I am try to inset multiple rows to new table fetched from other table, but problem is that only last single row is being inserted and no other now is getting insert, so please tell the issue where i am lacking
<?php
error_reporting(1);
session_start();
$s=$_SESSION['username'];
//connect database
$con=mysql_connect("localhost","root","") or die(mysql_error());
// select database
mysql_select_db("education",$con);
$date= date("Y/m/d");
//select all values from empInfo table
$data="SELECT * FROM student";
$val=mysql_query($data);
?>
<html>
<body>
<table>
</table>
<form action="submit.php" method="post" >
<table>
<tr>
<th>Teacher name</th>
<th>Date</th>
<th>Roll No</th>
<th>Student name</th>
<th>Father name</th>
<th>Addhaar No</th>
<th>Status(P)</th>
<th>Status(A)</th>
<th>Status(L)</th>
</tr>
<?php while($r=mysql_fetch_array($val))
{?>
<tr style="border:2px solid black;">
<td><input type="text" name="teacher" value="
<?php echo $s; ?>"></td>
<td><input type="text" name="date" value="
<?php echo $date; ?>"></td>
<td ><input name="roll_no" value="
<?php echo $r['roll_no']; ?>">
</td>
<td><input name="student_name" value="
<?php echo $r['student_name'] ?>">
</td>
<td><input name="father_name" value="
<?php echo $r['father_name'] ?>">
</td>
<td>
<input name="addhaar_no" value="
<?php echo $r['addhaar_no'] ?>">
</td>
<td>
<input type="checkbox" value="present" name="status"> Present
</td>
<td>
<input type="checkbox" name="status" value="absent">Absent
</td>
<td>
<input type="checkbox" name="status" value="leave">Leave
</td>
</tr>
</table>
<?
}
?>
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>
submit.php -
<?php
error_reporting(1);
$con=mysql_connect("localhost","root","") or die(mysql_error());
// select database
mysql_select_db("education",$con);
//get data from html form
$roll_no=$_POST['roll_no'];
$student_name=$_POST['student_name'];
$father_name=$_POST['father_name'];
$addhaar_no=$_POST['addhaar_no'];
$status=$_POST['status'];
//Insert values in empInfo table with column name
$query="INSERT INTO attandance
VALUES ('', '$roll_no','$student_name','$father_name','$addhaar_no','$status'),
VALUES ('', '$roll_no','$student_name','$father_name','$addhaar_no','$status')";
echo $query;
die();
mysql_query($query);
?>
page
You need to have unique name for each fields. What you can do is have a counter in loop and add it the names of the fields to make it unique.
Sample:
$ctr = 0;
while($r=mysql_fetch_array($val)){
echo "<input type="text" name='teacher_".$ctr."'>";
$ctr++;
}
Or make the names array, and loop through the values in saving the data.
while($r=mysql_fetch_array($val)){
echo "<input type="text" name='teacher[]'>";
}
I think you should study PHP a bit more... As i can see in your code, you haven't understood fundamentals of PHP.
1: Normally, you won't mix up HTML and PHP like you did in your first code. Its just confusing and really annoying to read the code later.
2: When you post your form, for example the variable $_POST['student_name']; will just contain the value of the last row (your problem). So, why? Because you can't assign more than one value to a variable. Or at least, not the way you tried it. Array would be a good keywoard for this problem.
3: Please check your SQL syntax... Thats why i'm saying you haven't understand fundamentals... http://www.w3schools.com/sql/sql_insert.asp
Why you're repeating your values? You think the second time the variables will contain the values of the next row? Thats just false. A Variable contains everytime the same value, as long as you don't assign a new value to it.
4: mysql is depracted. Use mysqli or PDO instead.
My tip: You need to have unique input names. Just take a look at PHP, how for/while loops work, study a bit more and try it again. It's not difficult to solve, but i think you'll learn a lot more if we don't give you the direct solution.
