I am dynamically generating a set of radio buttons from MySQL. The buttons are creating and the variables assigned to them are populating as I did an echo print_r and it shows the array for the variable. I now want to compare the values generated from this and if the vale is "0" I want to insert a score and present a green check graphic and the word correct. If the value is "1" I want it to input different values for the score and present Incorrect and a red X graphic. Here is what I have so far (Everything populates dynamically both the question and the answers as radio buttons):
<?php
echo '<form id="frmQuestion" name="frmQuestion" method="post" action="QuizQuestion1.php">';
// Connect to the Database
require_once('mysqli_connect.php');
//create the query for the question
$q = "SELECT `Question` FROM tbl_Question WHERE QuestionID = 1";
//Create the query for the Answers
$q2 = "SELECT `Answer`,`AnswerStatusID`,`AnswerResponse` FROM tbl_Answer WHERE QuestionID = 1";
//Run the query
$r = mysqli_query($conn,$q);
//run the answer query
$r2 = mysqli_query($conn,$q2);
while($row = mysqli_fetch_array($r,MYSQLI_ASSOC)){
echo '<div id="Question1"><p> ' . $row['Question'] . '</div></p>';
}
//Declare the variables as a array
$AnswerResponse = array();
$AnswerStatusID = array();
while($row2 = mysqli_fetch_array($r2,MYSQLI_ASSOC)){
echo '<div id="Question1"><input name="q1" type="radio" value="'.$AnswerStatusID.'"/>' . $row2['Answer'] . '</div><br/>';
//Assign the AnswerStatusID to a var
$AnswerStatusID[] = $row2['AnswerStatusID'];
//Assign the AnswerResponse to a var
$AnswerResponse[] = $row2['AnswerResponse'];
}
//Create the submit button
echo '<input type="submit" value="Submit Answer" name="submit"/>';
echo '</form>';
//Logic for correct or incorrect answers
if (isset($_POST['q1']) && ($_POST['q1'] == '0'))
{
//create the query for the score
$q3 = "INSERT INTO tbl_Score (`Score`,`QuestionID`) VALUES ('100%','1')";
//Run the query
$r = #mysqli_query ($conn,$q3);
if($r){
//Confirm message data was entered with a correct response and a graphic
echo '<h1>Correct!!</h1><img src="/images/green_Check_Low.jpg" alt="Green Check"/>';
echo 'Click here for the next question';
}
else
{
//there was an error
echo'<h1>System error</h1>';
//Debugging message
echo'<p>' . mysqli_error($conn) . '<br/><br/>Query:' . $q3 . '</p>';
}//End of nested IF
}
else{
//create the query for the score
$q4 = "INSERT INTO tbl_Score (`Score`,`QuestionID`) VALUES ('0%','1')";
//Run the query
$r2 = #mysqli_query ($conn,$q3);
if($r2){
//Confirm message data was entered with a correct response and a graphic
echo '<h1>Incorrect!!</h1><img src="/images/red_X_Low.jpg" alt="Red X"/>';
echo 'Click here for the next question';
}
else
{
//there was an error
echo'<h1>System error</h1>';
//Debugging message
echo'<p>' . mysqli_error($conn) . '<br/><br/>Query:' . $q3 . '</p>';
}//End of nested IF
}
//Free up the results for the Question query
mysqli_free_result($r);
//Free up the results from the Answer query
mysqli_free_result($r2);
//close the DB connection
mysqli_close($conn);
?>
This is the answer and work as intended. Thanks for everyones input.
