I am creating a deletion page for the removal of rows in the database. I have everything working up until the actual removal of the rows upon submit. The records display in the drop down without problem, any ideas of a solution are appreciated. (I am using deprecated SQL, I know.)
1ST PART OF PHP (FORM)
$query = 'SELECT event_name FROM event';
$result = mysql_query($query);
echo "<select name='events' value='-'>\n";
echo "<option value='Select event'>Select an event to be deleted\n";
while($row = mysql_fetch_row($result))
{
$eventSelect = $row[0];
echo "<option value='$eventSelect'>$eventSelect</option>\n";
}
echo "</select>"
2ND PART OF PHP (DELETION)
if (isset($_POST['eventSelect']))
{
$eventselection = $_POST['eventSelect'];
$query = "DELETE FROM event WHERE event_name = '$eventselection'";
$result = mysql_query($query);
}
Related
Update: I’m now getting an error telling me that roomID is null and this is why it’s not going into the database. Can you see why it would be null when it should be the option that the user selects in the select box?
I'm working on a hotel booking system as an exercise. I am trying to make a form that will insert a new booking into the database. I have the customer and room IDS in the mySQL database and I have put them into HTML select boxes for the user to select an option.
This works fine until I try to save the new booking to the database. I don't get any errors and it tells me the booking has been saved on my page but if I go into the database it hasn't been saved. I have tried inserting the customerID and roomID values from text boxes so just typing it in instead of the drop down menu and that inserts into the database just fine.
I'm pretty sure the problem is something to do with getting the value out of the select box and into the POST variable but I'm not 100% sure. I've only been coding for 6 months so I don't really know what I'm doing! I've tried to figure it out by looking at other examples on here but nothing seems to work.
Here's my code:
Getting the data to fill the select boxes:
//Locate room names
$query = $query = 'SELECT `roomID` FROM `room`';
$result2 = mysqli_query($DBC,$query);
$rowcount = mysqli_num_rows($result2);
if ($rowcount > 0) {
$row = mysqli_fetch_assoc($result2);}
//Locate customerIDs
$query = $query = 'SELECT `customerID` FROM `customer`';
$result = mysqli_query($DBC,$query);
$rowcount = mysqli_num_rows($result);
if ($rowcount > 0) {
$row = mysqli_fetch_assoc($result);}
Creating Select boxes:
<form method="POST" action="newbooking.php">
<p>
<label for="customerID">Customer ID: </label>
<?php
echo "<select name='customerID'id='customerID' input type='number'>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<option value='customerIDselect'>" . $row['customerID'] . "</option>";
//customerID
$customerID = cleanInput($_POST['customerIDselect']);
}
echo "</select>";
?>
</p>
<p>
<label for="rooms">Room (name, type, beds): </label>
<?php
echo "<select name='roomID'id='roomID' input type='number'>";
// output data of each row
while($row = $result2->fetch_assoc()) {
echo "<option value='roomIDselect'>" . $row['roomID'] . "</option>";
//roomID
$roomID = cleanInput($_POST['roomIDselect']);
}
echo "</select>";
?>
</p>
Inserting data into database:
//save the customer data if the error flag is still clear
if ($error == 0) {
$query = 'INSERT INTO `bookings` (`customerID`,`roomID`,`checkin`,`checkout`, `extras`) VALUES (?,?,?,?,?)';
$stmt = mysqli_prepare($DBC,$query); //prepare the query
mysqli_stmt_bind_param($stmt,'sssss', $customerID, $roomID, $checkin, $checkout, $extras);
mysqli_stmt_execute($stmt);
mysqli_stmt_close($stmt);
echo "<h2>booking saved</h2>";
} else {
echo "<h2>$msg</h2>".PHP_EOL;
}
EDIT: IGNORE ANY SQL INJECTIONS OR VULNERABLE CODE STATEMENTS :D
(School Project).
I wish to create a insert form on my webpage where I can select an artist from a table, including a song from a table and combine them for an insert into a combined foreign key table.
I have managed to do selects and insert with only individual artist and song drop-downs on my web-page, but would wish for combining the two ID's from each table to combine them to a many to many relative table. But when I press the submit button nothing happens, and I'm a beginner and don't know if I'm missing any important bits of actually Posting the information.
For troubleshooting I have tried my code, and tested it. I see if I remove my code theres no problem, so the problem persists on the syntax I believe, as the first dropdown shows, alongside the second dropdown and submit button, but the problem is within the actual processing and SQL query part, where it never goes to the DB..
