so i tried to get data from my database to checkbox and this is what i have tried :
<?php
$query = "SELECT * FROM siswa WHERE id_tingkatan='$idtingkatan'";
$result = $koneksi->query($query);
while($row=$result->fetch_assoc()){
?>
<input type="checkbox" name="murid[]" value="<?php echo $row['id_siswa']; ? >"><?php echo $row["nama_siswa"]; ?><br><br>
<?php } ?>
and save the value of checked checkbox into database, this is what i have tried :
if(isset($_POST["submit"]))
{
if(!empty ($_POST['murid']))
{
$i= 0;
foreach($_POST as $murid){
$kelas = $_POST['kelas'];
$murid = $_POST['murid'][$i];
$query = "INSERT INTO murid (id_kelas, id_siswa) VALUES ('$kelas', '$murid')";
$q = mysqli_query($koneksi, $query) or die (mysqli_error($koneksi));
$i++;
}
}
else
{
echo "<script type= 'text/javascript'>alert('Pilih minimal 1 siswa');</script>";
header('Location: ./kelas.php');
}
}
when i submit it does input to database but theres one extra row with id_siswa value as 0
The code inserts one record for each value in $_POST. However, $_POST contains all parameters sent (including submit), not just the checkbox array to be inserted. Iterate over the checkbox array $_POST['murid'] instead.
Change
foreach($_POST as $murid) ...
To
foreach($_POST['murid'] as $murid) ...
Related
I want to enroll the student and insert the student id into Mysql database after I check and submit the checkbox value, but I already tried so many ways but still cannot...
This is the php code
<?php
if (isset($_POST['submitxd'])) {
foreach ($_POST['enrol'] as $items) {
$insert = $link->query("INSERT INTO student_course(studentID) values ('$items')");}
}
?>
This is the html code
$result = $link->query("SELECT * FROM student WHERE programmeName = '$programme' AND intake = '$intake'");
while ($row = mysqli_fetch_array($result)) {
echo "<tr>
<td>".$row['studentID']."</td>
<td>".$row['studentName']."</td>
<td>".$row['studentGender']."</td>
<td>".$row['studentContact']."</td>
<td>
<input type='checkbox' name='enrol[]' value='".$row['studentID']."'>
</td>
</tr>";
}
check whether your array contains values or not:
echo "<pre>";
print_r($_POST['enrol']);
echo "</pre>";
if not, you should write html code properly i.e. check form tag and its action path carefully and before submitting the form, remember to check out the checkbox
I need a help with my code; somehow my code creates two rooms (it inserts two rows into a table at once), I don't know why.
(I need to require an id for every insert to know in which house we create a new room. My database contains table 'house' and table 'room'. Table 'room' has a field 'house_id' which is a foreign key with a field 'id' in table 'house'.)
That is my php page:
<?php
// turn autocommit off
mysqli_autocommit($con, FALSE);
// fetch the houses so that we have access to their names and id
$query = "SELECT name, id
FROM house";
$result = mysqli_query($con, $query);
// check query returned a result
if ($result === false) {
echo mysqli_error($con);
} else {
$options = "";
// create an option
while ($row = mysqli_fetch_assoc($result)) {
// $options .= "".$row['name']."";
$options .= "<option value='".$row['id']."'>";
$options .= $row['name'];
$options .= "</option>";
}
}
include('templates/add_room.html');
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$price = mysqli_real_escape_string($con, $_POST["price"]);
$house = mysqli_real_escape_string($con, $_POST["house_id"]);
$query = "INSERT INTO room (price, house_id)
VALUES ('$price', '$house')";
// run the query to insert the data
$result = mysqli_query($con, $query);
// check if the query went ok
if ( $con->query($query) ) {
echo "<script type= 'text/javascript'>alert('New room created successfully with the id of {$con->insert_id}');</script>";
mysqli_commit($con);
} else {
echo "There was a problem:<br />$query<br />{$con->error}";
mysqli_rollback($con);
}
}
//free result set
mysqli_free_result($result);
?>
and that is my html template with form:
<h2>Add new room</h2>
<form action='' method='POST'>
<fieldset>
<label for='price'>Price:</label>
<input type='number' name='price'>
</fieldset>
<fieldset>
<label for='house_id'>House:</label>
<select name='house_id' required>
<option value='' disabled selected>Select house</options>
<?php echo $options; ?>
</select>
</fieldset>
<button type='submit'>Add</button>
</form>
It inserts 2 rows because of your using the query function twice:
$result = mysqli_query($con, $query);
// check if the query went ok
if ( $con->query($query) ) {
So you'll need to change that conditional statement to:
if ($result) {
By the way, use a prepared statement, it's safer than real_escape_string():
https://en.wikipedia.org/wiki/Prepared_statement
You are inserting it twice
first here:
// run the query to insert the data
$result = mysqli_query($con, $query);
then here:
// check if the query went ok
if ( $con->query($query) ) {
Remove the first one and you should be fine, or check on the result of the first one and remove the second one.