//Declare the variables as a array
$AnswerResponse = array();
$AnswerStatusID = array();
while($row2 = mysqli_fetch_array($r2,MYSQLI_ASSOC)){
echo '<div id="Question1"><input name="q1" type="radio" value="'.$row2['AnswerStatusID'].'"/>' . $row2['Answer'] . '</div><br/>';
//Assign the AnswerStatusID to a var
$AnswerStatusID[] = $row2['AnswerStatusID'];
//Assign the AnswerResponse to a var
$AnswerResponse[] = $row2['AnswerResponse'];
}
//Create the submit button
echo '<input type="submit" value="Submit Answer" name="submit"/>';
echo '<input type="hidden"name="submitted"value="TRUE"/>';
echo '</form>';
if($_POST['submitted']) {
//Logic for correct or incorrect answers
if (isset($_POST['q1']) && ($_POST['q1'] == '0'))
{
//create the query for the score
$q3 = "INSERT INTO tbl_Score (`Score`,`QuestionID`) VALUES ('100%','1')";
//Run the query
$r3 = #mysqli_query ($conn,$q3);
//Confirm message data was entered with a correct response and a graphic
echo '<h1>Correct!!</h1><img src="/images/green_Check_Low.jpg" alt="Green Check"/>';
echo 'Click here for the next question';
}
else{
//create the query for the score
$q4 = "INSERT INTO tbl_Score (`Score`,`QuestionID`) VALUES ('0%','1')";
//Run the query
$r4 = #mysqli_query ($conn,$q4);
//Confirm message data was entered with a correct response and a graphic
echo '<h1>Incorrect!!</h1><img src="/images/red_X_Low.jpg" alt="Red X"/><br/>';
echo 'Click here to try again';
}
}
Related
In my code, I have two forms for users to select options. The first variable will save but as soon as the user submits the second form, the variable from the first form is no longer saved.
<div class = "school">
<h3>Please select the university you previously attended</h3>
<form action = "" method = "post" name = "school_form">
<select name="school" size ="10">
<?php
//shows options for $selected_school
$sql = "SELECT DISTINCT school FROM data;";
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
if ($resultCheck > 0){
while($row = mysqli_fetch_assoc($result)){
// inserts all data as array
echo "<option>". $row['school'] ."</option>";
}
}
?>
</select>
<br>
<input type ="submit" name = "submit_school" value = "Enter">
</form>
<?php
//saves selected option as $selected_school
if(isset($_POST['submit_school'])){
$selected_school = mysqli_real_escape_string($conn, $_POST['school']);
echo "You have selected: " .$selected_school;
}
?>
</div>
<div class ="courses">
<h3>Please select the courses you took</h3>
<form action = "" method ="post" name ="course_form">
<?php
//user shown options for courses
$sql2 = "SELECT transfer_course, transfer_title FROM data WHERE school = ? ORDER BY transfer_course ASC";
$stmt = mysqli_stmt_init($conn);
if(!mysqli_stmt_prepare($stmt, $sql2)) {
echo "SQL statement failed";
} else {
mysqli_stmt_bind_param($stmt, "s", $selected_school);
mysqli_stmt_execute($stmt);
$result2 = mysqli_stmt_get_result($stmt);
while($row2 = mysqli_fetch_assoc($result2)){
echo "<input type='checkbox' name ='boxes[]' value = '" . $row2['transfer_course'] . "' >" . $row2['transfer_course'] . "<br>";
}
}
?>
<br>
<input type ="submit" name = "submit_courses" value = "Enter">
</form>
<br>
<?php
//saved selected option(s) as $selected_course
if(isset($_POST['submit_courses'])){//to run PHP script on submit
if(!empty($_POST['boxes'])){
foreach($_POST['boxes'] as $selected_course){
echo "You have selected: " . $selected_course . "</br>";
}
}
}
?>
</div>
<div class = "output">
<h3>Course Equivalency</h3>
<?php
$sql3 = "SELECT arcadia_course, arcadia_title FROM data WHERE school = " . $selected_school . " AND transfer_course = " . $selected_course . "";
$result3 = mysqli_query($conn, $sql3);
if($result3)
{
while($row3 = mysqli_fetch_assoc($result3)){
echo $row3['arcadia_course'] . " " . $row3['arcadia_title'] . "<br>";
}
} else {
echo "failed";
echo $sql3;
}
?>
So by the time I get to my next sql statement
$sql3 = "SELECT arcadia_course, arcadia_title FROM data WHERE school = " . $selected_school . " AND transfer_course = " . $selected_course . "";
When the school is selected, it saves the variable, but when the course is selected, $selected_school becomes blank again.
