Stop Inserting Blank data in Refreshing browser - php

I'm new in PHP. some one please help to stop inserting blank data when refreshing the browser. I'm trying to insert data in SQL database from a HTML form & showing them in the same page in a table.
here is my from & PHP code
Thanks in advance....
</head>
<body>
<?php
$sub = mysql_connect ("localhost", "root","");
if (!$sub)
{
die('Could Not Connect:'.mysql_erro());
}
mysql_select_db("test", $sub);
$sql="INSERT INTO te(Dat,Sadi,Jam,Washi) VALUES('$_POST[Dat]','$_POST[Sadi]','$_POST[Jam]','$_POST[Washi]')";
if (!mysql_query($sql, $sub))
{
die('Error:'.mysql_error());
}
echo 'One Record added';
$result=mysql_query("SELECT * FROM te");
echo "<table border='1'>;
<tr>
<th>Date</th>
<th>Sadi</th>
<th>Jam</th>
<th>Washi</th>
</tr>";
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" .$row['Dat']."</td>";
echo "<td>" .$row['Sadi']."</td>";
echo "<td>" .$row['Jam']."</td>";
echo "<td>" .$row['Washi']."</td>";
echo "</tr>";
}
echo "</table>";
Mysql_close($sub);
?>
</body>
<form action="index.php" method="post">
<input type="text" name="Dat" placeholder="Date, DD-MM-YY">
<input type="text" name="Sadi" placeholder="Sadi">
<input type="text" name="Jam" placeholder="Jam">
<input type="text" name="Washi" placeholder="Washi">
<input type="submit" name="sub">
</form>

Just verify if the $_POST is not empty
if(!empty($_POST))
{
$sql="INSERT INTO te(Dat,Sadi,Jam,Washi) VALUES('$_POST[Dat]','$_POST[Sadi]','$_POST[Jam]','$_POST[Washi]')";
if (!mysql_query($sql, $sub))
{
die('Error:'.mysql_error());
}
}

