Guys I cant see why it is that my code will only Update the last row on the table. It will populate the entire HTML page with a table with the info from phpAdmin. I can then change this info, on the html page. It all works fine if there is only one record, anymore than one and it only takes effect on the last row. I am new to all this, so excuse the code, here it is......
<html>
<?php
$con = mysql_connect("localhost","root");
if(!$con){
die("Cant get there Bren: " . mysql_error());
}
mysql_select_db("Web_Data",$con);
if (isset($_POST['update'])){
$UpdateQuery = "UPDATE Vehicles SET Vehicle_Id='$_POST[Vehicle_Id]',
Registration='$_POST[Registration]',Make='$_POST[Make]',Model='$_POST[Model]',
Classification='$_POST[Classification]',Rental_Price='$_POST[Rental_Price]',
Current_Status='$_POST[Current_Status]',Mileage='$_POST[Mileage]'
WHERE Vehicle_Id='$_POST[hidden]'";
echo "<center> Vechicle Id '$_POST[hidden]' succesfully VEHICLE UPDATED </center>";
mysql_query($UpdateQuery, $con);
};
$sql = "Select * From Vehicles";
$myData = mysql_query($sql,$con);
echo" <center> <table border = 3>
<tr>
<th>Vehicle_Id</th>
<th>Registration</th>
<th>Make</th>
<th>Model</th>
<th>Classification</th>
<th>Rental_Price</th>
<th>Current_Status</th>
<th>Mileage</th>
</tr></center>";
while($record = mysql_fetch_array($myData)){
echo "<form action = UpdateWD.php method=post>";
echo "<tr>";
echo "<td>" . "<input type = text name = Vehicle_Id value=" . $record['Vehicle_Id'] . " </td>";
echo "<td>" . "<input type = text name = Registration value=" . $record['Registration'] . " </td>";
echo "<td>" . "<input type = text name = Make value=" . $record['Make'] . " </td>";
echo "<td>" . "<input type = text name = Model value=" . $record['Model'] . " </td>";
echo "<td>" . "<input type = text name = Classification value=" . $record['Classification'] . " </td>";
echo "<td>" . "<input type = text name = Rental_Price value=". $record['Rental_Price'] . " </td>";
echo "<td>" . "<input type = text name = Current_Status value=" . $record['Current_Status'] . " </td>";
echo "<td>" . "<input type = text name = Mileage value=" . $record['Mileage'] . " </td>";
echo "<td>" . "<input type = hidden name = hidden value=" . $record['Vehicle_Id'] . " </td>";
echo "<td>" . "<input type = submit name = update value= update" . " </td>";
echo "</from>";
}
echo"</table>";
mysql_close($con);
?>
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<footer>
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</footer> <!--footer-->
</html>
See here:
Use while loop to display all fetched records.
<?php
// Make a MySQL Connection
$sql = "Select * From Vehicles";
$myData = mysql_query($sql,$con) or die(mysql_error());
while($row = mysql_fetch_array($myData)){
echo $row['Vehicle_Id'];
}
?>
And mysql_query can't use multiple queries.
The easiest thing is to just run them separately. I believe you can do multi query but I haven't tried it.
Just get the idea how you can run multiple queries using foreach loop.
$updateArray = array(21=>300,23=>200,24=>100);
foreach($updateArray as $id=>$value)
{
$query = "UPDATE cart SET cart_qty='$value' WHERE cart_id = '$id'";
mysql_query($query,$link);// $link is specified above
}
here is my mistake
hours and hours and all it was
echo "<td>" . "<input type = 'submit' name = 'update' value= 'update'" . " />";
echo "</td>";
echo "</tr>";
echo "</form>";
had "form" spelt "from" & didnt close the </tr>
Related
I am working with three database tables.
users
class
user_class
I'm making a page so that student can be assigned to classes using a checkbox. How do I get it so that if the value of a student being in a class is currently in the database, the checkbox for that class will already have the student value checked.
