I am trying to edit form data by displaying the previous saved data on the form and then update it. It shows the data on the form which is saved in database but when i enter the new data it does not get the id of the row. I echo the update query, it shows the changed values but it shows id equals to empty. Here is my code for edit record and update; Edit record is working but update isn't:
<?php
include('connection.php');
$id = '';
if( isset( $_GET['id'])) {
$id = $_GET['id'];
}
$udfname = mysql_real_escape_string($_POST["udfname"]);
$udlname = mysql_real_escape_string($_POST["udlname"]);
$udpwd = mysql_real_escape_string($_POST["udpwd"]);
$udeml = mysql_real_escape_string($_POST["udeml"]);
$udnum = mysql_real_escape_string($_POST["udnum"]);
$query="UPDATE form
SET fname = '$udfname', lname = '$udlname', pwd = '$udpwd', eml = '$udeml', num = '$udnum'
WHERE id='$id'";
$res= mysql_query($query);
if($res){
echo "<p> Record Updated<p>";
}else{
echo "Problem updating record. MY SQL Error: " . mysql_error();
}
?>
Form for editing record:
<?php
include('connection.php');
$id = (int)$_GET['id'];
$query = mysql_query("SELECT * FROM form WHERE id = '$id'") or die(mysql_error());
while($row = mysql_fetch_array($query)) {
echo "";
$fname = $row['fname'];
$lname = $row['lname'];
$pwd = $row['pwd'];
$eml = $row['eml'];
$num = $row['num'];
}
?>
<html>
<head>
<title>Edit</title>
<script>
'
'
Jquery code here
'
'
</script>
</head>
<body>
<form action="update.php" method="post">
<input type="hidden" name="ID" value="<?=$id;?>">
First Name: <input type="text" name="udfname" value="<?=$fname;?>"><br>
Last Name: <input type="text" name="udlname" value="<?=$lname?>"><br>
Password: <input type="text" name="udpwd" value="<?=$pwd?>"><br>
Email: <input type="text" name="udeml" value="<?=$eml?>"><br>
Contact Number: <input type="text" name="udnum" value="<?=$num?>"><br>
<input type="Submit">
</form>
</body>
</html>
At update time your form is submitted using POST request. So you need to get ID using POST method. So to get ID of hidden field change your code as below:
$id = '';
if( isset( $_POST['ID'])) {
$id = $_POST['ID'];
}
Please try below code
if( isset( $_POST['id']) && $_POST['id']!=null) {
$id = $_POST['id'];
}
Dear i think the problem with your method you are sending the data using post method and its very simple instead of this code
if( isset( $_GET['id'])) {
$id = $_GET['id'];
}
write
if( isset( $_POST['id'])) {
$id = $_POST['id'];
}
and one more thing that is you are using the mysql deprecated function for database kindly use the pdo for this or new mysqli functions.
Related
I have been following a lesson on how to make an admin page. I got all the information out of my database to a table on the page. I have an update button and when I change the information and press the button I receive this error: Warning: undefined array key "WebID" in ..\Update.php on line 3
From my search online everyone is trying to change the code so that if array key does not exist: return null. I tried that and the error does not appear no more, but the table does not change.
Any thoughts?
