="<?php echo $fetch['pass_word'];?>" />
</p>
<p>
<label for="simple-input" >Email ID</label>
<input type="text" id="Email" class="round default-width-input" autofocus name="Email" value="<?php echo $fetch['Email'];?>" />
</p>
<p>
<label for="simple-input" >Website</label>
<input type="text" id="website" class="round default-width-input" autofocus name="website" value="<?php echo $fetch['website'];?>" />
</p>
</fieldset>
<input type="submit" class="btn btn-primary btn-large" name="form_submit" value="Update Data"/>
</form>
<?php
if(isset($_POST['form_submit']))
{
"UPDATE `table_users` SET `pass_word` = '".$_POST['pass_word']."',`Email` = '".$_POST['Email']."',`website` = '".$_POST['website']."', WHERE `SNo` = '$user_id'";
// sql query for update data into database
if(mysql_query($sql_query))
{
echo '<script type="text/javascript">';
echo 'alert("Data Are Updated Successfully");';
echo '</script>';
}
else
{
echo '<script type="text/javascript">';
echo 'alert("error occured while updating data");';
echo '</script>';
}
}
?>
</div>
</div>
</div>
been working with this for hours and still the data are not updated to the mysql database, been trying several way but still the textbox value cant update the database, please help
remove , before where in query
$sql_query="UPDATE `table_users` SET `pass_word` = '".$_POST['pass_word']."',`Email` = '".$_POST['Email']."',`website` = '".$_POST['website']."' WHERE `SNo` = '$user_id'";
store query in $sql_query because you not store update string into variable
without store in $sql_query you run sql query
if(mysql_query($sql_query)) so store update query in $sql_query
Related
I need to display only the id row of the database on a textbox, my problem was the data is displaying all id's rows and columns.
I was fetching all data in the database, but what I need is only the id on the database.
Here is my code in forms, input text fields and submit.
<form action="" method="POST">
<div class="row col-md-4">
<label>Amount</label>
<input type="text" name="id" class="form-control validate">
<br>
<input type="submit" class="form-control btn-warning" name="search" value="Search Data"></input><br>
<?php
$connection = mysqli_connect("localhost","root","");
$db = mysqli_select_db($connection, 'qrproject');
if(isset($_POST['search']))
{
$id = $_POST['id'];
$query = "SELECT * FROM scratch_cards WHERE amount='$id' ";
$query_run = mysqli_query($connection, $query);
while($row = mysqli_fetch_array($query_run))
{
?>
<form action="" method="POST">
<input type="text" name="code" value="<?php echo $row['code'] ?>" class="form-control validate" id="mapo">
<input type="text" name="pin" value="<?php echo $row['pin'] ?>" class="form-control validate" id="mact">
<input type="text" name="status" value="<?php echo $row['status'] ?>" class="validate form-control" id="soluong">
<input type="date" name="card_expiration" value="<?php echo $row['card_expiration'] ?>" class="validate form-control" id="cardex">
<input type="number" name="card_validity" value="<?php echo $row['card_validity'] ?>" class="validate form-control" id="cardval">
</form>
<?php
}
}
?>
</form>
From your comments, it seems what you want to do is output a single, random row which matches the $_POST['id'] value against the amount column in your table. You can do this by changing your query to this:
SELECT * FROM scratch_cards WHERE amount='$id' ORDER BY RAND() LIMIT 1
You can also change
while($row = mysqli_fetch_array($query_run))
to
if($row = mysqli_fetch_array($query_run))
although with the change to the query to limit the output to one row this is no longer absolutely necessary.
As has been pointed out, you are vulnerable to SQL injection, you should read this question to see how to resolve that.
Hello I have a database and its data is this please see this image
I need a solution to this problem, the solution is i need the data to be change and not repeating, so when i input a value that is written on the database the displayed value on the textbox will not REPEAT and Change everytime i input the same value.