<?php
$showAllStudents = "SELECT * FROM users";
mysqli_query($mysqli, $showAllStudents) or die ('Error finding Students');
$result = mysqli_query($mysqli, $showAllStudents);
echo"<table border='1' cellspacing='10' align='center'>";
echo "<tr><th></th><th>User ID</th><th>User Name</th><th>First Name</th>
<th>Second Name</th></tr>";
while ($row = $result->fetch_object()){
echo "<tr>";
echo "<td><input type='checkbox' id='" .$row->userID . "'
name='check_box[]' value='" .$row->userID . "' /></td>";
echo "<td>" .$row->userID . "</td>";
echo "<td>" .$row->username . "</td>";
echo "<td>" .$row->forename . "</td>";
echo "<td>" .$row->surname . "</td>";
echo "</tr>";
}
if (isset($_POST['submitClassStudent'])) {
//get ID from header
$classID = $_GET['id'];
//print_r ($_POST);
//for each ticked checkbox convert to a UserID variable
foreach ($_POST['check_box'] as $userID) {
$editClassStudentQuery = "INSERT INTO `user_class`(userID, classID)
VALUES('$userID', '$classID')";
$insert_row = $mysqli->query($editClassStudentQuery) or die($mysqli->error . __LINE__);
if ($insert_row) {
header('Location: classeditor.php');
} else {
echo "Error: " . $editClassStudentQuery . "<br>" . $mysqli->error;
}
}
}
?>
To mark a checkbox as already checked you just need to add checked as one of its attributes.
<input type="checkbox" id="already-checked" checked> Already checked
W3 Schools
if the user exists then echo 'checked' in tag
Example: echo "<input type ='checkbox'";if(YourConditionHere)echo "checked>";else echo ">";
I'm trying to make a HTML table as a frontend to a mySQL database. The table displays fine and I can type in the edits I want to make to each row of the table but when I press the submit button the changes aren't actually made. Can anyone see where I'm going wrong?
<?php
include("db.php");
$sql = "SELECT * FROM `artist`";
$result = mysqli_query($conn, $sql);
if (isset($_POST['update'])){
$artID = $_POST['artID'];
$artName = $_POST['artName'];
$key = $_POST['hidden'];
$UpdateQuery = "UPDATE `artist` SET `artID` = '$artID', `artName` = '$artName' WHERE `artist`.`artID` = '$key'";
mysqli_query($conn,$UpdateQuery);
header("Location: {$_SERVER['HTTP_REFERER']}");
exit;
};
echo "<table border='1'>";
echo "<tr>";
echo "<th>ID</th>";
echo "<th>Name</th>";
echo "</tr>";
if ($result->num_rows > 0) {
echo "<form id ='artisttable' action ='getartiststable.php' method ='post'>";
// output data of each row
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" ."<input type='text' name ='artID' value ='" . $row['artID'] . "' </td>";
echo "<td>" . "<input type='text' name ='artName' value ='" . $row["artName"] . "' </td>";
echo "<td>" . "<input type = 'hidden' name ='hidden' value='" . $row['artID'] . "' </td>";
echo "<td>" . "<input type='submit' name ='update'" . " </td>";
echo "</tr>";
}
echo "</form>";
echo "</table>";
} else {
echo "0 results";
}
$conn->close();
?>
The db.php file simply includes the connection info to the mySQL database and I'm 100% sure there's nothing wrong with it as it retrieves the table correctly it just doesn't update.
You are putting form tag inside tr which is not allowed td are only allowed
so you have to remove that tr from there.
You have to use jquery or you can replace the table with some other grid structure so that it can look the same and the form can be placed there as well
One more suggestion Don't mix the php and html together separate them for the clean code
If you do all these you code will be like this
Your form is being constructed with multiple elements with the same name. When you submit the form it is using the last elements as the values so regardless of the record you want updated the last record is being updated (or throwing an error because of string encapsulation). You should use parameterized queries as well.
So instead of:
if ($result->num_rows > 0) {
echo "<form id ='artisttable' action ='getartiststable.php' method ='post'>";
// output data of each row
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" ."<input type='text' name ='artID' value ='" . $row['artID'] . "' </td>";
echo "<td>" . "<input type='text' name ='artName' value ='" . $row["artName"] . "' </td>";
echo "<td>" . "<input type = 'hidden' name ='hidden' value='" . $row['artID'] . "' </td>";
echo "<td>" . "<input type='submit' name ='update'" . " </td>";
echo "</tr>";
}
echo "</form>";
echo "</table>";
Use:
if ($result->num_rows > 0) {
// output data of each row
while($row = mysqli_fetch_array($result)) {?>
<form class='artisttable' action ='getartiststable.php' method ='post'>
<tr>
<td><input type='text' name ='artID' value ='<?php echo $row['artID'];?>' /></td>
<td><input type='text' name ='artName' value ='<?php echo $row["artName"];?>' /></td>
<td><input type = 'hidden' name ='hidden' value='<?php echo $row['artID'];?>' /></td>
<td><input type='submit' name ='update'" . " </td>
</tr>
</form>
<?php } ?>
</table>
So you get a form for each data set. Here's a link on prepared statements with mysqli: http://php.net/manual/en/mysqli.quickstart.prepared-statements.php. You also should update your mark up. Tables for formatting aren't the best approach. Your inputs also weren't closed missing >.