I already have session_start() at the top of the page.
You can used session variable ,it will help to make data accessible across the various pages .
So,whenever form get submitted you can save that value in session and retrieve it anytime.In top of your php file you need to start session i.e session_start(); .Then in your code
<?php
//saves selected option as $selected_school
if(isset($_POST['submit_school'])){
$_SESSION['selected_school ']=$selected_school;// here you are storing value to session
$selected_school = mysqli_real_escape_string($conn, $_POST['school']);
echo "You have selected: " .$selected_school;
}
?>
Same you can do with your $selected_course also .Then you can passed value to your query like below :
$sql3 = "SELECT arcadia_course, arcadia_title FROM data WHERE school = " .$_SESSION['selected_school ']. " AND transfer_course = " .$_SESSION['selected_course']. "";
For more information refer here
It looks like your option doesn't have a value it is passing. Try this in your first form while loop:
echo '<option value="' . $row['school'] . '">' . $row['school'] . '</option>';
It looks like there may be some more issues you are having as well. If this doesn't fix your issue, I'll dig deeper.
EDIT: Then, yes, as others have suggested, you probably want to add a hidden input field to pass that variable value on the second form submit as well.
What we are saying about the hidden input field is this:
<input type="hidden" name="selected_school" value="<?php if(isset($selected_school) echo $selected_school; ?>">
Hi I am trying to do a Registration that the users will put their name password and their answers to some questions and then an admin will manually answer to it if it's accepted.I did the system that loads their name password and answers in the database,and I also ran the things that will show the answers to the admin,but I can't figure a way to change a value just for one user not for all of them,I will leave you my codes and everything over here.
Here is my admin.viewapplications.php code
(Here,it shows everything fine,but I can't figure a way that the button to act just for one id not for all)
<?php
//include(__DIR__ . "/signup.php");
include("../resources/config.php");
//$name = $_POST['Name'];
//$mg = $_POST['MG'];
//$pg = $_POST['PG'];
//$rk = $_POST['RK'];
$sql = "SELECT id, name, tutorial, MG, PG, RK FROM rp_users WHERE tutorial = 2";
//$tutorial = "SELECT tutorial FROM rp_users";
$result = mysql_query($sql);
//$result2 = mysql_query($tutorial);
//$value = mysql_fetch_object($result2)
/*if($result)
{
echo "Succes";
}
else
{
die(mysql_error());
}*/
//if($value > 1)
//
while($row = mysql_fetch_array($result))
{
//$tutorial = row["tutorial"];
//f($tutorial == 2)
//}
$id = $row["id"];
$name = $row["name"];
$mg = $row["MG"];
$pg = $row["PG"];
$rk = $row["RK"];
echo "ID: " . $id."<br> <br>";
echo "Nume: " . $name."<br> <br>";
echo "MG: " . $mg."<br> <br>";
echo "PG: " . $pg."<br> <br>";
echo "RK: " . $rk."<br> <br>";
echo '<form action="./?p=applicationaccept" method="POST">';
echo '<input type="submit" name="accept" value="Accepta">';
echo '</form><br>';
echo '<form action="./?p=applicationdeny" method="POST">';
echo '<input type="submit" name="deny" value="Respinge">';
echo '</form><br> <br> <br>';
}
//}
//
?>
And here is my applicationaccept.php
<?php
include("../admin/admin.viewapplications.php");
include("../resources/config.php");
$iduser = $id;
$sql = "UPDATE rp_users SET tutorial=0";
$result = mysql_query($sql);
if($result)
{
echo "Succes";
}
else
{
die(mysql_error());
}
/*while($row = mysql_fetch_array($result))
{
}*/
?>
I think what you want to do is a simple UPDATE to your MySQL database..
but make sure you format the PHP code you're using otherwise it'll give you an ERROR!
Also you have to use 'mysqli' now in PHP!
<?php
$someID = '1';
$sql = "UPDATE `rp_users` SET `tutorial`= '0' WHERE `id` = $someID";
$result = mysqli_query($link, $sql);
if($result)
{
echo "Success";
}
else
{
echo ("Error");
}
?>
BTW I forgot to mntion the '$link' is the connection to your database!