Not 100% certain, but it looks like you run INSERT query twice. Once here:
$result = mysqli_query($con, $query);
and once a moment later when you try to check for something. you inadvertently use the OOP style when you are apparently trying to check for something
if ( $con->query($query) ) {
I am currently running into an issue, where I have this form consisting of checkboxes. I get the values of user preferences for the checkboxes from a database. Everything works great, and does what is supposed to do, however after I change and check some boxes and then hit the submit button, it will still show the old values to the form again. If I click again in the page again it will show the new values.
The code is shown below with comments.
<form action="myprofile.php" method="post">
<?php $usr_cats=array();
$qry_usrcat="SELECT category_id_fk
FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';";
$result = mysqli_query($conn,$qry_usrcat);
while($row = mysqli_fetch_array($result)){
$usr_cats[] = $row[0]; // getting user categories from db stored in array
}
$query_allcats="SELECT category_id,category_name, portal_name
FROM categories
INNER JOIN portals on categories.portal_id=portals.portal_id
ORDER BY category_id;"; // select all category queries
$result = mysqli_query($conn,$query_allcats);
while($row = mysqli_fetch_array($result)){
echo $row['portal_name'] . "<input "; //print categories
if(in_array($row['category_id'], $usr_cats)){ // if in array from db, check the checkbox
echo "checked ";
}
echo "type='checkbox' name='categories[]' value='";
echo $row['category_id']."'> ". $row['category_name']."</br>\n\t\t\t\t\t\t";
}
?>
<input type="submit" name="submit" value="Submit"/>
<?php
$qry_del_usrcats="DELETE FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';"; //delete all query
if(isset($_POST['submit'])){
if(!empty($_POST['categories'])){
$cats= $_POST['categories'];
$result = mysqli_query($conn,$qry_del_usrcats); //delete all
for ($x = 0; $x < count($cats); $x++) {
$qry_add_usrcats="INSERT INTO `user_categories` (`user_id_fk`, `category_id_fk`)
VALUES ('".$_SESSION['user_id']."', '".$cats[$x]."');";
$result = mysqli_query($conn,$qry_add_usrcats);
}
echo "success";
}
elseif(empty($_POST['categories'])){ //if nothing is selected delete all
$result = mysqli_query($conn,$qry_del_usrcats);
}
unset($usr_cats);
unset($cats);
}
?>
I am not sure what is causing to do that. Something is causing not to update the form after the submission. However, as i said everything works great meaning after i submit the values are stored and saved in the DB, but not shown/updated on the form. Let me know if you need any clarifications.
Thank you
Your procedural logic is backwards and you're doing a bunch of INSERT queries you don't need. As #sean said, change the order.
<?php
if(isset($_POST['submit'])){
if(isset($_POST['categories'])){
$cats= $_POST['categories'];
// don't do an INSERT for each category, build the values and do only one INSERT query with multiple values
$values = '';
for($x = 0; $x < count($cats); $x++) {
// add each value...
$values .= "('".$_SESSION['user_id']."', '".$cats[$x]."'),";
}
// trim the trailing apostrophe and add the values to the query
$qry_add_usrcats="INSERT INTO `user_categories` (`user_id_fk`, `category_id_fk`) VALUES ". rtrim($values,',');
$result = mysqli_query($conn,$qry_add_usrcats);
echo "success";
}
elseif(!isset($_POST['categories'])){ //if nothing is selected delete all
// you may want to put this query first, so if something is checked you delete all, so the db is clean and ready for the new data.
// and if nothing is checked, you're still deleting....
$qry_del_usrcats="DELETE FROM user_categories WHERE user_id_fk='".$_SESSION['user_id']."';"; //delete all query
$result = mysqli_query($conn,$qry_del_usrcats);
}
unset($usr_cats);
unset($cats);
}
?>
<form action="myprofile.php" method="post">
<?php $usr_cats=array();
$qry_usrcat="SELECT category_id_fk FROM user_categories WHERE user_id_fk='".$_SESSION['user_id']."';";
$result = mysqli_query($conn,$qry_usrcat);
while($row = mysqli_fetch_array($result)){
$usr_cats[] = $row[0]; // getting user categories from db stored in array
}
$query_allcats="SELECT category_id,category_name, portal_name FROM categories INNER JOIN portals on categories.portal_id=portals.portal_id ORDER BY category_id;"; // select all category queries
$result = mysqli_query($conn,$query_allcats);
while($row = mysqli_fetch_array($result)){
echo $row['portal_name'] . "<input "; //print categories
if(in_array($row['category_id'], $usr_cats)){ // if in array from db, check the checkbox
echo "checked ";
}
echo "type='checkbox' name='categories[]' value='";
echo $row['category_id']."'> ". $row['category_name']."</br>\n\t\t\t\t\t\t";
}
?>
<input type="submit" name="submit" value="Submit"/>
Typically this occurs due to the order of your queries within the script.