Related

how to display auto increment data form mysql in php

In my university database project I have an auto increment field roll number in my SQL. What I want is that when new admission take place and a student record is inserted it displays all finds on same page including roll number. However despite my best efforts all it returns 0 in roll number.
Here is the code:
<?php
$con = mysqli_connect("localhost", "root", "") or die("conection error");
mysqli_select_db($con, "hamdard university") or die("dbase error");
if (isset($_POST['subbtn'])) {
$r = "SELECT RollNo FROM admission_form";
$result = mysqli_query($con, $r);
if (mysqli_query($con, $r)) {
$last_id = mysqli_insert_id($con);
}
$n = $_POST['txtname'];
$f = $_POST['txtfac'];
$s = $_POST['txtsem'];
$sql = "insert into admission_form(name,faculty,semester)values ('$n','$f','$s')";
mysqli_query($con, $sql);
echo "<table border=1>
<th>RollNo</th>
<th>Name</th>
<th>Faculty</th>
<th>Semester</th>";
echo "<tr>";
echo "<td>";
echo $last_id;
echo "</td>";
echo "<td>";
echo $n;
echo "</td>";
echo "<td>";
echo $f;
echo "</td>";
echo "<td>";
echo $s;
echo "</td>";
echo "<br>";
}
?>
<html>
<head></head>
<body>
<form name="f1" action="" method="POST">
RollNo:
<input type="text" name="txtroll" readonly> Name:
<input type="text" name="txtname"> Faculty:
<input type="text" name="txtfac"> Semester:
<input type="text" name="txtsem">
<input type="submit" value="done" name="subbtn">
</form>
</body>
</html>
You need get $last_id after INSERT query
$sql="insert into admission_form(name,faculty,semester)values ...
mysqli_query($con,$sql);
$last_id = mysqli_insert_id($con);
You have misplaced insert query. Change it as:
<?php
$con=mysqli_connect("localhost","root","")or die("conection error");
mysqli_select_db($con,"hamdard university")or die("dbase error");
if(isset($_POST['subbtn']))
{
$sql="insert into admission_form(name,faculty,semester)values ('$n','$f','$s')";
if (mysqli_query($con, $sql))
{
$last_id = mysqli_insert_id($con);
}
$n=$_POST['txtname'];
$f=$_POST['txtfac'];
$s=$_POST['txtsem'];
$r="SELECT RollNo FROM admission_form";
$result=mysqli_query($con, $r);
echo "<table border=1>
<th>RollNo</th>
<th>Name</th>
<th>Faculty</th>
<th>Semester</th>";
echo "<tr>";
echo "<td>";
echo $last_id;
echo "</td>";
echo "<td>";
echo $n;
echo "</td>";
echo "<td>";
echo $f;
echo "</td>";
echo "<td>";
echo $s;
echo "</td>";
echo "<br>";
}
?>
<html>
<head></head>
<body>
<form name="f1" action="" method="POST">
RollNo:<input type="text" name="txtroll" readonly>
Name:<input type="text" name="txtname">
Faculty:<input type="text" name="txtfac">
Semester:<input type="text" name="txtsem">
<input type="submit" value="done" name="subbtn">
</form>
</body>
</html>
You have some problems in this page.
As stated before, the INSER query must be before the select.
If you do "SELECT RollNo FROM admission_form" you will get ALL existing RollNo in admission_form and not the lastest one. If this is an int with auto increment (that seams to be) you should do "SELECT max(RollNo) FROM admission_form".
You don't need to do the above select, as mysqli_insert_id get the ID inserted in the last query.
You might not see the result in your table because it is on HTML head, but it should be in the body of the page.
The table you create is not "closed".
You should be good to go with the code bellow.
<HTML>
<HEAD></HEAD>
<BODY>
<?php
$con=mysqli_connect("localhost","root","")or die("conection error");
mysqli_select_db($con,"hamdard university")or die("dbase error");
if(isset($_POST['subbtn']))
{
$n=$_POST['txtname'];
$f=$_POST['txtfac'];
$s=$_POST['txtsem'];
$sql="insert into admission_form(name,faculty,semester) values ('$n','$f','$s')";
if (mysqli_query($con, $r))
{
$last_id = mysqli_insert_id($con);
}
echo "<table border=\"1\"><th>RollNo</th> <th>Name</th> <th>Faculty</th> <th>Semester</th>";
echo "<tr>";
echo "<td>";
echo $last_id;
echo "</td>";
echo "<td>";
echo $n;
echo "</td>";
echo "<td>";
echo $f;
echo "</td>";
echo "<td>";
echo $s;
echo "</td>";
echo "</tr></table>";
}
?>
<form name="f1" action="" method="POST">
RollNo:<input type="text" name="txtroll" readonly>
Name:<input type="text" name="txtname">
Faculty:<input type="text" name="txtfac">
Semester:<input type="text" name="txtsem">
<input type="submit" value="done" name="subbtn">
</form>
</body>
</html>
Try this:
$mysqli = new mysqli(SQLI_SERVER, MYSQLI_USER, MYSQLI_PWD, MYSQLI_DBNAME);
if ($result = $mysqli->query("INSERT INTO admission_form(name, facility,semester) VALUES..) {
echo 'The ID is: '.$mysqli->insert_id;
}