The problem:
As you can see below I have a the text Song Name appear with a drop-down menu in the bottom left corner including the Artist Name with a submit button. But my problem persists as the select and then insert from the two drop downs into the combined table does not work, it does not actually submit, I want it to post into the DB what can I do. But somethings off? I would appreciate any questions or help, this community is so amazing and wonderful to operate in!
Database
PHP
<form method='POST'>
<?php
include('connect_mysql.php');
if(isset($_POST["mangetilmange"])) {
$song_id = $_POST["song_id"];
$artist_id = $_POST["artist_id"];
$sql ="INSERT INTO artist_has_song (song_id, artist_id) VALUES
('$song_id', '$artist_id')";
if($conn->query($sql)) {
echo "Completed";
} else {
echo "Blablalbablablablablablablabl $sql
($conn->error.";
}
}
?>
Song Name
<?php
$sql = "SELECT * FROM song";
$resultat = $conn->query($sql);
echo "<select name='song_id'>";
while ($rad = $resultat->fetch_assoc()) {
$song_id = $rad["song_id"];
$songname = $rad["songname"];
echo "<option value='$song_id'>$songname</option>";
}
echo "</select>";
?>
Artist Name
<?php
$sql = "SELECT * FROM artist";
$resultat = $conn->query($sql);
echo "<select name='artist_id'>";
while ($rad = $resultat->fetch_assoc()) {
$artist_id = $rad["artist_id"];
$artistname = $rad["artistname"];
echo "<option value='$artist_id'>$artistname</option>";
}
echo "</select>";
?>
</form>
<input type="submit" name="mangetilmange" value ="Submit">
change you code to this:
<form method='POST'>
<?php
include('connect_mysql.php');
if(isset($_POST["mangetilmange"])) {
$song_id = $_POST["song_id"];
$artist_id = $_POST["artist_id"];
$sql ="INSERT INTO artist_has_song (song_id, artist_id) VALUES
('$song_id', '$artist_id')";
if($conn->query($sql)) {
echo "Completed";
} else {
echo "Blablalbablablablablablablabl";
}
}
?>
Song Name
<?php
$sql = "SELECT * FROM song";
$resultat = $conn->query($sql);
echo "<select name='song_id'>";
while ($rad = $resultat->fetch_assoc()) {
$song_id = $rad["song_id"];
$songname = $rad["songname"];
echo "<option value='$song_id'>$songname</option>";
}
echo "</select>";
?>
Artist Name
<?php
$sql = "SELECT * FROM artist";
$resultat = $conn->query($sql);
echo "<select name='artist_id'>";
while ($rad = $resultat->fetch_assoc()) {
$artist_id = $rad["artist_id"];
$artistname = $rad["artistname"];
echo "<option value='$artist_id'>$artistname</option>";
}
echo "</select>";
?>
<input type="submit" name="mangetilmange" value ="Submit">
</form>
Newby to php / sql so be kind to my ignorance.
On page index.php I have a navigation menu that retrieves options from table car_category, column car_type.
If user selects option from navigation, it takes him to another page where he should see list of entries selected from column car_review_details, coming from another table, car_review.
First function executes on index.php and retrieves the data from table car_category:
function list_type (){
global $conn, $sql, $test;
$test = "SELECT car_type FROM `car_category` WHERE car_type IS NOT NULL AND car_type <> ''";
$result = mysqli_query($conn, $test);
if (!$result){
die("failed");
}
while ($row = mysqli_fetch_assoc($result)){
echo "<a href='car_review.php?subject=";
echo urlencode ($row["car_type"]);
echo "'>";
echo $row ["car_type"];
echo "</a>";
echo "<br>";
}
}
The function above works fine.
Now, on page care_review.php I would like display the data from column car_review_details. The tables car_category and car_review have the column catID which should determine which data gets selected from table car_review, based on the selected link in page index.php
Code for function on page car_review.php looks as following:
function car_style(){
global $conn;
$details = "SELECT *
FROM `car_category`
INNER JOIN `car_review`
ON `car_category`.`catID` = `car_review`.`catID`
WHERE `car_category`.`catID` = $row ['catDesc']";
$result = mysqli_query($conn, $details);
if (!$result){
die("failed");
}
while ($row_1 = mysqli_fetch_assoc($result)){
echo ($row_1["car_review_details"]);
echo "<br>";
}
}
How do I go about echoing data from column car_review_details based on selection from function list_type ().
Many thanks for your feedback.