This is the code:
<?php
require_once("DB/DB.php");
$SearchQueryParameter = $_GET["WebID"];
if (isset($_POST["Update"])) {
$Ename = $_POST["Ename"];
$Eid = $_POST["Eid"];
$Erank = $_POST["Erank"];
$Eemail = $_POST["Eemail"];
$Edate = $_POST["Edate"];
$Epassword = $_POST["Epassword"];
$Specialisms = $_POST["Specialisms"];
global $ConnectingDB;
$sql ="UPDATE emp_data SET Ename='$Ename', Eid='$Eid', Erank='$Erank', Eemail='$Eemail', Edate='$Edate', Epassword='$Epassword',
Specialisms='$Specialisms' WHERE WebID='$SearchQueryParameter'";
$Execute = $ConnectingDB->query($sql);
if ($Execute) {
echo '<script>window.open("adminpage.php?WebID=Recored Updated","_self")</script>';
}
}
?>
<?php
<?php
global $ConnectingDB;
$sql = "SELECT * FROM emp_data WHERE WebID='$SearchQueryParameter'";
$stmt = $ConnectingDB->query($sql);
while ($DataRows = $stmt->fetch()) {
$WebID = $DataRows["WebID"];
$Ename = $DataRows["Ename"];
$Eid = $DataRows["Eid"];
$Erank = $DataRows["Erank"];
$Eemail = $DataRows["Eemail"];
$Edate = $DataRows["Edate"];
$Epassword = $DataRows["Epassword"];
$Specialisms = $DataRows["Specialisms"];
}
?>
Html file used to update:
<form id="UpdateForm" method="post" action="Update.php?WebID<?php echo $SearchQueryParameter; ?>">
<div class="form-group">
<button type="submit" name="Update" class="form-control-submit-button">Update</button>
</div>
you have to write the form action like this.. you missed the = sign
action="Update.php?WebID=<?php echo $SearchQueryParameter; ?>"
<form id="UpdateForm" method="post" action="Update.php?WebID=<?php echo $SearchQueryParameter; ?>">
You missed the = sign, in the url
When I click my 'Create' button I want the record to be added to my category table, however for some reason it is being added twice - even though I just click the button once. Any ideas why that may be? I can't see where else the
if (isset($_POST['create'])) { could be called from. I only have 4 pages in my whole project.
<?php require('dbConnect.php');
//use the variables we created in volleyLogin.php
session_start();
$username = $_SESSION['username'];
$user_id = $_SESSION['user_id'];
echo "user name is " . $username . "<br>";
echo "user id is " . $user_id . "<br>";
if (isset($_POST['create'])) {
$category = ($_POST['category']);
$name = ($_POST['name']);
$phonenumber = ($_POST['phonenumber']);
$address = ($_POST['address']);
$comment = ($_POST['comment']);
//check if the category being entered is already there
$check="SELECT COUNT(*) FROM category WHERE cat_name = '$_POST[category]'";
$get_value = mysqli_query($con,$check);
//check the number of values of the category being posted
$data = mysqli_fetch_array($get_value, MYSQLI_NUM);
//if the category name already exists in the category table
if($data[0] >= 1) {
echo "This Already Exists<br/>";
}
else if ($data[0] < 1)
{
//if it's not in there, then add the category in the category table.
$sql = "INSERT INTO category VALUES(NULL, '{$category}', '$user_id')";
$rs1=mysqli_query($con, $sql);
if ($con->query($sql) === TRUE) {
echo "Yes, it's been added correctly";
} else {
echo "Error: " . $sql . "<br>" . $con->error;
}
}
$con->close();
}
?>
<!doctype html>
<html>
<body>
<h2>Create new Contact</h2>
<form method="post" action="" name="frmAdd">
<p><input type="text" name = "category" id = "category" placeholder = "category"></p>
<p><input type="text" name = "name" id = "name" placeholder = "name"></p>
<p><input type="text" name = "phonenumber" id = "phonenumber" placeholder = "phone number"></p>
<p><input type="text" name = "address" id = "address" placeholder = "address"></p>
<p><input type="text" name = "comment" id = "comment" placeholder = "comment"></p>
<p><input type="submit" name = "create" id = "create" value = "Create new Contact"></p>
Exit
</form>
</body>
</html>
You're running the $sql query twice, with two different methods:
$rs1=mysqli_query($con, $sql);
if ($con->query($sql) === TRUE) {
That's why you're getting duplicate entries.
You should either remove $rs1 as it's not being used, or verify it's value on the conditional instead of running the function again.
I am posting a shortened version of the form and updating lines. I will truly appreciate any help. I have spent the last 48 hours trying all I could think of and it's driving me insane. If I remove the line if($_SERVER["REQUEST_METHOD"]=="POST"), the program runs on loading the page and does update the table at the ID in the url with a blank field. Thanks in advance. Here's the code:
<?php
$id = $_GET['id'];
$user = $_SESSION['user'];
Echo '<form action="editone.php" method="POST">
Enter new name:<input type="text" name="namex" />
<input type="submit" name="Submit" value="Update List" /> </form>';
if($_SERVER["REQUEST_METHOD"]=="POST")
{
$dblink = "nn000185_manager";
$cxn = new mysqli("localhost","user","password", $dblink);
$details = mysqli_real_escape_string($cxn, $_POST['namex']);
$numb = mysqli_real_escape_string($cxn, $id);
$query = "UPDATE EDITORES SET nom_edit = '$details' WHERE edit_id = $numb";
mysqli_query($cxn, $query);
echo $query;
}
?>
I think your form action didn't pass id.