Edit sql data in php page

I need to edit my database table using submitted data.
This is the form:
mysql_query("set names 'utf8'");
$query = "SELECT * FROM sec1octa";
$result = mysql_query($query);
?>
<div align="center">
<form method="get" action="edit_data.php">
<table width="104" border="1" class="center1">
<tr>
<th width="94">first</th>
<th width="94">second</th>
<th width="94">status</th>
</tr>
<tr>
<?php
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result)) {
?>
<tr>
<td><input type="text" name="id" value="<?php echo $row ['stu_no']; ?> " size=10></td>
<td><input type="text" name="name" value="<?php echo $row ['stu_name']; ?> " size=10></td>
<td><?php
echo '<select name="status">'; {
echo '<option value="open">'.$row['stu_status'].'</option>';
echo '<option value="close">'.prevent.'</option>';
}
echo '</select>';
?></td>
</tr>
<?php
}
}
?>
</tr>
</table>
<input type="submit" name="submit" value="done" />
</form>
The problem is in the edit_data.php page.
I can't UPDATE.
I use this code but it's not working.
require_once('../Connections/config.php');
$id= $_GET['id'];
$status= $_GET['status'];
$query= mysql_query("UPDATE `goh`.`sec1octa` SET `stu_status` = '$status'
WHERE stu_no='".$id."'") or die (mysql_error ());
if($query){echo $status ."done ";}
The reason you are only getting the last values in your edit_data.php $_GET is because you are not setting the input/select names as arrays.
<input type="text" name="id" value="some_stu_no">
is happening over and over and over and every new one overwrites the previous.
Instead, you should use:
<input type="text" name="id[]" value="some_stu_no">
This will allow you to pass multiple id's in a single form submission.
Your form:
<form method="POST" action="edit_data.php">
....
echo "<tr>";
echo "<th>id</th>";
echo "<th>name</th>";
echo "<th>status</th>";
echo "</tr>";
if(mysql_num_rows($result)>0){
while($row=mysql_fetch_array($result)){
echo "<tr>";
echo "<td><input type=\"text\" name=\"id[]\" value=\"{$row['stu_no']}\" size=\"10\"></td>";
echo "<td>{$row['stu_name']}</td>";
echo "<td>";
echo "<select name=\"status[]\">"; // I don't like your option set up here, but I don't fully understand it either.
echo "<option value=\"open\">{$row['stu_status']}</option>";
echo "<option value=\"close\">.prevent.</option>";
echo "</select>";
echo "</td>";
echo "</tr>";
}
}
....
<input type="submit" value="Submit All">
</form>
edit_data.php
// create a mysqli connection called $db
if(isset($_POST['id'])){
$tally=0;
// build all queries for the batch
foreach($_POST['id'] as $index=>$id){
$queries[]="UPDATE `goh`.`sec1octa` SET `stu_status`='".mysqli_real_escape_string($db,$_POST['status'][$index])."' WHERE `stu_no`='".mysqli_real_escape_string($db,$id)."'";
}
// run all queries
if(mysqli_multi_query($db,implode(';',$queries)){
do{
$tally+=mysqli_affected_rows($db);
} while(mysqli_more_results($db) && mysqli_next_result($db));
}
// assess the outcome
if($error_mess=mysqli_error($db)){
echo "Syntax Error: $error_mess";
}else{
echo "$tally row",($tally!=1?"s":"")," updated";
}
mysqli_close($con);
}

How do I create an html form with multiple checkboxes

This is my code for creating an html form that reads from a database and will allow the user to check and uncheck boxes for each of the 640 items. This is the form.php:
// execute query
$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error());
// see if any rows were returned
if (mysql_num_rows($result) > 0) {
// yes
// print them one after another
echo "<html><body> <table cellpadding=10 border=1>";
while($row = mysql_fetch_assoc($result)) {
echo "<tr>";
echo "<td>".$row['stickerID']."</td>";
echo "<td>" .$row['stickerName']."</td>";
echo "<td>".$row['stickerSection']."</td>";
echo "<td>"?>
<form name="some form" action="editform.php" method="post">
<input type="checkbox" name="<?php echo $row['stickerID'] ?>" value=" <?php echo $row['stickerStatus'] ?> ">
<?php "</td>";
echo "</tr>";
}
echo "</table></body></html>";
echo " " ?>
<input type="submit" name="editWish" value="Edit">
</form>
<?php " ";
} else {
// no
// print status message
echo "No rows found!";
}
The user must then be able to click on submit and have those values updated in the mysql database.
Right now when I click the submit button, it posts to edit form.php which has this:
<?php
//echo results
foreach($_POST['stickerID'] as $k=>$v ){
echo $k;
echo $v;
}
?>
But I don't get anything echoed. I was thinking the problem could be that Im actually creating a form for every row instead of 1 form with many rows/checkboxes. But when I move the form code after the and the tag to the line where line, I can't even load the form.php, it just loads blank.
Where is my problem? :) Thx
Name your checkbox like this:
<input type="checkbox" name="stickerID[]" value=" <?php echo $row['stickerStatus']; ?> ">
And as Amal already said update your code to PDO or MySQLi
you can do this with a tag :-
echo "<td>" .$row['stickerName']."</td>";
echo "<td>".$row['stickerSection']."</td>";
echo "<td>"?>
<form name="some form" action="editform.php" method="post">
<input type="checkbox" name="checkbox[]" value=" <?php echo $row['stickerStatus'] ?> ">
<?php "</td>";
echo "</tr>";
on your php code you get :-
$all_checkes_checkbox = $_POST['checkbox'];
here is your all checked checkbox:-
and this array also bale key and value

