I have this simple form that updates the values in database, the url of the page is
www.example.com?id=1
After i submit the form the values get updated and i get a success message but the url gets changed, it becomes
www.example.com
Can anyone tell how i can keep the url same i.e: www.example.com?id=1
The code is as follows
<?
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
$text = mysqli_real_escape_string($con, $_POST['textvalue']);
$title = mysqli_real_escape_string($con, $_POST['title']);
$blogid = mysqli_real_escape_string($con, $_POST['blogid']);
$sql = "UPDATE blog SET text='".$text."', title='".$title."' WHERE id='".$blogid."'";
if (mysqli_query($con, $sql))
{
$msg = "Blog updated";
}
else
{
echo "There was an error";
}
}
?>
<div>
<?
if($msg!="")
{
echo $msg;
}
?>
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" method="post" enctype="multipart/form-data" >
<input type="text" name="title" value="<? echo $title; ?>" style="width:100%;"/>
<textarea name="textvalue" ><? echo $text; ?></textarea>
<input type="hidden" name="blogid" value="<? echo $blogid; ?>"/>
<input type="submit" name="edit" alt="edit" value="Edit"/>
</form>
</div>
Related
I am a newbie to PHP. & My PHP Code doesn't work, I want to update some date using MySQL but it seems that first IF condition is 'false' i don't why, I am using PHP 7 & XAMP as a local host, Dreamweaver as an IDE & this is my code:
if(isset($_POST["btn_edit"]))
{
$name = $_POST["name"];
$email = $_POST["email"];
$password = $_POST["password"];
if(!empty($_FILES["img"]["name"]))
{
$img = $_FILES["img"]["name"];
$img_temp = $_FILES["img"]["tmp_name"];
if(move_uploaded_file($img_temp, "assets/images/".$img))
{
$query = mysqli_query($Connection, "UPDATE entry_data SET names='$name',emails='$name',passwords='$password',images='$img' WHERE id='$ID'");
if($query)
{
$result = header("Location:index.php");
}
else
{
echo mysql_error();
}
}
}
else
{
$query = mysqli_query($Connection, "UPDATE entry_data SET names='$name',emails='$name',passwords='$password',images='$img' WHERE id='$ID'");
if($query)
{
echo "<h5>Updated</h5>";
}
}
}
it showing me nothing just refresh the page & this is HTML CODE:
<form method="post" enctype="multipart/form-data">
<input name="name" value="<?php echo $name ?>" />
<input name="email" value="<?php echo $email ?>" />
<input name="password" value="<?php echo $password ?>" />
<img width="50" height="50" src="<?php echo 'assets/images/'.$row[4] ?>" />
<input name="img" type="file" class="text-info" required="required" />
<br/>
<input name"btn_edit" type="submit" />
<?php if(isset($_POST["btn_edit"])) echo $result ?>
You have syntax issue in your button HTML.
This:-
<input name"btn_edit" type="submit" />
Need to be:-
<input name="btn_edit" type="submit" /><!-- = is missing in name -->
I have written the below HTML code:
<form action="index.php" method="POST">
<input type="text" name="title" required>
<input type="text" name="brief_text" required>
<textarea name="text" required></textarea>
<input type="submit" name="add" value="Add">
</form>
My PHP code:
<?php
require_once('db.php');
if(isset($_POST['add'])){
$title = $_POST['title'];
$brief_text = $_POST['brief_text'];
$text = $_POST['text'];
$blog_cat_id = $_POST['blog_cat_id'];
if($title AND $brief_text AND $text AND $blog_cat_id){
$insert_blog = "insert into blog values ('','$title','$brief_text','$text','$blog_cat_id',NOW())";
$run_insertion = mysqli_query($con, $insert_blog);
if($run_insertion){
echo "Blog has been added!";
}
else{
echo "Error adding blog!!!";
}
}
else{
echo "All fields are required!";
}
}
else{
echo "GOODBYE";
}
?>
Every time I refresh the page, it only shows the form and "GOODBYE" and does not even insert the data into database table.