<form action="" method="POST">
<div class="row col-md-4">
<label>Amount</label>
<input type="text" name="id" class="form-control validate">
<br>
<input type="submit" class="form-control btn-warning" name="search" value="Search Data"></input><br>
//HERE IS WHERE I SUBMIT THE DATA RIGHT NOW IT IS NOT RANDOMIZED WHEN I SUBMIT AGAIN THE SAME VALUE APPEARS
<?php
$connection = mysqli_connect("localhost","root","");
$db = mysqli_select_db($connection, 'qrproject');
if(isset($_POST['search'])){
$id = $_POST['id'];
$query = "SELECT * FROM scratch_cards WHERE amount='$id' ";
$query_run = mysqli_query($connection, $query);
while($row = mysqli_fetch_array($query_run)) {
?>
<form action="" method="POST">
<input type="text" name="code" value="<?php echo $row['code'] ?>" class="form-control validate" id="mapo">
<input type="text" name="pin" value="<?php echo $row['pin'] ?>" class="form-control validate" id="mact">
<input type="text" name="status" value="<?php echo $row['status'] ?>" class="validate form-control" id="soluong">
<input type="date" name="card_expiration" value="<?php echo $row['card_expiration'] ?>" class="validate form-control" id="cardex">
<input type="number" name="card_validity" value="<?php echo $row['card_validity'] ?>" class="validate form-control" id="cardval">
</form>
<?php
}
}
?>
</form>
Maybe as per your comment, you need unique records, so you can use DISTINCT in your query so by giving DISTINCT to any column, you won't get repeated records.
Here in your case:
SELECT DISTINCT id, amount FROM scratch_cards WHERE amount='$id'
Maybe this can help you to get unique records
I'm newbie in php and i make a simple CRUD app. Unfortunately, I'm stacked with this problem, I don't know what's wrong with my code in my update.php. When i click update in my index.php it Undefined variable. I think my value in form is wrong. Any help is appreciated.
update.php
<?php
include("connection.php");
if (isset($_POST['customerNumber'])) {
$customerNumber = $_POST['customerNumber'];
$q = "SELECT customerNumber, checkNumber, paymentDate, amount FROM payments WHERE customerNumber='$customerNumber'";
$rq = mysqli_query($conn, $q);
while ($row = mysqli_feth_assoc($rq)) {
$customerNumber = $row['customerNumber'];
$checkNumber = $row['checkNumber'];
$paymentDate = $row['paymentDate'];
$amount = $row['amount'];
}
}
?>
<!-- from the index.php update -->
<form action="update.php?customerNumber=$customerNumber" method="post">
<label>
<input type="text" name="customerNumber" value="<?php echo $row['customerNumber']; ?>" placeholder="Customer Number" required>
</label>
<label>
<input type="text" name="checkNumber" value="<?php echo $row['checkNumber']; ?>" placeholder="Check Number" required>
</label>
<label>
<input type="text" name="paymentDate" value="<?php echo $row['paymentDate']; ?>" placeholder="Payment Date" required>
</label>
<label>
<input type="number" name="amount" value="<?php echo $row['amount']; ?>" placeholder="Amount">
</label>
<input type="submit" name="submit" value="update">
</form>
<?php
include('connection.php');
if (isset($_POST['submit'])) {
$customerNumber = $_POST['customerNumber'];
$checkNumber = $_POST['checkNumber'];
$paymentDate = $_POST['paymentDate'];
$amount = $_POST['amount'];
$q = "UPDATE payments SET customerNumber='$customerNumber', checkNumber='$checkNumber', paymentDate='$paymentDate', amount='$amount' WHERE customerNumber='$customerNumber' ";
$rq = mysqli_query($conn, $q);
if($rq){
header('Location: index.php');
}else{
echo "Something went wrong";
}
}
?>
I am trying to fill a html form with data being received out of my mysql database. However I cannot set the forms to display on-load the variables being extracted from the database. I would like the form on-load to hold the data last entered into the forms which have been added to the database previously.
$query = "SELECT FROM character_tbl WHERE character_player
='".$_SESSION["user"]."' character_tbl";
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
$name = $row['character_name'];
$race = $row['character_race'];
$class = $row['character_class'];
$alignment = $row['character_alignment'];
$hp = $row['character_hp'];
$str = $row['character_str'];
$dex = $row['character_dex'];
$con = $row['character_con'];
$int = $row['character_int'];
$wis = $row['character_wis'];
$cha = $row['character_cha'];
$ac = $row['character_ac'];
$touch = $row['character_touch'];
$flat = $row['character_flat'];
$fort = $row['character_fort'];
$ref = $row['character_ref'];
$will = $row['character_will'];
}
echo $will;
mysql_close();
?>
<!DOCTYPE html>
<html>
<body>
<div id="nav">
<form action="user.php">
<input type="submit" value="Back">
</form>
</div>
<div id="section">
<form action="update.php" method="POST">
Character Name:<br>
<input type="text" name="name" value="<?php echo $name;?>">
<br>
Race<br>
<input type="text" name="race" value="<?php echo $race;?>">
<br>
Class<br>
<input type="text" name="class" value="<?php echo $class;?>">
<br>
Alignment<br>
<input type="text" name="alignment" value="<?php echo $alignment;?>">
<br>
HP<br>
<input type="text" name="hp" value="<?php echo $hp;?>">
<br>
STR<br>
<input type="number" name="str" value="<?php echo $str;?>">
<br>
DEX<br>
<input type="number" name="dex" value="<?php echo $dex;?>">
<br>
CON<br>
<input type="text" name="con" value="<?php echo $con;?>">
<br>
INT<br>
<input type="text" name="int" value="<?php echo $int;?>">
<br>
WIS<br>
<input type="text" name="wis" value="<?php echo $wis;?>">
<br>
CHA<br>
<input type="text" name="cha" value="<?php echo $cha;?>">
<br>
AC<br>
<input type="text" name="ac" value="<?php echo $ac;?>">
<br>
Touch AC<br>
<input type="text" name="touch" value="<?php echo $touch;?>">
<br>
Flat-Footed AC<br>
<input type="text" name="flat" value="<?php echo $flat;?>">
<br>
Fortitude<br>
<input type="text" name="fort" value="<?php echo $fort;?>">
<br>
Reflex<br>
<input type="text" name="ref" value="<?php echo $ref;?>">
<br>
Will<br>
<input type="text" name="will" value="<?php echo $will;?>">
</br>
<input type="submit" value="Update">
</form>
I think the SQL has error:
SELECT FROM character_tbl WHERE character_player
try:
SELECT * FROM character_tbl WHERE character_player
You have syntax error in your mysql query. You have not place field or columns name or (*) for all columns to extract.