Now mysql is depracted. So, you can use mysqli or PDO instead.
I can use now PDO. Please follow bellow code carefully:
<?php
$user = 'root';
$pass = '';
$dbh = new PDO('mysql:host=localhost;dbname=education', $user, $pass);
try {
$select = $dbh->query('SELECT * from student');
} catch (PDOException $e) {
print "Error!: " . $e->getMessage() . "<br/>";
die();
}
?>
<html>
<body>
<form action="submit.php" method="post" >
<table>
<tr>
<th>Teacher name</th>
<th>Date</th>
<th>Roll No</th>
<th>Student name</th>
<th>Father name</th>
<th>Addhaar No</th>
<th>Status(P)</th>
<th>Status(A)</th>
<th>Status(L)</th>
</tr>
<?php
foreach($select as $val) {
?>
<tr style="border:2px solid black;">
<td><input type="text" name="teacher" value="<?php echo $s; ?>"></td>
<td><input type="text" name="date" value="<?php echo $date; ?>"></td>
<td><input name="roll_no" value="<?php echo $r['roll_no']; ?>"></td>
<td><input name="student_name" value="<?php echo $r['student_name'] ?>"></td>
<td><input name="father_name" value="<?php echo $r['father_name'] ?>"></td>
<td><input name="addhaar_no" value="<?php echo $r['addhaar_no'] ?>"></td>
<td><input type="checkbox" value="present" name="status"> Present</td>
<td><input type="checkbox" name="status" value="absent">Absent</td>
<td><input type="checkbox" name="status" value="leave">Leave</td>
</tr>
<?php
}
?>
</table>
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>
For submit.php code bellow
<?php
$user = 'root';
$pass = '';
$dbh = new PDO('mysql:host=localhost;dbname=education', $user, $pass);
$stmt = $dbh->prepare("INSERT INTO attandance (roll_no, student_name, father_name, addhaar_no, status) VALUES (?, ?, ?, ?, ?)");
$stmt->bindParam(1, $roll_no);
$stmt->bindParam(2, $student_name);
$stmt->bindParam(2, $father_name);
$stmt->bindParam(2, $addhaar_no);
$stmt->bindParam(2, $status);
//if you insert 2 time then
for($x=0; $x<2; $x++) {
$roll_no = $_POST['roll_no'];
$student_name = $_POST['student_name'];
$father_name = $_POST['father_name'];
$addhaar_no = $_POST['addhaar_no'];
$status = $_POST['status'];
$stmt->execute();
}
?>
Related
I have read articles and questions regarding this issue but I'm still finding it difficult to go about it.
I'm trying to achieve course registration. So I have this table populated from my db.
<form method="post" action="courses.php">
<table>
<thead>
<th>Action<th>
<th>Course Title<th>
<th>Course Code<th>
<th>Course Unit<th>
</thead>
<tbody>
<!--Query that fetches the data from the db (included dbcon.php).. the courses are created by admin-->
<tr>
<td><input type="checkbox" name="isChecked[]" value="<?php echo $course_id?>"></td>
<td><input type="hidden" name="ctitle[]" value="<?php echo $title?>"> <?php echo $title?></td>
<td><input type="hidden" name="ccode[]" value="<?php echo $code?>"> <?php echo $code?></td>
<td><input type="hidden" name="cunit[]" value="<?php echo $unit?>"> <?php echo $unit?></td>
</tr>
</tbody>
</table>
<input type="submit" name="enroll">
</form>
If a checkbox is checked, How can I get all the values of the array (title, code, unit) and insert them on its corresponding column in my enroll table also note that the courses are more than 1?
here is the little I have done but I'm beginning to think that I'm not going to go through with it (imaging creating another foreach of the array? initializing a counter foreach??)