Help me out please.
Is it still showing 'GOODBYE' now you've changed
$_POST['add_blog']
to
$_POST['add']
when you click submit?
You have few mistakes,
1) should be: $_POST['add'] instead of $_POST['add_blog']
2) don't have $_POST['blog_cat_id'] as not in form
EDIT
Copied your code and made some changes:
code:
if(isset($_POST['add'])){
print_r($_POST);
$title = $_POST['title'];
$brief_text = $_POST['brief_text'];
$text = $_POST['text'];
$blog_cat_id = $_POST['blog_cat_id'];
if($title AND $brief_text AND $text AND $blog_cat_id){
echo "inside condition";
$insert_blog = "insert into blog values ('','$title','$brief_text','$text','$blog_cat_id',NOW())";
$run_insertion = mysqli_query($con, $insert_blog);
if($run_insertion){
echo "Blog has been added!";
}
else{
echo "Error adding blog!!!";
}
}
else{
echo "All fields are required!";
}
}
else{
echo "GOODBYE";
}
HTML:
<form action="index.php" method="POST">
<input type="text" name="title" required>
<input type="text" name="brief_text" required>
<input type="text" name="blog_cat_id" required>
<textarea name="text" required></textarea>
<input type="submit" name="add" value="Add">
</form>
output
Array ( [title] => test [brief_text] => test [blog_cat_id] => 1 [text] => testing [add] => Add )
inside condition
Now, check your query if doesn't work.
Hope this will help you.
Your query is wrong, columns are not specified and you are open to sql injection you should learn to use parameterized query. but for this time you can use the following.
Try this:
htmlcode
<form action="index.php" method="POST">
<input type="text" name="title" required>
<input type="text" name="brief_text" required>
<input type="text" name="blog_cat_id" required>
<textarea name="text" required></textarea>
<input type="submit" name="add" value="Add">
</form>
index.php
<?php
require_once('db.php');
if(isset($_POST['add'])){
$title = $_POST['title'];
$brief_text = $_POST['brief_text'];
$text = $_POST['text'];
$blog_cat_id = $_POST['blog_cat_id'];
if($title AND $brief_text AND $text AND $blog_cat_id){
$insert_blog = "insert into blog('col1','col2','col3','col4','col5','col6') values ('','$title','$brief_text','$text','$blog_cat_id',NOW())";
$run_insertion = mysqli_query($con, $insert_blog);
if($run_insertion){
echo "Blog has been added!";
}
else{
echo "Error adding blog!!!";
}
}
else{
echo "All fields are required!";
}
}
?>
Note : col1, col2, col3, col4, col5 and col6 will be your column name.
i have a code for updating data to myql. It looks doesn't have a problem but it ain't changed
my update code :
//previous data//
....
if (isset($_POST['update'])) {
$nim = mysqli_real_escape_string($connection, ($_POST['nim']));
$name = mysqli_real_escape_string($connection, ($_POST['name']));
$class1 = mysqli_real_escape_string($connection, ($_POST['class2']));
$class2 = mysqli_real_escape_string($connection, ($_POST['class1']));
if (!preg_match("/^[1-9][0-9]*$/",$nim)) {
$error = true;
$nim_error = "NIM only contain numbers";
}
if (!preg_match("/[^a-zA-Z]/",$name)) {
$error = true;
$name_error = "NIM only contain numbers";
}
if (!preg_match("/^[1-9][0-9]*$/",$class1)) {
$error = true;
$class1_error = "Class only contain numbers";
}
if (!preg_match("/^[1-9][0-9]*$/",$class1)) {
$error = true;
$class2_error = "Class only contain numbers";
}
$result = "UPDATE users SET nim='$nim', name='$name', class1='$class1', class1='$class1' WHERE id='$id'";
mysqli_query($connection, $result);
}
?>
and this is my html code :
<div id="popup2" class="overlay">
<div class="popup">
<h2 class="range2">Edit</h2>
<a class="close" href="#">×</a>
<div class="content">
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input class="input" type="text" name="nim" placeholder="NIM" required/>
<input class="input" type="text" name="name" placeholder="Name" required/>
<i>SK</i>
<input class="input1" type="text" name="class1" placeholder="00" required/>
<i>-</i>
<input class="input1" type="text" name="class2" placeholder="00" required/>
<input name="update" type="submit" class="button" id="submit" value="Submit">
</form>
</div>
</div>
</div>
is there any wrong code ? Thank you..
