how do fetching and then inserting from multi check boxes in php - php

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.

Related

php page to list and update sqlite

I have the following code to display and modify a simple sqlite table
<?php
$db = new SQLite3("my_sqlite.db");
$query = "SELECT rowid, * FROM students";
$result = $db->query($query);
if( isset($_POST['submit_data']) ){
// Gets the data from post
$name = $_POST['name'];
$email = $_POST['email'];
$query = "UPDATE students set name='$name', email='$email'";
if( $db->exec($query) ){
echo "Data is updated successfully.";
}else{
echo "Sorry, Data is not updated.";
}
}
?>
<table border="1">
<form action="" method="post">
<tr>
<td>Name</td>
<td>Email</td>
</tr>
<?php while($row = $result->fetchArray()) {?>
<tr>
<td><input name="name" type="text" value="<?php echo $row['name'];?>"></td>
<td><input name="email" type="text" value="<?php echo $row['email'];?>"></td>
</tr>
<?php } ?>
<input name="submit_data" type="submit" value="Update Data">
</form>
</table>
PROBLEM: When I change some of the information and update, the whole column changes into the same change. E.g.: if I write a the name Nick, every name changes into Nick.
First, you should only do updates for one record at a time so each record needs its own update button. Attached is the corresponding rơwid of the record. you can use:
<input type="hidden" name="rowid" value="$row['rowid]">
You should add a WHERE clause to the update statement to know exactly which records should be updated.If you omit the WHERE clause, ALL records will be updated!

How to insert multiple row to new table using single submit button

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();
}
?>

database field data not appearing in form textbox in PHP

i have this code in PHP and a database sql.. the situation is .. if i type the 1, 2 or 3 (productID) .. the textbox will be populated and field with database values.. but when i run the program.. fortunately it has no errors.. but when i type the id or 1 and click the submit button.. it doesnt get the neccessary values.. sorry for this im a complete newbie and im practicing PHP for a while now.. any help will do.. thank you..
<?php
session_start();
include_once 'dbconnect.php';
if(!isset($_SESSION['user'])){
header("Location: index.php");
}
$res = mysql_query("SELECT * FROM users WHERE user_id=".$_SESSION['user']);
$userRow = mysql_fetch_array($res);
?>
<?php
require('dbconnect.php');
$id = (isset($_REQUEST['productID']));
$result = mysql_query("SELECT * FROM tblstore WHERE productID = '$id'");
$sql = mysql_fetch_array($result);
if(!$result){
die("Error: Data not found");
} else {
$brandname = $sql['brandname'];
$price = $sql['price'];
$stocks = $sql['stocks'];
}
?>
<html>
<body>
<p>
hi' <?php echo $userRow['username']; ?> Sign Out
</p>
<form method="post">
<table align="center">
<tr>
<td>Search Apparel:</td>
<td><input type="text" name="search" name="productID" /></td>
</tr>
<tr>
<td>Brandname:</td>
<td><input type="text" name="brandname" value="<?php echo $brandname; ?>"/ </td>
</tr>
<tr>
<td>Price:</td>
<td><input type="text" name="price" value="<?php echo $price; ?>"/></td>
</tr>
<tr>
<td>Stocks:</td>
<td><input type="text" name="stocks" value="<?php echo $stocks; ?>"/></td>
</tr>
<tr>
<td> </td>
<td><input type="submit" name="submit" value="Search" /></td>
</tr>
</table>
</form>
</body>
</html>
your getting the id incorrectly, you have:
<?php
$_REQUEST['productID']=8; //for testing
$id = (isset($_REQUEST['productID']));
if you check it you will find the output is true\false as returned by isset
var_dump($id); //true
what you should use is:
<?php
if(isset($_REQUEST['productID'])){ //maybe also check its a number and or valid range
$id=$_REQUEST['productID'];
}

