So, I am trying to design a php website, and so far it works well in terms of adding an entry to the list table.
The problem is, it isnt able to update the table using edit.php. When the edit link is clicked it shows a message:
"There is no data to be edited."
But if I try to manually put localhost/edit.php**?id=1** it shows the id numbered list and works fine. Please help.
home.php
<html>
<head>
<title>My first PHP Website</title>
</head>
<?php
session_start(); //starts the session
if($_SESSION['user']){ // checks if the user is logged in
}
else{
header("location: index.php"); // redirects if user is not logged in
}
$user = $_SESSION['user']; //assigns user value
?>
<body>
<h2>Home Page</h2>
<hello>!
<!--Display's user name-->
Click here to go logout<br/><br/>
<form action="add.php" method="POST">
Add more to list: <input type="text" name="details" /> <br/>
Public post? <input type="checkbox" name="public[]" value="yes" /> <br/>
<input type="submit" value="Add to list"/>
</form>
<h2 align="center">My list</h2>
<table border="1px" width="100%">
<tr>
<th>Id</th>
<th>Details</th>
<th>Post Time</th>
<th>Edit Time</th>
<th>Edit</th>
<th>Delete</th>
<th>Public Post</th>
</tr>
<?php
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("first_db") or die("Cannot connect to database");
$query = mysql_query("select * from list");
while($row = mysql_fetch_array($query))
{
print "<tr>";
print '<td align="center">'. $row['id'] . "</td>";
print '<td align="center">'. $row['details'] . "</td>";
print '<td align="center">'. $row['date_posted'] . " - " . $row['time_posted'] . "</td>";
print '<td align="center">'. $row['date_edited'] . " - " . $row['time_edited'] . "</td>";
print '<td align="center">edit</td>';
print '<td align="center">delete</td>';
print '<td align="center">'. $row['public'] . "</td>";
print "</tr>";
}
?>
</table>
</body>
</html>
)
edit.php
<html>
<head>
<title>My first PHP website</title>
</head>
<?php
session_start(); //starts the session
if($_SESSION['user']){ //checks if user is logged in
}
else{
header("location:index.php"); // redirects if user is not logged in
}
$user = $_SESSION['user']; //assigns user value
$id_exists = false;
?>
<body>
<h2>Home Page</h2>
<p>Hello <?php Print "$user"?>!</p> <!--Displays user's name-->
Click here to logout<br/><br/>
Return to Home page
<h2 align="center">Currently Selected</h2>
<table border="1px" width="100%">
<tr>
<th>Id</th>
<th>Details</th>
<th>Post Time</th>
<th>Edit Time</th>
<th>Public Post</th>
</tr>
<?php
if(!empty($_GET['id']))
{
$id = $_GET['id'];
$_SESSION['id'] = $id;
$id_exists = true;
mysql_connect("localhost", "root","") or die(mysql_error()); //Connect to server
mysql_select_db("first_db") or die("Cannot connect to database"); //connect to database
$query = mysql_query("Select * from list Where id='$id'"); // SQL Query
$count = mysql_num_rows($query);
if($count > 0)
{
while($row = mysql_fetch_array($query))
{
Print "<tr>";
Print '<td align="center">'. $row['id'] . "</td>";
Print '<td align="center">'. $row['details'] . "</td>";
Print '<td align="center">'. $row['date_posted']. " - ". $row['time_posted']."</td>";
Print '<td align="center">'. $row['date_edited']. " - ". $row['time_edited']. "</td>";
Print '<td align="center">'. $row['public']. "</td>";
Print "</tr>";
}
}
else
{
$id_exists = false;
}
}
?>
</table>
<br/>
<?php
if($id_exists)
{
Print '
<form action="edit.php" method="POST">
Enter new detail: <input type="text" name="details"/><br/>
public post? <input type="checkbox" name="public[]" value="yes"/><br/>
<input type="submit" value="Update List"/>
</form>
';
}
else
{
Print '<h2 align="center">There is no data to be edited.</h2>';
}
?>
</body>
</html>
<?php
if($_SERVER['REQUEST_METHOD'] == "POST")
{
mysql_connect("localhost", "root","") or die(mysql_error()); //Connect to server
mysql_select_db("first_db") or die("Cannot connect to database"); //Connect to database
$details = mysql_real_escape_string($_POST['details']);
$public = "no";
$id = $_SESSION['id'];
$time = strftime("%X");//time
$date = strftime("%B %d, %Y");//date
foreach($_POST['public'] as $list)
{
if($list != null)
{
$public = "yes";
}
}
mysql_query("UPDATE list SET details='$details', public='$public', date_edited='$date', time_edited='$time' WHERE id='$id'") ;
header("location: home.php");
}
?>
and here's the one with ?id=1 in the url
http;//s15,postimg,org/yoabiq0p7/screenshot_21,png (change the commas with fullstops).
You are printing only the edit.php, you need to print the entire edit link.
print '<td align="center">edit</td>';
Replace this line with:
print '<td align="center">edit</td>';
This will solve the problem.