Also this changed artisttable from an id to class because there will be multiples. Update CSS/JS accordingly.
I have a table which displays
-Staff ID (Primary Key)
-Staff Name
-Staff Position
All the data loads in to my grid, the grid has an update button witch should let me to update it but it returns original result after clicking update.
<html>
<head>
</head>
<body>
<?php
$conn = mysql_connect("localhost", "root", "");
if (!$conn){
die("Can not connect: " . mysql_error());
}
mysql_select_db("pizza_shop",$conn);
if (isset($_POST['submit']) && $_POST['submit'] == 'update'){
$UpdateQuery = "UPDATE staff SET StaffName='$_POST[staffname]', Position='$_POST[staffposition]' WHERE StaffID='$_POST[hiddenid]'";
mysql_query($UpdateQuery);
}
$sql = "SELECT * FROM staff";
$myData = mysql_query($sql, $conn);
echo "<table border=1>
<tr>
<th>Staff Name<th>
<th>Staff Position<th>
</tr>";
while($record = mysql_fetch_array($myData)){
echo "<form action=#edit_staff.php method=post>";
echo "<tr>";
echo "<td>" . "<input type=text name =staffname value=" . $record['StaffName'] ." </td>";
echo "<td>" . "<input type=text name =staffposition value=" . $record['Position'] ." </td>";
echo "<td>" . "<input type=hidden name=hiddenid value=" . $record['StaffID'] . "</td>";
echo "<td>" . "<input type=submit name = update values=Update" . "</td>";
echo "</form>";
}
echo "</table>";
$conn = null;
?>
</body>
</html>
You need to change your update query from
$UpdateQuery = "UPDATE staff SET StaffName='$_POST[staffname]', Position='$_POST[staffposition]' WHERE StaffID='$_POST[hiddenid]'";
to
$UpdateQuery = "UPDATE staff SET StaffName='".$_POST['staffname']."', Position='".$_POST['staffposition']."' WHERE StaffID='".$_POST['hiddenid']."'";
What you were doing is $_POST[staffname] which must be like as $_POST['staffname'] and always try to check using error_reporting(E_ALL) function and need to check that your values are set or not
I'm just a beginner in terms of programming, so I'm just referring all my codes through tutorials. Luckily, I found this online tutorial in youtube where users are allowed to add, update, and delete data in mysql using php. I follow all his instructions, I got it working but then it stopped when I added css on it.
This is not a general issue, I just need some help. If anyone can help me, much appreciated. Thank you so much.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "rssfeed";
$connect = mysql_connect($servername, $username, $password, $dbname);
if (!$connect) {
die("Connection failed. Error" . mysql_error());
}
$database = mysql_select_db($dbname, $connect);
if (!$database) {
die("Can't select database");
}
$sql = "SELECT * FROM record";
$data = mysql_query($sql, $connect);
if (isset($_POST['update'])){
$updateQuery = "UPDATE record SET name = '$_POST[name]', url = '$_POST[url]', description = '$_POST[desc]' WHERE name = '$_POST[hidden]'";
mysql_query($updateQuery, $connect);
header("Location: maintenance.php");
};
if (isset($_POST['delete'])){
$deleteQuery = "DELETE FROM record WHERE name = '$_POST[hidden]'";
mysql_query($deleteQuery, $connect);
header("Location: maintenance.php");
};
if (isset($_POST['add'])){
$addQuery = "INSERT INTO record (name, url, description) VALUES ('$_POST[iName]', '$_POST[iUrl]', '$_POST[iDesc]')";
mysql_query($addQuery, $connect);
header("Location: maintenance.php");
};
echo "<div class=center>
<table id=myTable border=1>
<tr>
<th> Name </th>
<th> URL </th>
<th> Description </th>
</tr>";
while($record = mysql_fetch_array($data)) {
echo "<form method=post action=maintenance.php>";
echo "<tr>";
echo "<td>" . "<input type=text name=name id=name value=" . $record['name'] . " </td>";
echo "<td>" . "<input type=text name=url id=url value=" . $record['url'] . " </td>";
echo "<td>" . "<textarea rows=1 cols=50 wrap=physical name=desc id=desc>" . strip_tags($record['description']) . "</textarea></td>";
echo "<input type=hidden name=hidden value=" . $record['name'] . ">";
echo "<td>" . "<input type=submit name=update id=update value=update" . " </td>";
echo "<td>" . "<input type=submit name=delete id=delete value=delete" . " </td>";
echo "</tr>";
echo "</form>";
}
echo "</table>";
echo "<table border=1>";
echo "<form method=post action=maintenance.php>";
echo "<tr>";
echo "<td><input type=text name=iName></td>";
echo "<td><input type=text id=url name=iUrl></input></td>";
echo "<td><textarea rows=1 cols=50 name=iDesc></textarea></td>";
echo "<td>" . "<input type=submit name=add id=add value=add" . " </td>";
echo "</tr>";
echo "</form>";
echo "</table>";
echo "</div>";
mysql_close($connect);
?>
In your queries, $_POST[name] should be $_POST[\"name\"]. BUT, this is terrible, you are very open to SQL injections.