Blank screen when updating data in php table

I have been working on a project and i am at the final stages of the project. My problem is whenever i try to update data in my database table into returns a blank screen with no error messages. Please find the php script and html form (the form responsible for updating the database table) below, i have divided it into about four sections:
Thanks in advance
Update Form:
<a name="inventoryEditForm" id="inventoryEditForm"></a>
<h3>↓Add New Question Form↓</h3>
<form action="inventory_edit.php" enctype="multipart/from-data" name="myForm" id="myForm" method="post">
<table width="80%" border="0" cellspacing="3" cellpadding="7">
<tr>
<td width="20%"> </td>
<td width="80%"> </td>
</tr>
<tr>
<td>Question</td>
<td><textarea rows="" name="question" cols=""><?php echo $question; ?></textarea></td>
</tr>
<tr>
<td> </td>
</tr>
<tr>
<td>Venue</td>
<td><input type="text" name="venue" maxlength="50" value="<?php echo $venue; ?>"></td>
</tr>
<tr>
<td> </td>
</tr>
<tr>
<td>Date</td>
<td><input type="date" name="questiondate" value="<?php echo $date; ?>"></td>
</tr>
</table>
<br>
<input name="thisID" type="hidden" value="<?php echo $targetID; ?>"/>
<input type="submit" name="submit" value="Update Question">
<input type="reset" name="clear" value="Clear Form">
</form>
PHP Script:
<?php
//Error reporting due to long script
error_reporting(E_ALL);
ini_set('display_errors', '1');
?>
<?php
error_reporting(E_PARSE);
//Update question table
If (isset($_POST['question'])) {
$id = mysqli_real_escape_string($link, $_POST['thisID']);
$question = mysqli_real_escape_string($link, $_POST['question']);
$venue = mysqli_real_escape_string($link, $_POST['venue']);
$date = mysqli_real_escape_string($link, $_POST['questiondate']);
//Update question in the table
$sql = mysqli_query($link, "UPDATE DebateQuestion SET question='$question',venue='$venue',date='$date' WHERE qQuestionNo='$id'LIMIT 1") or die(mysql_error());
header("location: inventory.php");
exit();
}
?>
<?php
error_reporting(E_PARSE);
//Gather this questions full information and insert automatically into the edit form
if (isset($_GET['qid'])) {
$targetID = $_GET['qid'];
$sql = mysqli_query($link, "SELECT * FROM DebateQuestion WHERE qQuestionNo='$targetID'LIMIT 1") or die(mysql_error());
$questionCount = mysqli_num_rows($sql); // count the output amount
if ($questionCount > 0) {
while ($row = mysqli_fetch_array($sql, MYSQLI_ASSOC)) {
$id = $row["qQuestionNo"];
$question = $row["qQuestion"];
$venue = $row["qDebateVenue"];
$date = strftime("%b %d, %Y", strtotime($row["qDate"]));
}
} else {
echo "Oops, no questions like that exists. Check <a href='inventory.php'>inventory</a>again";
exit();
}
}
?>
In your update query you have the data column without using ` back ticks , date is also a mysql's function try to wrap up your column names with back ticks if you are not sure whether they conflict with mysql's reserved keywords
$sql = mysqli_query($link,"UPDATE DebateQuestion SET
`question`='$question',`venue`='$venue',`date`='$date'
WHERE qQuestionNo='$id'LIMIT 1")
"SELECT * FROM DebateQuestion WHERE qQuestionNo='$targetID'LIMIT 1"
Here is qQuestionNo column a string type?if not remove quotes around $targetID.
Note : I have not tested the code - just read it on screen.
I've never seen an IF statement capitalized before :
If (isset($_POST['question'])) {
I'd guess this makes a difference however.
There's lots of other weird things going on in your files, but none that should give you white screen. Try lowercase 'I' in your if statement first.
ALSO - re: the UPDATE statement, you are missing a space between the $id and the LIMIT :
**qQuestionNo='$id'LIMIT 1**

Input values doesnt get display in the browser instead shows blank field

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`.

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