WHy is my drop down list not populating with the table data? (dropdown box is empty)
And what is used to display data upon selection of an item in that drop down - is it a "VIEW" (please do provide a study link so I can learn)
My Code
<?php
$con=mysqli_connect("localhost","root","","ismat_db");
//check connection
if(mysqli_errno($con))
{
echo "Can't Connect to mySQL:".mysqli_connect_error();
}
else
{
echo "Connected to mySQL</br>";
}
//$query = 'SELECT FirstName FROM persons';
//$result = mysqli_query($con,$query);
$query = mysqli_query($con,"SELECT 'FirstName' FROM persons");
//print_r($query);
//echo '<select name="FirstName">';
echo "<select name= 'FirstName'>";
//while($row=mysqli_fetch_array($result))
while($row=mysqli_fetch_array($query))
{
echo $row;
//echo "<option value='".$row['FirstName']."'>".'</option>';
}
echo '</select>';
?>
You had 2 errors:
I pointed the first in the comment: to print an option you must use this code:
echo "<option value='". $row['FirstName']."'>".$row['FirstName']
. '</option>';
The second is in your SQL: you are not selecting the FirstName field from the database, but a string 'FirstName' instead. That's why it is printed twice as you said. Use this SQL to get the field:
$query = mysqli_query($con,"SELECT FirstName FROM persons");
Also usually people put an id of the record and not a field, that may have possible duplicates into the value of an <option>. So, I would have used:
echo "<option value='". $row['id']."'>".$row['FirstName']
. '</option>';
selecting the id from the database together with first name.
Try this:
echo "<option value='".$row['FirstName']."'>".$row['FirstName']."</option>";
Also seems that you are having an issue with the database query. Swap your while loop with the following and see if it works
if ($result = $mysqli->query($query)) {
while ($row = $result->fetch_assoc()) {
echo "<option value='".$row['FirstName']."'>".$row['FirstName']."</option>";
}
$result->free();
}
Just like the title says i'm having difficulties in achieving it.
Here's my dropdownlist:
<?php
$query = "SELECT data, rel_id FROM $tbl_rel_balansas INNER JOIN $tbl_balansas ON $tbl_rel_balansas.rel_id = $tbl_balansas.id WHERE $tbl_rel_balansas.member_id = '$_SESSION[id]' group by data";
$result = mysql_query ($query);
echo "<select name=data value=''>Data</option>";
while($nt=mysql_fetch_array($result)){
echo "<option value=$nt[data] name=\"blabla\">$nt[data]</option>";
}
echo "</select>";
?>
Here's the buttonclick:
<?php
if(isset($_POST['Submit']))
{
$query = "SELECT SUM(suma), paskirtis FROM $tbl_rel_balansas INNER JOIN $tbl_balansas ON $tbl_rel_balansas.rel_id = $tbl_balansas.id WHERE $tbl_rel_balansas.member_id = '$_SESSION[id]' AND data ='".$_POST['data']."' group by paskirtis";
$result = mysql_query ($query);
echo "<tr><td>Paskirtis:</td><td>Biudzetas:</td><td>Isleista:</td><td>Likutis:</td></tr>";
while($nt=mysql_fetch_array($result)){
if($nt['SUM(suma)'] != null){
$suma = $nt['SUM(suma)'];
}
echo "<tr><td>$nt[paskirtis]</td>
<td><input type=\"text\" name=\"isleista[]\" value=\"Skiriamų pinigų kiekis...\" method=\"post\"></td><td>".$suma." Lt</td><td>--</td></tr> <br>";
}
}
?>
After I press it, it retrieves the data I want from the date I've chosen from the drop down list and also reset whole drop down list showing the first value of the dates from sql database, not the one I selected. If anyone knows how to keep the selected value in the list any help is greatly appriciated!
Try this, you need to place select="selected" in the while loop. See below code how I placed the $selected
<?php
$query = "SELECT data, rel_id FROM $tbl_rel_balansas INNER JOIN $tbl_balansas ON $tbl_rel_balansas.rel_id = $tbl_balansas.id WHERE $tbl_rel_balansas.member_id = '$_SESSION[id]' group by data";
$result = mysql_query ($query);
echo "<select name=data value=''>Data</option>";
while($nt=mysql_fetch_array($result)){
$selected = ($_POST['blabla'] == $nt[data])?'selected="selected"':NULL;
echo "<option value=$nt[data] name=\"blabla\" $selected >$nt[data]</option>";
}
echo "</select>";
?>