It is really hard to explain: Take a look.
If you want to update a single data you will need a identity(Primary
key). That mean which data you want to update.
Below Example: check index.php file
In file index.php change dbname to your database name in connection.
browse project_url/index.php?id=1 [here use any id from your database]
Then update your data.
index.php
//Show existed data againist id
if(isset($_GET['id'])){
$id = $_GET['id'];
$stmt = $pdo->prepare('SELECT * FROM users WHERE id = :id');
$stmt->execute(array('id'=>$id));
$data = $stmt->fetch();
if (empty($data)) {
echo "No data found in user table. Use proper ID.";
}
}
//Update query
$msg = array();
if (isset($_POST['id']) && $_POST['id']!='') { //operation is update, because id exist
if($_POST['nim']!=0 && is_numeric($_POST['nim'])){
$nim = $_POST['nim'];
}else{
$msg[]="Nim only can be number";
}
if($_POST['name']!=''){
$name = $_POST['name'];
}else{
$msg[]="came only can not be empty";
}
if(is_numeric($_POST['class1'])){
$class1 = $_POST['class1'];
}else{
$msg[]="Class1 only can be number";
}
if(is_numeric($_POST['class2'])){
$class2 = $_POST['class2'];
}else{
$msg[]="Class1 only can be number";
}
$id = $_POST['id'];
if(count($msg)==0){
$stmt = $pdo->prepare('UPDATE users SET nim=:nim, name=:name, class1=:class1, class2=:class2 WHERE id=:id');
$result = $stmt->execute(array(
'nim' => $nim,
'name' => $name,
'class1'=> $class1,
'class2'=> $class2,
'id' => $id,
));
if($result){
echo "successfully updated.";
}else{
echo "update failed";
}
}
}else{
//You can run here insert operation because id not exist.
echo "Id not set";
}
?>
<div id="popup2" class="overlay">
<div class="popup">
<h2 class="range2">Edit</h2>
<a class="close" href="#">×</a>
<div class="content">
<?php foreach ($msg as $value) {
echo $value."<br>";
}?>
<form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<?php if(isset($data)){?>
<input class="input" type="hidden" name="id" value="<?php echo $data['id']; ?>" />
<?php } ?>
<input class="input" type="text" name="nim" value="<?php echo isset($data)?$data['nim']:''?>" placeholder="NIM" required/>
<input class="input" type="text" name="name" value="<?php echo isset($data)?$data['name']:''?>" placeholder="Name" required/>
<i>SK</i>
<input class="input1" type="text" name="class1" value="<?php echo isset($data)?$data['class1']:''?>" placeholder="00" required/>
<i>-</i>
<input class="input1" type="text" name="class2" value="<?php echo isset($data)?$data['class2']:''?>" placeholder="00" required/>
<input name="update" type="submit" class="button" id="submit" value="Submit">
</form>
</div>
</div>
</div>
My friend,
only do one thing to resolve this
echo $result = "UPDATE users SET nim='$nim', name='$name', class1='$class1', class1='$class1' WHERE id='$id'";
die;
then submit your form again and you will get your static query into your page then just copy that query and try to run into phpmyadmin then you will get your actual error.