If you want to show your updated results after submission, you should make your update or insert queries to be conditional, and have the script call itself. The order of your scripts is fine, but you just need to do the following:
Take this query:
$qry_del_usrcats="DELETE FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';"
and put it inside the if statement so it looks like this:
if (isset($_POST['submit'] {
$qry_del_usrcats="DELETE FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';"
$result = mysqli_query($conn,$qry_del_usrcats);
[along with the other updates you have]
}
Also, you will need to move this entire conditional above the form itself; typically any updates, inserts, or deletes should appear year the top of the form, and then call the selects afterward (outside of the conditional)
http://i.stack.imgur.com/Gy3o0.png
That is what the site looks like now. What I want to do is when you click on the approve registration on the table, it will extract the value of the id no and the name of that particular record. I thought i was on the right track. I knew how to get the id no. But it doesn't get the value of the name at the same time.
This is how the code looks like:
while($row = mysql_fetch_array($mayor))
{
$id = $row['identification_no'];
$name = $row['lastname'].", ".$row['firstname'];
echo "<tr>";
echo "<td><form method=post action=approvedmayor.php><input type='radio' name=id value='$id'></td>";
}
approvedmayor.php
$query = mysql_query("insert into tbcandidates VALUES ($id, '$name', 'mayor')");
if ($query)
{
echo "You appproved ";
echo $name;
}
else
echo "error";
you can try like this...
<?php
while($row = mysql_fetch_array($mayor))
{
$id = $row['identification_no'];
$name = $row['lastname'].", ".$row['firstname'];
echo "<tr><td><a href='approvedmayor.php?id=$id&name=$name'>Approve</a></td></tr>";
}
?>
in this type don't use the form, and Approve button... try this alone
Actually it is bad practice to insert that kind of data directly from POST data.
If you have all these candidates already stored in the database, you should run a SELECT query in your approvedmayor.php first, and if the candidate still exists, insert it's data to another table.
$query = mysql_query('SELECT * FROM `candidates` WHERE `id` = '.$id.' LIMIT 1');
if(mysql_num_rows($query)) {
$candidate = mysql_fetch_assoc($query);
$insertQuery = mysql_query("insert into tbcandidates VALUES ($candidate['id'], $candidate['name'], $candidate['mayor'])");
if ($insertQuery) {
echo "You appproved ";
echo $name;
} else echo "error";
} else echo 'This candidate is no longer available';
I understand your question,
But thats not the best way go ahead, Before we move let us understand some little elements functions
Radio Button : Its an input type element, that allows the user to choose only one [ 1 ] of option given list.
Check Boxes : Its an input type element, that allows the user to select n number of options or selections from give list.
Fore info - http://www.w3schools.com/html/html_forms.asp
Now comming to your question..
You need to modify your code to check boxes as below
<input type='checkbox' name=id[] value='$id'>
Notice : elements name is in Array mode.. ie whenever a user one or more than one, the values are stored in array.
Once the values are stored in array, call it / use if however you want.
For your mentioned code
echo "<form method=post action=approvedmayor.php>';
while($row = mysql_fetch_array($mayor))
{
$id = $row['identification_no'];
$name = $row['lastname'].", ".$row['firstname'];
echo "<tr>";
echo "<td><input type='radio' name=id[] value='$id'></td>";
}
echo "</form>";
approvedmayor.php
$temp_app_arr = $_POST['id'];
foreach ($temp_app_arr as $pos => $val) {
$query = mysql_query("insert into tbcandidates VALUES ('$val', '$name', 'mayor')");
if ($query) {
echo "You appproved ";
echo $name;
} else {
echo "error";
}
}
And i believe this should gonna be good code / algorithm for your project.
I have a php form with some textbox and checkboxes which is connected to a database
I want to enter all the details into the database from the form.All the textbox values are getting entered except the checkbox values.I have a single column in my table for entering the checkbox values and the column name is URLID.I want to enter all the selected checkbox(URLID)values to that single column only ,separated by commas.I have attached the codings used.can anyone find the error and correct me?
NOTE: (The URLID values in the form is brought from the previous php page.those codings are also included.need not care about it)
URLID[]:
<BR><?php
$query = "SELECT URLID FROM webmeasurements";
$result = mysql_query($query);
while($row = mysql_fetch_row($result))
{
$URLID = $row[0];
echo "<input type=\"checkbox\" name=\"checkbox_URLID[]\" value=\"$row[0]\" />$row[0]<br />";
$checkbox_values = implode(',', $_POST['checkbox_URLID[]']);
}
?>
$URLID='';
if(isset($_POST['URLID']))
{
foreach ($_POST['URLID'] as $statename)
{
$URLID=implode(',',$statename)
}
}
$q="insert into table (URLID) values ($URLID)";
You really should separate your model from the view. It's 2011, not 2004 ;o
if (isset($_POST['checkbox_URLID']))
{
// Notice the lack of '[]' here.
$checkbox_values = implode(',', $_POST['checkbox_URLID']);
}
$URLIDs = array();
while($row = mysql_fetch_row($result))
{
$URLIDs[] = $row[0];
}
foreach ($URLIDs as $id)
{
?>
<input type="checkbox" name="checkbox_URLID[]" value="<?php echo $id; ?>" /><?php echo $id; ?><br />
<?php
}
?>
<input type="submit"/>