As of my understanding of your question if your form action is applicationaccept.php and you are trying to update for one user in applicationaccept.php file, try this:
<?php
include("../admin/admin.viewapplications.php");
include("../resources/config.php");
$iduser = $_POST["id"]; // pass id as parameter in form
$sql = "UPDATE rp_users SET tutorial=0";// change this line to following line
$sql = "UPDATE rp_users SET tutorial=0 where id=$iduser";
$result = mysql_query($sql);
if($result)
{
echo "Succes";
}
else
{
die(mysql_error());
}
?>
Be aware your code is vulnerable
I have the facility to update what I call 'documents' (ver similar to creating a post) on my cms which works fine but I have introduced categories where the documents are associated to them. Now I have managed to bind them when creating the doc from new but when trying update them I am getting a bit stuck. I am using checkboxes to show the list of categories and when selected it updates a join table which uses the doc_id and the cat_id.
Here is the script for updating the doc:
<?php
include ('includes/header.php');
require ('../../db_con.php');
echo '<h1>Document Edit</h1>';
// Check for a valid document ID, through GET or POST:
if ( (isset($_GET['id'])) && (is_numeric($_GET['id'])) ) { // From view_docs.php
$id = $_GET['id'];
} elseif ( (isset($_POST['id'])) && (is_numeric($_POST['id'])) ) { // Form submission.
$id = $_POST['id'];
} else { // No valid ID, kill the script.
echo '<p class="error">This page has been accessed in error.</p>';
exit();
}
// Check if the form has been submitted:
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
$errors = array();
// Check for a document name:
if (empty($_POST['doc_name'])) {
$errors[] = 'You forgot to enter your document name.';
} else {
$dn = mysqli_real_escape_string($dbc, trim($_POST['doc_name']));
}
// Check for a document content:
if (empty($_POST['doc_content'])) {
$errors[] = 'You forgot to enter your last name.';
} else {
$dc = mysqli_real_escape_string($dbc, trim($_POST['doc_content']));
}
if (empty($errors)) { // If everything's OK.
// Test for unique doc title:
$q = "SELECT doc_id FROM docs WHERE doc_name='$dn' AND doc_id != $id";
$r = mysqli_query($dbc, $q);
if (mysqli_num_rows($r) == 0) {
// Make the query:
$q = "UPDATE docs SET doc_name='$dn', doc_content='$dc', doc_name='$dn' WHERE doc_id=$id LIMIT 1";
$r = mysqli_query ($dbc, $q);
if (mysqli_affected_rows($dbc) == 1) { // If it ran OK.
$doc_id = mysqli_insert_id($dbc);
$query = "UPDATE doc_cat_join (cat_id,doc_id) VALUES ";
$cat_ids = $_POST['cat_id'];
$length = count($cat_ids);
for ($i = 0; $i < count($cat_ids); $i++) {
$query.='(' . $cat_ids[$i] . ',' . $doc_id . ')';
if ($i < $length - 1)
$query.=',';
}
// Print a message:
echo '<p>The document has been edited.</p>';
} else { // If it did not run OK.
echo '<p class="error">The document could not be edited due to a system error. We apologize for any inconvenience.</p>'; // Public message.
echo '<p>' . mysqli_error($dbc) . '<br />Query: ' . $q . '</p>'; // Debugging message.
}
} else { // Already used.
echo '<p class="error">The document name has already been used.</p>';
}
} else { // Report the errors.
echo '<p class="error">The following error(s) occurred:<br />';
foreach ($errors as $msg) { // Print each error.
echo " - $msg<br />\n";
}
echo '</p><p>Please try again.</p>';
} // End of if (empty($errors)) IF.
} // End of submit conditional.
// Always show the form...
// Retrieve the document's information:
$q = "SELECT * FROM docs WHERE doc_id=$id";
$r = mysqli_query ($dbc, $q);
if (mysqli_num_rows($r) == 1) { // Valid document ID, show the form.