<form action="editone.php" method="POST">
If you're using this single file as form editor and action, your form editor URL should be http://localhost/editone.php?id=1
Try to change your form action to
<form action="editone.php?id='.$_GET['id'].'" method="POST">
or just leave the action blank
<form action="" method="POST">
Ok - maybe I'm way off base here but I see the following problems.
1) Your method is POST however your id is coming from GET.
2) I don't see where the id is coming from. It could be coming from somewhere and not posted but I don't see it.
Have you checked to verify the value is actually being passed through to the php?
try this
echo "GET = " . var_dump($_GET);
echo "<br><br>";
echo "POST = " . var_dump($_POST);
exit();
Post the results and then post where the id is coming from if you can't figure it out still. :)
Use the below code:
$query = "SELECT now_edit, FROM EDITORIES WHERE edit_id='$numb' LIMIT 1";
I assume your page is being called initially from an anchor link on another page which is why you are getting the id from $_GET['id'].
When the user presses the submit button of course the form is being submitted as a POST so all the data will be in $_POST, therefore $_GET['id'] will fail and should be generating an error message.
You need to save the $_GET['id'] from the first instantiation so you can use it when the form is posted to you. So put it in a hidden field that will be posted to you with the post
<?php
session_start();
$user = $_SESSION['user'];
if($_SERVER["REQUEST_METHOD"]=="GET") {
if ( isset($_GET['id']) ) {
$id = $_GET['id']);
} else {
// no param passed, could be a hack
header('Location: some_error_page.php');
exit;
}
echo '<form action="editone.php" method="POST">';
echo '<input type="hidden" name="id" value="' . $id . '">';
echo 'Enter new name:<input type="text" name="namex" />';
echo '<input type="submit" name="Submit" value="Update List" /></form>';
}
if($_SERVER["REQUEST_METHOD"]=="POST") {
$dblink = "nn000185_manager";
$cxn = new mysqli("localhost","user","password", $dblink);
$details = mysqli_real_escape_string($cxn, $_POST['namex']);
$numb = mysqli_real_escape_string($cxn, $_POST['id']);
$query = "UPDATE EDITORES SET nom_edit = '$details' WHERE edit_id = $numb";
mysqli_query($cxn, $query);
echo $query;
}
?>
<html><head>
<title>Add record to my_database/my_table</title></head>
<body>
<?php
$self = $_SERVER['PHP_SELF'];
$id = $_POST['id'];
$fname = $_POST['fname'];
$lname = $_POST['lname'];
?>
<form action="<?php echo( $self ); ?>" method="post">
ID: <input type="text" name="id" size="3">
First Name: <input type="text" name="fname" size="8">
Last Name: <input type="text" name="lname" size="8"><br>
<input type="submit" value="Submit">
</form>
<?php
if( $id and $fname and $lname)
{
$conn=#mysql_connect( "localhost", "root", "" ) or die( "Err:Conn" );
select the specified database
$rs = #mysql_select_db( "add_record", $conn) or die( "Err:Db" );
create the query
$sql = "insert into my_table ( id, first_name, last_name ) values ( $id, \"$fname\", \"$lname\" )";
execute query
$rs = mysql_query( $sql, $conn );
if( $rs )
{
echo( "Record added:$id $fname $lname" );
}
}
?>
</body></html>
here am getting erro as undefined index id,fname,lastname and when i enter values in this am getting db error
At first when your page load $_POST['id'] value is empty because u ve'nt posted any value in $_POST[];
if(isset($_POST['submit'])){
//all your php code here like below
$self = mysql_real_escape_string($_SERVER['PHP_SELF']);
$id = mysql_real_escape_string($_POST['id']);
$fname = mysql_real_escape_string($_POST['fname']);
$lname = mysql_real_escape_string($_POST['lname']);
}
AND
$sql = "insert into my_table ( id, first_name, last_name ) values ( '$id', '$fname', '$lname' )";
By the way what is your db error??