<?php
if(isset($_POST['enroll'])){
$i=0;
if(!empty ($_POST['isChecked'])){
foreach($_POST['isChecked'] as $course){
$course = $_POST['isChecked'][$i];
$query = "INSERT INTO enrolled (course_id)
VALUES ('$course')";
$insert = $db->query($query);
$i++;
}
}?>
The above insert the course id to my DB quite fine. But what I want is to actually insert all the data i.e (id, title, code, and unit to the database )
That is because your code is wrong. Try this:
<tr>
<td><input type="checkbox" name="isChecked['course_id']" value="<?php echo $course_id?>"></td>
<td><input type="hidden" name="isChecked['title']" value="<?php echo $title?>"> <?php echo $title?></td>
<td><input type="hidden" name="isChecked['code']" value="<?php echo $code?>"> <?php echo $code?></td>
<td><input type="hidden" name="isChecked['unit']" value="<?php echo $unit?>"> <?php echo $unit?></td>
</tr>
Do not forget to BIND your values and check them before to execute your SQL code.
I'm new to php programming. I'm kind of confused and I can't find any helpful information online. I'm trying to build a school manangement system from scratch. What I need is to get all the 'offered courses' from the database and put it in a form and allow the student to add the classes directly from the row. How can I do that?
I created a form and an input where the student might enter the course number and register for class. And it works fine. But I feel like it's not practical.
Here is my code
<?php
session_start();
// include("config.php");
include("functions.php");
// $sql="SELECT `course_num`, `professors`.`name` AS pName, `courses`.`name`AS cName , max_students FROM `courses`, `student_courses`,`professors` WHERE `professors`.`id`=`courses`.`professor_teaching` AND `student_id`= '".$_SESSION['student_id']."'";
// $result = mysqli_query($link, $sql);
$result = viewAllCourses();
?>
<form method="post" action="functions.php">
<table>
<tr>
<th><label>Course No.</label></th>
<th><label>Course Name</label></th>
<th><label>Professor</label></th>
<th><label>Max. Students</label></th>
<th><label>Action</label></th>
</tr>
<?php
if(mysqli_num_rows($result)){
while ($row = mysqli_fetch_assoc($result)) {
?>
<tr>
<td><label name="course_num"><?php echo $row['course_num'];?>
<input type="hidden" name ="coursenumber" value=<?php $row['course_num']?>>
</label></td>
<td><label><?php echo $row['cName'];?></label>
</td>
<td><label><?php echo $row['pName']; ?></label></td>
<td><label><?php echo $row['max_students']; ?></label></td>
<td><input type="submit" name="add" value="add"></td>
</tr>
</form>
</table>
<?php
}
}
?>
and then in the functions.php I have this code:
if (isset($_GET['add'])) {
$link = conn();
echo "TST";
exit;
$courseNum= $_POST['coursenumber'];
$record = mysqli_query($link, "INSERT INTO `student_courses`
(`student_id`, `course_id_num`)
VALUES ('".$_SESSION['student_id']."', '$courseNum')");
}
But it does nothing.
I tried adding an input tag for the course_number and passing it from there. But it doesn't work. What is the right way to do this?
You're the same names for the inputs in all the rows. When you submit the form, $_POST['coursenumber'] will just be the last course number in the table, not the one the user clicked on. You can put the course number in the value of the add button, rather than a hidden input. When a form has multiple submit buttons, the value comes from the one that was clicked.
You need to fix the order of the </form> and </table> tags, so they nest properly.
<?php
if(mysqli_num_rows($result)){
while ($row = mysqli_fetch_assoc($result)) {
?>
<tr>
<td><label name="course_num"><?php echo $row['course_num'];?>
</label></td>
<td><label><?php echo $row['cName'];?></label>
</td>
<td><label><?php echo $row['pName']; ?></label></td>
<td><label><?php echo $row['max_students']; ?></label></td>
<td><input type="submit" name="add" value="<?php echo $row['course_num'];?>"></td>
</tr>
</table>
</form>
<?php
}
}
?>
Also, since you're submitting the form with method="POST", the button will be $_POST['add'], not $_GET['add'].