// Get the document's information:
$row = mysqli_fetch_array ($r, MYSQLI_NUM);
// Create the form:
echo '<form action="edit_doc.php" method="post">
<p>Document Name: <input type="text" name="doc_name" size="15" maxlength="15" value="' . $row[1] . '" /></p>
<textarea name="doc_content" id="doc_content" placeholder="Document Content" style="display: none;"></textarea>
<iframe name="editor" id="editor" ></iframe>'
?>
<div class="row">
<div class="col-group-1">
<?php
$q = "SELECT * FROM cats";
$r = mysqli_query ($dbc, $q); // Run the query.
echo '<div class="view_body">';
// FETCH AND PRINT ALL THE RECORDS
while ($row = mysqli_fetch_array($r)) {
echo '<br><label><input type="checkbox" name="cat_id[]" value="' . $row['cat_id'] . '">' . $row["cat_name"] . '</label>';
}
echo '</div>';
?>
</div>
</div>
<br><br>
<input onclick="formsubmit()" type="submit" value="Update Document" name="submit"/>
<?php echo'
<input type="hidden" name="id" value="' . $id . '" />
</form>
<br><br>Back to docs list';
} else { // Not a valid document ID.
echo '<p class="error">This page has been accessed in error.</p>';
}
?>
<?php
mysqli_close($dbc);
?>
So I have three tables:
docs
doc_id
doc_name
doc_content
cats
cat_id
cat_name
doc_cat_join
doc_id
cat_id
the join table related the doc_id and cat_id which then associates them together. I am guessing in my script when I update a doc it will need to delete the rows and then re-insert them? I just need to know a way or the easiest way of updating the join table as I am a tad stuck...
In case of checkbox update you need to delete previous stored checkbox of with appropriate id and insert new one you can't update checkbox as we can't predict how many checkbox will be selected by user....
Case:
It may happen that user remove one checkbox at update time so you will never know which one to be deleted.......
In your code...
docs
doc_id
doc_name
doc_content
cats
cat_id
cat_name
doc_cat_join
id
doc_id
cat_id
here you have to delete old checkbox of updation doc,
DELETE FROM doc_cat_join WHERE cat_id = some_id
next you can insert selected checkbox as you are inserting first time...
I have two tables:
'courses' with these fields:COURSE_ID(auto
increment),start_date,end_date,title
'courses_students' with these fields:ID(auto
increment),COURSE_ID,STUDENT_ID.
I want to insert some values in my mysql table called "courses_students" from my other table called "courses".
Users can see in a page the data from 'courses'(courses names,starting dates,ending dates) and they must select which course they want to attend,by clicking the button 'attent course'.
Everytime someone clicks the submit button,the values are inserted in courses_students table,but not correctly.The problem is that everytime,the COURSE_ID from 'courses_students' has the value of the last COURSE_ID from 'courses'.And,other strange problem is that the values are inserted twice,everytime.
This is the code:
<?php
$link = mysql_connect('localhost','root','');
if(!$link){
die('Could not connect: '.mysql_error());
}
mysql_selectdb("db");
?>
<ul>
<?php
$sql = "SELECT * FROM courses";
$result = mysql_query($sql);
while($file = mysql_fetch_array($result)){
echo '<ul>';
echo '<li>';
$STUDENT_ID = $_SESSION['ID'];
$COURSE_ID = $file['COURSE_ID']; //**It dislays the CORRECT ID for each course!**
echo 'the course id: ' .$COURSE_ID;
echo 'course name: ' .$file['title'];
echo 'Starting: ' .$file['start_date'];
echo ' ending: '.$file['end_date'];
echo '<form action="lista_cursuri.php" method="post"> <input type="submit" name="submit" value="attent course!"> </form>';
echo '</li>';
}
?>
</ul>
<?php
if(isset($_POST['submit'])){
$sql1 = "INSERT INTO `courses_students` (COURSE_ID,STUDENT_ID) VALUES ($COURSE_ID,$STUDENT_ID)";
$result = mysql_query($sql1);
}
?>
I can't manage to see where the problem is.Maybe this is not the correct procedure.