member.php
<?php
if(isset($_POST['submit']))
{
$membername = $_POST['membername'];
if(empty($membername))
{
$errors .= "Please enter member name<br />";
}
if($errors)
{
$action = $_POST['submit'];
echo $errors;
displayForm();
}
else
{
?>
<input type="hidden" name="mname" value="<?php echo $_POST['membername']; ?>" />
<?php
$action = $_POST['submit'];
header("Location: commit.php?action=$action");
exit();
}
}
else
{
displayForm();
}
?>
<?php
function displayForm()
{
?>
<form action = "member.php" action="post">
Member Name <input type="text" name="membername" value="<?php if(isset($row['name'])) echo $row['name'];
else echo ''; ?>" /><br /><input type="text" name="membername" value=
<input type="submit" name="submit" name="add" />
</form>
<?php
}
?>
Commit.php
<?php
echo $_POST['mname']; //HERE
?>
I want to pass the hidden value from member.php. When I run commmit.php, I want to get hidden field value. However, the error is the following:
**Undefined index: mname in member.php in commit.php.
What am I doing wrong?
I'm building a form, and I want that all the inserted values will be kept, in case of form submit failure. This is my code:
<?php
$error = "";
$name = "";
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$name = $_POST["name"];
// Verify $_POST['name'] greater than 4 chars
if ( strlen($name) < 4 ){
$error= 'Name too short!';
}
}
?>
<html>
<head>
</head>
<body>
<form method="post" action="<?php echo htmlspecialchars($_SERVER['PHP_SELF']); ?>" name="myForm" id="idForm">
<input type="text" placeholder="Name" id="name" name="name" value=""/>
<input type="submit" value="submit"/>
</form>
<?php
echo "<h2>Input:</h2>";
echo $name;
if($error) {
// No Update AND refresh page with $name in text field
echo "<br/>" . $error;
} else {
// Update name in DB
}
?>
</body>
</html>
I would like that name field keeps the inserted input text, after submit. I tried to do with php code in input value but doesn't work. Any ideas?
Solved. This is the solution that I was looking for.
I added in value tag of input the following:
<?php if (isset($_POST['name'])) echo $_POST['name']; ?>
Therefore input field would look like:
<input type="text" placeholder="Name" id="name" name="name" value="<?php if (isset($_POST['name'])) echo $_POST['name']; ?>"/>
Thanks for your responses, helped me.
<?php
$error = "";
$name = isset($_POST["name"])?$_POST["name"]:""; //Added condition
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$name = $_POST["name"];
// Verify $_POST['name'] greater than 4 chars
if ( strlen($name) < 4 ){
$error= 'Name too short!';
}
}
?>
<html>
<head>
</head>
<body>
<form method="post" action="<?php echo htmlspecialchars($_SERVER['PHP_SELF']); ?>" name="myForm" id="idForm">
<input type="text" placeholder="Name" id="name" name="name" value="<?php echo $name; ?>"/>
<input type="submit" value="submit"/>
</form>
<?php
echo "<h2>Input:</h2>";
echo $name;
if($error) {
// No Update AND refresh page with $name in text field
echo "<br/>" . $error;
} else {
// Update name in DB
}
?>
</body>
</html>
You can just echo $_POST['name'] in the value attribute of the input.
Make sure you check POST values to avoid XSS.
I also put up the update DB function, as you don't need to show the form to the user if the name in longer the 4 chars!
<?php
$error = "";
$name = "";
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (isset($_POST['name'])){ //change name content only if the post value is set!
$name = filter_input (INPUT_POST, 'name', FILTER_SANITIZE_STRING); //filter value
}
// Verify $_POST['name'] greater than 4 chars
if ( strlen($name) < 4 ){
$error= 'Name too short!';
} else {
// Update name in DB
// Redirect
}
}
?>
<html>
<head>
</head>
<body>
<form method="post" action="<?php echo htmlspecialchars($_SERVER['PHP_SELF']); ?>" name="myForm" id="idForm">
<input type="text" placeholder="Name" id="name" name="name" value="<?php echo $name; ?>"/>
<input type="submit" value="submit"/>
</form>
<?php
echo "<h2>Input:</h2>";
echo $name;
if($error) {
// No Update AND refresh page with $name in text field
echo "<br/>" . $error;
};
?>
</body>
</html>
If you want to keep all values/inputs for further use you could achieve that with php session values.
Besides - you should use $_SERVER['SCRIPT_NAME'] instead of $_SERVER['PHP_SELF']
Mars