You should use a prepared statement to protect against SQL-injection.
if (isset($_POST['add'])) {
$link = conn();
$courseNum= $_POST['add'];
$stmt = mysqli_prepare("INSERT INTO student_courses (student_id, course_id_num) VALUES (?, ?)");
mysqli_stmt_bind_param($stmt, "ii", $_SESSION['student_id'], $courseNum);
mysqli_stmt_execute($stmt);
}
If you want to pass multiple fields, you can put a separate form with multiple hidden inputs into each row, rather than making the whole table a form.
<?php
if(mysqli_num_rows($result)){
while ($row = mysqli_fetch_assoc($result)) {
?>
<tr>
<td><label name="course_num"><?php echo $row['course_num'];?>
</label></td>
<td><label><?php echo $row['cName'];?></label>
</td>
<td><label><?php echo $row['pName']; ?></label></td>
<td><label><?php echo $row['max_students']; ?></label></td>
<td><form action="functions.php" method="post">
<input type="hidden" name="coursenumber" value="<?php echo $row['course_num'];?>">
<niput type="hidden" name="something" value="<?php echo $row['something'];?>">
<input type="submit" name="add" value="add">
</form></td>
</tr>
</table>
<?php
}
}
?>
Then the functions.php script can use $_POST['course_num'] and $_POST['something'] to get these parameters.
I am going to fetching table values in a html table along checkbox in each row and then inserting values in another database table from multi check boxes in php.
Only the values of checked boxes should be submitted to that table.
db name "laboratory":
test: fetching values.
package: inserting table.
view
Status
Active
Inactive
<?php
$conn=mysqli_connect("localhost","root","","laboratory") or die(mysql_error());
mysql_select_db("test") or die(mysql_error());
$query="SELECT * FROM test";
$result=mysqli_query($conn,$query);
if ($result) {
while ($record=mysqli_fetch_array($result)) {
Please try to follow this code and implement in your program . Hope that this will cooperate you much
if(isset($_POST['name'])){
$name = $_POST['name'];
$status = $_POST['status'];
if(empty($name) || empty($status)){
echo "Field Must Not be empty";
} else{
$conn=new mysqli("localhost","root","","test");
if($conn){
$query = "SELECT * FROM userdata limit 5";
$stmt = $conn->query($query);
$val = '<form action="" method=""> ';
$val .= '<table> ';
if ($stmt) { ?>
<form action="" method="post">
<table>
<?php while ($result=$stmt->fetch_assoc()) { ?>
<tr>
<td><?php echo $result['post']; ?></td>
<td><input value="<?php echo $result['post']; ?>" type="checkbox" name="check[]" /></td>
</tr>
<?php } ?>
<tr>
<td>Actual Price </td>
<td>Discount</td>
<td>Final Price</td>
</tr>
<tr>
<td><input type="text" name="actual"/></td>
<td><input type="text" name="discount"/></td>
<td><input type="text" name="final"/></td>
</tr>
<tr>
<td>Description</td>
<td><textarea name="description" id="" cols="30" rows="10"></textarea></td>
</tr>
<tr>
<td><input type="submit" value="Submit" /></td>
<td><input type="reset" value="Cancel" /></td>
</tr>
</table>
</form>
<?php }} }}?>
<?php
if(isset($_POST)){
echo "<pre>";
print_r($_POST);
echo "<pre>";
}
?>`enter code here`
First of all you have to decide that what are you using either mysqli or mysql, if you are using mysqli then you have to improve your code
$query="SELECT * FROM test";
$result=mysqli_query($conn,$query);
if ($result) {
while ($record=mysqli_fetch_array($result)) {
and when you want to insert the checked data will be inserted in package table. If package table in another database then you have to give us the full detail i mean tell us the database name of package table.