Salut
Change mysql_fetch_array which is a multidimensional array once numerically indexed and once by field name which gives you double results to mysql_fetch_assoc
and $COURSE_ID is always last because for each loop it is overwritten
Try this:
Part 1:
while($file = mysql_fetch_assoc($result)){
echo '<ul>';
echo '<li>';
$STUDENT_ID = $_SESSION['ID'];
$COURSE_ID = $file['COURSE_ID'];
echo 'the course id: ' .$COURSE_ID;
echo 'course name: ' .$file['title'];
echo 'Starting: ' .$file['start_date'];
echo ' ending: '.$file['end_date'];
echo "<form action='lista_cursuri.php' method='post'>
<input type='hidden' name='courseid' value='".$COURSE_ID."' >
<input type='submit' name='submit' value='attent course!''> </form>";
echo '</li>';
}
Part 2:
if(isset($_POST['submit'])){
$course = $_POST['courseid'];
$sql1 = "INSERT INTO `courses_students` (COURSE_ID,STUDENT_ID) VALUES ($course,$STUDENT_ID)";
$result = mysql_query($sql1);
}
Try this:
$sql1 = "INSERT INTO `courses_students` (COURSE_ID,STUDENT_ID)
VALUES ('".$COURSE_ID."','".$STUDENT_ID."')"; $result = mysql_query($sql1);
I am having a big issue.
This is the first time I sue a foreach and I do not even know if it's the right thing to use.
I have a textarea where my members can add some text.
They also have all the accounts where to send the posted text.
Accounts are of two types F and T.
They are shown as checkboxes.
So when a member types "submit" the text should be INSERTED in a specific table for EACH of the selected accounts. I thought php foreach was the right thing. But I am not sure anymore.
Please take in mind I do not know anything about foreach and arrays. So please when helping me, consider to provide the modded code =D . Thank you so much!
<?php
require_once('dbconnection.php');
$MembID = (int)$_COOKIE['Loggedin'];
?>
<form action="" method="post">
<p align="center"><textarea id="countable1" name="addit" cols="48" rows="10" style="border-color: #ccc; border-style: solid;"></textarea>
<br>
<?
$DB = new DBConfig();
$DB -> config();
$DB -> conn();
$on="on";
$queryF ="SELECT * FROM `TableF` WHERE `Active`='$on' AND `memberID`='$MembID' ORDER BY ID ASC";
$result=mysql_query($queryF) or die("Errore select from TableF: ".mysql_error());
$count = mysql_num_rows($result);
if ($count > 0)
{
while($row = mysql_fetch_array($result)) {
?><div style="width:400px; height:100px;margin-bottom:50px;"><?
$rowid = $row['ID'];
echo $row['Name'] . '</br>';
$checkit = "checked";
if ($row['Main'] == "")
$checkit = "";
if ($row['Locale'] =="")
$row['Locale'] ="None" . '</br>';
echo $row['Locale'] . '</br>';
if ($row['Link'] =="")
$row['Link'] ="javaScript:void(0);";
?>
<!-- HERE WE HAVE THE "F" CHECKBOXES. $rowid SHOULD BE TAKEN AND INSERTED IN THE FOREACH BELOW FOR ANY SELECTED CHECKBOX IN THE FIELD "Type".SEE BELOW -->
<input type="checkbox" name="f[<?php echo $rowid?>]" <?php echo $checkit;?>>
</div>
<?
}//END WHILE MYSQL
}else{
echo "you do not have any Active account";
}
$DB = new DBConfig();
$DB -> config();
$DB -> conn();
$queryTW ="SELECT * FROM `TableT` WHERE `Active`='$on' AND `memberID`='$MembID' ORDER BY ID ASC";
$result=mysql_query($queryTW) or die("Errore select TableT: ".mysql_error());
$count = mysql_num_rows($result);
if ($count > 0)
{
while($row = mysql_fetch_array($result)) {
?><div style="width:400px; height:100px;margin-bottom:50px;"><?
