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`.
Related
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 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'];
$stmt->execute();
}
?>
i have this code in PHP and a database sql.. the situation is .. if i type the 1, 2 or 3 (productID) .. the textbox will be populated and field with database values.. but when i run the program.. fortunately it has no errors.. but when i type the id or 1 and click the submit button.. it doesnt get the neccessary values.. sorry for this im a complete newbie and im practicing PHP for a while now.. any help will do.. thank you..
<?php
session_start();
include_once 'dbconnect.php';
if(!isset($_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'];
}
I have a table called "project_name" in my database called "encrypt_decrypt". The table contains only 1 column called "name" which contains different values of name (ex : p1, p2, p3...). I have to retrieve this values(p1,p2,p3..) from my database and display it on a registration form with each value displaying one below the other having a checkbox with it so that user can select any of the name while registering! How do i do this in php ???
Thanks in advance!
<html>
<body>
<form name="reg" action="code_exec2.php" onsubmit="return validateForm()" method="post">
<table width="274" border="0" align="center" cellpadding="2" cellspacing="0">
<tr>
<td><div align="right" style="white-space:nowrap" >Please select the project:</td>
</tr>
<tr>
<td><div align="right"><input type="checkbox" name="project[]" ></div></td>
<td><?php
session_start();
include('connection2.php');
$row=mysql_query("select * from project_name");
$array= array();
$output = mysql_fetch_assoc($row);
while($output){
$array[] = $output;
}
print_r($array);
?></td>
</tr>
<tr>
<td><div align="right"><input type="checkbox" name="Select all"
value="select all" onclick="toggle(this)"></div></td>
<td>Select all</td>
</tr>
<tr>
<td><div align="right"></div></td>
<td><input name="submit" type="submit" value="Submit" /></td>
</tr>
</table>
</form>
</body>
Try this.
$result=mysql_query("select * from project_name");
$checkboxes=array();
while($r=mysql_fetch_assoc($result)){
$checkboxes[]='<input type="checkbox" name="names[]" value="'.$r['name'].'">'.$r['name'].'<br />';
}
Then echo below wherever you want the checkboxes to appear.
echo implode("\n",$checkboxes);
Assuming, that your connection to db is correct, you need to change the way you display results:
<?php
session_start();
include('connection2.php');
$row=mysql_query("select * from project_name");
$array= array();
while($output = mysql_fetch_assoc($row)){
//now you have row with name, and you need to display each name with new tr:
?>
<tr>
<td><div align="right"><input type="checkbox" name="project[]" value="<?php echo $output['name'] ?>"></div></td>
<td><?php echo $output['name'] ?></td>
</tr>
<?php } // closing while loop
?>
Note that I added value to checkbox - if you want to do something with checked project, you have to know which one it was.
And remember taht mysql_ functions are depreated! You should use PDO or mysqli_ instead
i tried this and i got: any ways thanks for all your help :)
<?php
include('connection.php');
$r=mysql_query("select distinct name from project_name");
$ProdArray=array();
while ($row = mysql_fetch_object($r)) {
array_push($ProdArray,$row->name);
}
foreach($ProdArray as $p) {
echo "<tr>";
echo "<td><div align='right'>";
echo "<input type='checkbox' name='project[]' value=" .$p. " />";
echo "</div></td>";
echo "<td>$p</td>";
echo "</tr>";
}
?>