I just need to insert my form data into mysql database & display it in browser. But, when I fill up the form & click submit , the row gets added but with no data except for ID field which is autoincremented.. even the table in phpmyadmin looks same with the row added & empty fileds.
any suggestions will be highly appreciated...
my html form looks like this,
<table border="1">
<tr>
<td align="center">Form Input Students Data</td>
</tr>
<tr>
<td>
<table>
<form method="POST" action="data_insert_htmlform.php/">
<tr>
<td><label for="Name">Name</label></td>
<td><input type="text" name="name" size="20">
</td>
</tr>
<tr>
<td><label for="Age">Age</label></td>
<td><input type="text" name="age" size="20">
</td>
</tr>
<tr>
<td><label for="Birth_Date">Birth_Date</label></td>
<td><input type="text" name="Birth_Date" size="20">
</td>
</tr>
<tr>
<td><label for="Address"Address</label></td>
<td><input type="text" name="address" size="40">
</td>
</tr>
<tr>
<td></td>
<td align="center">
<input type="submit" name="submit" value="Sent">
</td>
</tr>
</form>
</table>
</td>
</tr>
</table>
and my php code,
<?php
error_reporting(E_ERROR | E_PARSE);
$database = 'students';
$continued=mysql_connect("localhost" , "root", "");
if(mysql_select_db($database))
echo ("<br><br>connection to the database succeeds");
else
echo ("connection failed");
/*$name = $_POST['name'];
$age = $_POST['age'];
$birth_date = $_POST['Birth_date'];
$address = $_POST['address'];*/
$insert = "INSERT INTO students_basicinfo(Name, Age, Birth_Date, Address) VALUES ('{$_POST['name']}','{$_POST['age']}' , '{$_POST['Birth_date']}' , '{$_POST['address']}')";
$abc = mysql_query($insert);
if($abc){
echo("<br>Input data is succeed");
}else{
echo("<br>Input data is fail");
}
$order = "SELECT * FROM students_basicinfo";
$result = mysql_query($order);
if($result === FALSE) {
die(mysql_error()); // TODO: better error handling
}
echo "<table border='1'>";
while($data = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>".$data[ID]."</td>";
echo "<td>".$data[Name]."</td>";
echo "<td>".$data[Age]."</td>";
echo "<td>".$data[Birth_Date]."</td>";
echo "<td>".$data[Address]."</td>";
echo "</tr>";
}
echo "</table>";
?>
This issue about input name case sensitive
Change $_POST['Birth_date'] to $_POST['Birth_Date'] with uppercase D
Try following query, this will work.
$insert = "INSERT INTO students_basicinfo(Name, Age, Birth_Date, Address) VALUES ('{$_POST['name']}','{$_POST['age']}' , '{$_POST['Birth_Date']}' , '{$_POST['address']}')";
Replace
'{$_POST['name']}','{$_POST['age']}' , '{$_POST['Birth_date']}' , '{$_POST['address']}'
with
'".$_POST['name']."','".$_POST['age']."' , '".$_POST['Birth_date']."' , '".$_POST['address']."'
Try :
echo "<tr>";
echo "<td>".$data['ID']."</td>";
echo "<td>".$data['Name']."</td>";
echo "<td>".$data['Age']."</td>";
echo "<td>".$data['Birth_Date']."</td>";
echo "<td>".$data['Address']."</td>";
echo "</tr>";
(Use prepared statements.)
Dump the statement, in a HTML comment <!-- ... ---> so you can try it yourself.
Use echo mysql_error() to check for errors.
I mistrust the date field, DATE? Use '2013-08-31or '2013-08-31 14_:07 / '2013-08-31T14_:07`.
my problem goes like this:
my home page has tables with rows pulled from the database (while loops)
each row has a - cell in which he can add an event to that specific row
in order to do that i send the row id as a $_GET variable from the home page table
and in the "add event" page i store it as a variable
but when i submit my addevent form without filling it properly (as i coded) it simply refreshes the form only without the row id in the url therefor also the query i do in the beginning of the page for pulling the row data can no longer execute and that pops a PHP error
for the id variable which i sign it the $_GET and the query (mysql fetch array).
also of course all the data which i display in the form from that query is gone.
any suggestions on how to approach this ? thanks in advance, Regards.