Try like this..
$query = "SELECT * FROM character_tbl WHERE character_player ='".$_SESSION['user']."'";
<?php
//include 'includes/connectie.php';
if (isset($_GET['id'])){
$product_id=$_GET['id'];
} else {
$product_id=$_POST['id'];
}
$user = 'userID';
$pass = 'mypassword';
$dbh = new PDO( 'mysql:host=localhost;dbname=webshop', $user, $pass );
$sql = "SELECT * FROM `producten` WHERE product_id='$product_id'";
$sql_result = $dbh->query($sql);
foreach($sql_result as $row)
{
$prijs=$row['prijs'];
$product_naam=$row['product_naam'];
$product_categorie=$row['product_categorie'];
$product_specificaties=$row['product_specificaties'];
$foto=$row['foto'];
$product_id=$row['product_id'];
$product_soort=$row['product_soort'];
echo "Product id nummer:", $product_id;
}
//$_SESSION['prijs'] = $prijs;
if ($_SERVER["REQUEST_METHOD"] == "POST"){
//if (!empty($product_naam) && !empty($product_specifcaties) && !empty($product_categorie) && !empty($prijs)
//&& !empty($product_soort))
If (isset($_POST['submit']))
{
$sql = "UPDATE producten
SET prijs='$prijs', product_naam='$product_naam', product_specificaties='$product_specificaties',
product_categorie='$product_categorie', product_soort='$product_soort',
WHERE product_id='$product_id'";
$query = $dbh->prepare( $sql );
$result = $query->execute();
if ($result){
echo "Product aangepast!!!!! in id:";
echo $product_id;
} else {
echo "Product NIET aangepast!!!!";
}
}
}
?>
<form name="admin" action="producten_echt_aanpassen.php" method="POST" enctype="multipart/form-data">
<p>
<label for 'product_id'>Product ID: </label><br>
<input type="text" name="id" value="<?php print $product_id; ?>"/>
</p>
<p>
<label for 'product_naam'>Naam: </label><br>
<input type="text" name="product_naam" value="<?php print $product_naam; ?>"/>
</p>
<p> <label for 'product_specificaties'>Specificaties: </label><br>
<textarea rows= "4" cols="50" name="product_specificaties"><?php print $product_specificaties; ?>
</textarea>
</p>
<p>
<label for 'prijs'>Prijs: </label><br>
<input type="text" name="prijs" value="<?php print $prijs; ?>"/>
</p>
<p>
<label for 'product_categorie'>Iphone: </label><br>
<input type="text" name="product_categorie" value="<?php print $product_categorie; ?>"/>
</p>
<p>
<label for 'product_soort'>Soort: </label><br>
<input type="text" name="product_soort" value="<?php print $product_soort; ?>"/>
</p>
<br/>
<label for 'uploadfile'>Kies foto <img src="<?php print $foto; ?>"></label><br>
<input type="file" name="file" ><br><br>
<input type="submit" name="submit" value="Submit">
</form>
I have a form in which I load properties of products like the product name, price, photo etc. The properties are then possible to change and then updated in the database. But the sql update statement does not execute. Can anybody help me out?
there is a , before the where on the update that should not be there. Try to activate error reporting like this: How to get useful error messages in PHP? so that you know wwhy things are failing