$rowid = $row['ID'];
echo $row['Name'] . '</br>';
$checkit = "checked";
if ($row['Main'] == "")
$checkit = "";
if ($row['Locale'] =="")
$row['Locale'] ="None" . '</br>';
echo $row['Locale'] . '</br>';
if ($row['Link'] =="")
$row['Link'] ="javaScript:void(0);";
?>
<!-- HERE WE HAVE THE "T" CHECKBOXES. $rowid SHOULD BE TAKEN AND INSERTED IN THE FOREACH BELOW FOR ANY SELECTED CHECKBOX IN THE FIELD "Type".SEE BELOW -->
<input type="checkbox" name="t[<?php echo $rowid?>]" <?php echo $checkit;?> >
</div>
<?
}//END 2° WHILE MYSQL
}else{
echo "you do not have any Active account";
}
?>
<input type="submit" value="submit">
</form>
<?
//WHEN CHECKBOXES "F" ARE FOUND, FOR EACH CHECKBOX IT SHOULD INSERT INTO TableG THE VALUES BELOW, FOR EACH SELECTED "F" CHECKBOX
if(!empty($_POST['addit']) && !empty($_POST['f'])){
$thispostF = $_POST['f'];
$f="F";
foreach ($thispostF as $valF)
//THE MOST IMPORTANT FIELD HERE IS "Type". THE ARRAY INSERT "on", I NEED INSTEAD THE VALUE $rowid AS ABOVE
$queryaddF="INSERT INTO TableG (ID,memberID,Type,IDAccount,Tuitting) VALUES (NULL,".$MembID.",'".$f."','".$valF."', '".$_POST['addit']."')";
$resultaddF=mysql_query($queryaddF) or die("Errore insert G: ".mysql_error());
}
//WHEN CHECKBOXES "T" ARE FOUND, FOR EACH CHECKBOX IT SHOULD INSERT INTO TableG THE VALUES BELOW, FOR EACH SELECTED "T" CHECKBOX
if(!empty($_POST['addit']) && !empty($_POST['t'])){
$thispostT = $_POST['t'];
$t="T";
foreach ($thispostT as $valF)
//THE MOST IMPORTANT VALUE HERE IS "Type". THE ARRAY GIVES "on", I NEED INSTEAD THE VALUE $rowid AS ABOVE
$queryaddT="INSERT INTO TableG (ID,memberID,Type,IDAccount,Tuitting) VALUES (NULL,".$MembID.",'".$t."','".$valF."', '".$_POST['addit']."')";
$resultaddF=mysql_query($queryaddF) or die("Errore insert G: ".mysql_error());
}
?>
foreach ($thispostT as $valF)
{
$queryaddT="INSERT INTO TableG (ID,memberID,Type,IDAccount,Tuitting) VALUES (NULL,".$MembID.",'".$t."','".$valF."', '".$_POST['addit']."')";
$resultaddF=mysql_query($queryaddF) or die("Errore insert G: ".mysql_error());
}
please put start and ending bracket to your foreach loop and try i have not read the whole code but just found you missing the brackets. hope that helps you.
I think I know what you're doing...
You're going to need to do:
foreach($_POST as $key => $value) {
$type = substr($key,0,1);
$id = substr($key,1);
if($type == 't') {
// do insert for t table here
} else if($type == 'f') {
// do insert for f table here
}
}
I didn't test it but it should be something like this.
My suggestion is
create field name as t[] (array)
onchecked value will be passed on the next page
the form checkbox field should be like that
<input type="checkbox" name="t[]" value="< ?php echo $rowid?>" <?php echo $checkit;? > >
and when you Submit the form
GET THE VALUE and insert in to database;
< ?
if($_POST['t'])
{
foreach($_POST['t'] as $v)
{
queryaddT="INSERT INTO TableG (ID,memberID,Type,IDAccount,Tuitting) VALUES (NULL,".$MembID.",'".$t."','".$valF."', '".$_POST['addit']."')";
$resultaddF=mysql_query($queryaddF) or die("Errore insert G: ".mysql_error());
}
}
? >