Please have a read of this and stop using mysql_query (it's deprecated)
you have several errors in your echo output..
echo "<td>" . "<input type=text name=name id=name value=" . $record['name'] . " </td>";
You have forgot the single quotes for each html-element-attribute and the bigger then at the end of multiple input elements...
try this:
echo "<td><input type='text' name='name' id='name' value='" . $record['name'] . "'> </td>";
so I have this code that other Stack members have helped me fine tune and correct some errors, the code all works as it should but there is one small detail, after successfully editing one record and clicking the update button ALL of the existing records that are in the database load into the page. Here is my code below:
<?php
$con = mysql_connect("localhost", "root", "M1q2w3e4r");
if (!$con) {
die("Can not connect: " . mysql_error());
}
mysql_select_db("inventory",$con);
if (isset($_POST['update'])){
$UpdateQuery = "UPDATE invoice SET `inv_number`='$_POST[inv_number]', `from_date`='$_POST[from_date]', `to_date`='$_POST[to_date]',`date_type`='$_POST[date_type]', `notes`='$_POST[notes]' WHERE id='$_POST[id]'";
mysql_query($UpdateQuery, $con);
};
$where = '';
if(!empty($_GET) && !empty($_GET['edit'])) {
$where = ' where id='.$_GET['edit'];
}
$sql = "SELECT * FROM invoice".$where;
$myData = mysql_query($sql,$con);
echo "<table border=1>
<tr>
<th>Inv #</th>
<th>From</th>
<th>To</th>
<th>Type</th>
<th>Notes</th>
</tr>";
while($record = mysql_fetch_array($myData)){
echo "<form action='edit.php' method='post'>";
echo "<tr>";
echo "<td>" . "<input type='text' name='inv_number' value='" . $record['inv_number'] . "'> </td>";
echo "<td>" . "<input type='text' id='from' name='from_date' value='" . $record['from_date'] . "'> </td>";
echo "<td>" . "<input type='text' id='to' name='to_date' value='" . $record['to_date'] . "'> </td>";
echo "<td>" . "<input type='text' name='date_type' value='" . $record['date_type'] . "'> </td>";
echo "<td>" . "<input type='text' name='notes' value='" . $record['notes'] . "'> </td>";
echo "<td>" . "<input type='hidden' name='id' value='" . $record['id'] . "'> </td>";
echo "<td>" . "<input type='hidden' name='hidden' value='" . $record['id'] . "'> </td>";
echo "<td>" . "<input type='submit' name='update' value='update'>" . " </td>";
echo "</tr>";
echo "</form>";
}
echo "</table>";
mysql_close($con);
?>
I know it has to do with the form action="edit.php", as it refreshes the page the id number in the url is pulled out. So I tried this:
echo "<form action='edit.php?edit=<?php echo $_REQUEST["id"]; ?>' method='post'>";
but this only led to my edit.php to display as a blank page. If anyone can help me figure out how to prevent all the database records from being displayed in the page after clicking the update button it would really help.
I might do this, for example, if I just wanted to show the record that was just updated:
if (isset($_POST['update'])){
$UpdateQuery = "UPDATE invoice SET `inv_number`='$_POST[inv_number]', `from_date`='$_POST[from_date]', `to_date`='$_POST[to_date]',`date_type`='$_POST[date_type]', `notes`='$_POST[notes]' WHERE id='$_POST[id]'";
mysql_query($UpdateQuery, $con);
$where = ' where id='.$_POST[id];
}
else {
$where = '';
if(!empty($_GET) && !empty($_GET['edit'])) {
$where = ' where id='.$_GET['edit'];
}
}
You could also use REQUEST instead of GET and make a hidden input field with the name "edit" in your form.