Those POST values will only be set when the form is POSTed. You can use isset()
$id = isset($_POST['id'])? $_POST['id'] : NULL;
Same for others.
This happens because you have no conditions on that PHP code that will prevent it from executing the first time when the form is loaded. They should only execute when the form is submitted. You can wrap that PHP with
if(isset($_POST))
{
// Your existing database code here
}
I am trying to build admin side of small website which consists of 2 pages: index.php and update php. On index.php I run query, that per-fills html form with data from database, which works fine.
Then I send data via $_POST to update.php page, where I try to get those values into variables and then make an update query. Which fails. I suspect something is wrong with $_POST array - some values are messed up or empty, but I don't understand why.
Here is the code for index.php:
<?php
if (!isset($page_id)) {
echo " <p>Please select page to be edited:</p>";
$query = mysql_query("SELECT page_id, title FROM pages");
$res = mysql_fetch_array($query);
do {
printf("<p><a href='index.php?page_id=%s'>%s</a></p>", $res['page_id'], $res['title']);
} while ($res = mysql_fetch_array($query));
} else { $query = mysql_query("SELECT * FROM pages WHERE page_id = '$page_id'");
$res = mysql_fetch_array($query);
require_once 'parts/form.php';}
?>
This is code for update.php:
<?php
//Here I try to get POST values and assign them to variables for update
//Ths is validation that those values are not empty,
require_once 'parts/guard.php';
if (isset($_POST['page_id'])) {
$page_id = $_POST['page_id'];
}
if (isset($_POST['title'])) {
$title = $_POST['title'];
}
if ($title == '') {
unset($title);
}
if (isset($_POST['description'])) {
$description = $_POST['description'];
}
if ($description == '') {
unset($description);
}
if (isset($_POST['keywords'])) {
$keywords = $_POST['keywords'];
}
if ($keywords == '') {
unset($keywords);
}
if (isset($_POST['text'])) {
$text = $_POST['text'];
}
if ($text == '') {
unset($text);
}
//variables are set
require_once 'parts/meta.php';
?>
<?php
//Here is all the values exist, the query is executed.
//Obviousely this query works in phpmyadmin, but not here - some fields come empty or messed up????
if (isset($title) && isset($keywords) && isset($description) && isset($text) && isset($page_id)) {
$query = mysql_query("UPDATE pages SET title = '$title', description = '$description', keywords = '$keywords', text = '$text' WHERE page_id = '$page_id' ");
if ($query == TRUE) {
echo "<p>Page Updated</p>";
echo "<p><a href = 'http://localhost:8888/travel.ru/admin/index.php'>
Edit Another Page</a></p>";
} else {
echo "<p>Page Is Not Updataed</p>";
}
} else {
echo "<p>You Left Some Fields Empty. Page Will Not Be Updated.</p>";
}
?>
And this is the form I use:
<form name="update" action = "update.php" method= "post">
<p> Page Name<br>
<input value = "<?php echo $res['title']; ?>" type = "text" name = "title"></p>
<p> Page Description<br>
<input value = "<?php echo $res['description']; ?>" type = "text" name = "title"></p>
<p> Page Keywords<br>
<input value = "<?php echo $res['keywords']; ?>" type = "text" name = "title"></p>
<p> Page Content<br>
<textarea type = "text" name ="text" cols = "68" rows = "15"><?php echo $res['text']; ?>
</textarea></p>
<input type = "hidden" name="page_id" value =$res[page_id]>
<p><input type = "submit" name ="submit" value ="Save Changes" id="submit"</p>
</form>
Any help will be most appreciated as I dont have a clue why I have this problem?
Most of your form fields are named title. Thus you don't actually have a field called description or page_id or keywords.
Mate also raises a valid point.
Try added php tag to your input value
<input type = "hidden" name="page_id" value ="<?php echo $res['page_id']; ?>" />
As mentioned Amadan , also check the names for all controls in your form.