how can i Select data from database using input type box and print on same page

THIS IS MY CODE HERE I AM GETTING PROBLEM THIS IS A INPUT TYPE TAG AND NOT SELECTED DATA FROM DATABASE USING PHP.
here is first part of this code is html formatted ad second part is php
i want select data from database to data slect to same page as we see on the sopping cart sites
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>phpSelect</title>
</head>
<body>
Insert Age for search
<form action="#" method="post" >
<input type="text" id="val" name="resValue" />
<input type="submit" value="submit" /></form>
<?php
if(isset($_POST['submit']))
{
$res=$_POST['resValue'];
echo $res;
}
//echo $res;
$con=mysqli_connect("localhost","root","","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Persons where Age=25");
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</body>
</html>
Try changing the SQL to this
SELECT * FROM Persons where Age='".mysql_escape_string($formValue['val'])."'
First issue I found on your code is here:
<input type="submit" value="submit" />
it should be:
<input type="submit" value="submit" name="submit" />
To be able to get the results. Below are the codes:
<?php
$query = "";
if(isset($_POST['submit']))
{
$res=$_POST['resValue'];
$query = " where Age='$res'"
}
//echo $res;
$con=mysqli_connect("localhost","root","","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Persons $query");
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
Change your select query to this one.
SELECT * FROM Persons where Age='".mysql_escape_string($_POST['val'])."'
In your problem the all value of form get by the name of all fields of form.\
so here should be
<input type="submit" name="submit" value="submit"/>
Because in $_POST['submit'] submit is same as button's name.
$result = mysqli_query($con,"SELECT * FROM Persons where Age="25");

Notice: Undefined variable: Lastname in C:\wamp\www\COP3718\tt2^homas12\delete.php on line 15

I am trying to create a database and within the tables I have two options, insert and delete. I have the insert option working but can not get the delete function working for me. (I need to insert an update option once I have this working) I am including my code and am hopeful someone will see my error on this, I have been working on getting this project done for two weeks now. Thank You in advance.
<!DOCTYPE html>
<html>
<body>
<h1>Franchise Call Log</h1>
<?php
$con=mysqli_connect("localhost","tt2^homas12","c3o7P1518","tt2^homas12");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM caller_info");
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Franchise</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['Firstname'] . "</td>";
echo "<td>" . $row['Lastname'] . "</td>";
echo "<td>" . $row['Franchise'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</body>
</html>
<h1>Insert a New Caller</h1>
<form action="insert.php" method="post">
Firstname: <input type="text" name="firstname">
Lastname: <input type="text" name="lastname">
Franchise: <input type="text" name="franchise">
<input type="submit" name="submit">
</form>
</body>
</html>
<html>
<body>
<h1>Delete a Caller</h1>
<form action="delete.php" method="post">
Lastname: <input type="text" name="lastname">
<input type="submit" name="submit">
</form>
</body>
</html>
<!DOCTYPE html>
<html>
<body>
<h1>Record has been updated</h1>
<?php
$con=mysqli_connect("localhost","tt2^homas12","c3o7P1518","tt2^homas12");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql= ("DELETE FROM caller_info WHERE Lastname = '$Lastname'");
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record deleted";
mysqli_close($con);
?>
</body>
</html>
On delete.php you are not getting the data from $_POST
if(isset($_POST['lastname'])){
$Lastname=$_POST['lastname'];
}else{
echo "Error";die();
}
$sql= ("DELETE FROM caller_info WHERE Lastname = '$Lastname'");
Learn about prepared statements and use PDO or MySQLi with bind params - this article will help you decide which. If you choose PDO, here is a good tutorial.

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