EDIT:** kill the new guy! -Sorry i guess
home page where i send the id :
$sql = "SELECT * FROM alarms WHERE alarmstatus = 'OFF' and starttime='::' ORDER BY clientid ASC";
$query = mysql_query($sql);
echo "<table cellpadding='1px' border='1px' bordercolor='#0066FF' cellspacing='0'>
<form action='hpage.php' method='get'>";
while($fetch = mysql_fetch_array($query)) {
echo "<tr>
<td>
".$fetch['clientid']."</td>
<td>".$fetch['controller']."</td>
<td>".$fetch['typeid']."</td>
<td style='color: red'>".$fetch['alarmstatus']."</td>
<td>".$fetch['starttime']."</td>
<td>".$fetch['endtime']."</td>
<td><a href='includes/editalarm.php?id=".$fetch['id']."'>Edit</a></td>
<td><a href='includes/addevent.php?id=".$fetch['id']."'>Add event</a></td>
<td><a href='includes/deletealarm.php?id=".$fetch['id']."'>Delete</a></td>
</tr>";
}
the add event where i get the variable and make the query:
$alarmid = $_GET['id'];
$sql = "SELECT * FROM alarms WHERE id=".$alarmid;
$query = mysql_query($sql);
$fetch = mysql_fetch_array($query);
?>
the form:
<table cellpadding="2px" cellspacing="0" >
<form action="addevent.php" method="post">
<tr>
<td>סניף:</td>
<td><input style="width:200px; background-color: #d6d6d6;" readonly name="client" value="<?php echo $fetch['clientid']; ?>" /></td>
</tr>
<tr>
<td>בקר:</td>
<td><input style="width:200px; background-color: #d6d6d6;" readonly name="controller" value="<?php echo $fetch['controller']; ?>" /></td>
</tr>
<tr>
<td>אזעקה:</td>
<td><input style="width:200px; background-color: #d6d6d6;" readonly name="controller" value="<?php echo $fetch['typeid']; ?>" /></td>
</tr>
<tr>
<td>מוקדן:</td>
<td>
<?php
$sql = "SELECT * FROM users WHERE privilege = '2'";
$query = mysql_query($sql);
echo "<select name='user' style='width:207px;'>";
echo "<option>..</option>";
while ($fetch2 = mysql_fetch_array($query)){
echo "<option>".$fetch2['username']."</option>";
}
echo "</select>";
?>
</td>
</tr>
<tr>
<td>איש קשר:</td>
<td><input type="text" name="contact" /></td>
</tr>
<tr>
<td>הודעה:</td>
<td><input type="text" style="width:200px; height:100px" name="message" /></td>
</tr>
<tr>
<td>תשובה:</td>
<td><input type="text" style="width:200px; height:100px" name="answer" /></td>
</tr>
<tr>
<td>שעה:</td>
<td>
<select name="eventhour">
<option value ="default"></option>
<?php
for($i = 0; $i<60; $i++){
$value = $i;
if($i<=9){
$value= "0".$i;
}
echo "<option>".$value."</option>";
}
?>
</select>
<select name="eventminute">
<option value ="default"></option>
<?php
for($i = 0; $i<24; $i++){
$value = $i;
if($i<=9){
$value= "0".$i;
}
echo "<option>".$value."</option>";
}
?>
</select>
</td>
</tr>
<tr>
<td>
<input type="submit" name="save" value="שמור" />
<input type="submit" name="cancell" value="בטל" />
</td>
<td></td>
</tr>
</form>
Your form action is POST. If you change that to GET then you will have the form as $_GET.