P.S: Be careful, your code is open for SQL Injection! Make sure to use mysql_real_escape_string() in this place:
$id = mysql_real_escape_string($_GET['id']);
If the id is only number, you can do the following too to avoid SQL Injection:
$id = intval($_GET["id"]);
The SQL Injection thing is very serious and you need to filter what comes from outside. I recommend using prepared statement PDO too.
you arent passing the data in the home.php
you are doing normal link to edit.php and you arent passing the data ?id=1
try edit this:
print '<td align="center">edit</td>';
to this:
print '<td align="center">edit</td>';
Related
I have a select box that shows the names of all the users in the database, however, I need, using a "Find Button" on the selected user on the combo box, that the data attached to that user shows up on the table
Table that currently shows the data of all users
<table class="table table-hover">
<thead class="thead-dark"></thead>
<tr>
<th scope="col">Shift ID</th>
<th scope="col">Name</th>
<th scope="col">Origin</th>
<th scope="col">Destination</th>
<th scope="col">Date</th>
</tr>
</thead>
<?php
global $result, $query;
$sql = "SELECT * FROM shifts";
$result = $db->query($sql);
//Fetch Data form database
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>" . $row["shift_id"]. "</td><td>" . $row["name"] . "</td><td>" . $row["origin"] . "</td><td>" . $row["destination"] . "</td><td>" . $row["date"] . "</td><td>"
. $row["password"]. "</td></tr>";
}
echo "</table>";
} else { echo "0 results"; }
?>
</table>
And here's the form that shows the users in the select box
<form name="form1" method="POST" action="">
<select name="getUser">
<?php
$res = mysqli_query($db, "SELECT * FROM shifts");
while ($row = mysqli_fetch_array($res))
{
?>
<option><?php echo $row ["name"]; ?></option>
<?php
}
?>
</select>
<button class="btn-primary rounded">Find</button>
</form>
I'm trying to make it that so when the selected user in the combo box and the find button is pressed, that the data found goes all into the table described above.
I was maybe gonna try to attach a variable to the select box and compare it with the names field on the database.
Something like this
$query = "SELECT * FROM shifts WHERE $name == $nameSelected ";
Thanks.
first echo the user id into the option's value
<option value-"<?echo your id?>"><?php echo $row ["name"]; ?></option>
then when your form submits you get get the value from the $_POST
$userId = $_POST['getUser'];
not you can use the variable to query the database, but you should NEVER put it straight in, you should use PDO prepared statements to prevent injection.
$servername = "localhost";
$username = "username";
$password = "password";
try {
$conn = new PDO("mysql:host=$servername;dbname=myDB", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
echo "Connected successfully";
}
catch(PDOException $e)
{
echo "Connection failed: " . $e->getMessage();
}
//something like this
$query = $conn->prepare("SELECT * FROM shifts WHERE id = :id");
$query->bindParam(':id',$userId,PDO::PARAM_INT);
$query->execute()
return $query->fetchAll();// I realised you wanted to get all the shifts so you don want fetchAll(),
notice that in mysql we only use a single = for our comparison unlike php. Also i've changed name to the unique row in the database, as unless your name field is unique how do you know which use called Dan you want?
If you want to do this without re-loading the whole page you will need to look into using Ajax and passing the value of the option tag via jQuery.
Here are some places to start:
https://www.w3schools.com/php/php_mysql_connect.asp
https://www.w3schools.com/xml/ajax_intro.asp
if you are not comfortable with javascript (AJAX), try on your form
<?php $res = mysqli_query($db, "SELECT * FROM shifts"); ?>
<form name="form1" method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>"
<select name="getUser">
<option value='All'>All</options>
<?php
while ($row = mysqli_fetch_array($res)) { ?>
<option value='$row ["name"]'><?php echo $row ["name"]; ?></option>
<?php } ?>
</select>
<button class="btn-primary rounded">Find</button>
</form>
And in your table
<table class="table table-hover">
<thead class="thead-dark"></thead>
<tr>
<th scope="col">Shift ID</th>
<th scope="col">Name</th>
<th scope="col">Origin</th>
<th scope="col">Destination</th>
<th scope="col">Date</th>
</tr>
</thead>
<?php
global $result, $query;
if ($_POST['getUser'] == 'All'){
$sql = "SELECT * FROM shifts";
} else {
$sql = "SELECT * FROM shifts WHERE name = " . $_POST['getUser'];
}
$result = $db->query($sql);
//Fetch Data form database
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>" . $row["shift_id"]. "</td><td>" . $row["name"] . "</td><td>" . $row["origin"] . "</td><td>" . $row["destination"] . "</td><td>" . $row["date"] . "</td><td>"
. $row["password"]. "</td></tr>";
}
echo "</table>";
} else { echo "0 results"; }
?>
</table>
I built a simple CRUD using PHP but the home file is displaying all the info I have in the database, even the info added by other users. How can I filter this to show only the logged users info?
Here is the home.php file:
<?php
session_start();
if(isset($_SESSION['user'])){
echo "Logado como ". $_SESSION['user'];
}
else {
echo"<script language='javascript' type='text/javascript'>alert('Voce deve estar logado');window.location.href='index.php';</script>";
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Perfil</title>
</head>
<body>
</br>
Sair</br>
<h2 align="center">Lista de contatos</h2>
<table align="center" border="1">
<tr>
<th>Nome</th>
<th>Telefone</th>
<th>Endereco</th>
<th>Editar</th>
<th>Deletar</th>
</tr>
<?php
mysql_connect("localhost", "root","") or die(mysql_error()); //Connect to server
mysql_select_db("forms") or die("Cannot connect to database"); //connect to database
$query = mysql_query("Select * from contatos"); // SQL Query
while($row = mysql_fetch_array($query))
{
Print "<tr>";
Print '<td align="center">'. $row['nome'] . "</td>";
Print '<td align="center">'. $row['telefone'] . "</td>";
Print '<td align="center">'. $row['endereco'] . "</td>";
Print '<td align="center">Editar </td>';
Print '<td align="center">Deletar </td>';
Print "</tr>";
}
?>
</table>
<div align="center">
Adicionar contato</br>
</div>
</body>
</html>
Wherever you are assigning $_SESSION['user'] also assign the user record id to the session.. this way you can add a where clause to your sql and fetch only the desired record..
Quick example:
if(isset($_SESSION['user_id'])) {
$sql = "SELECT * FROM contatos WHERE id = {$_SESSION['user_id']}";
// ....
}
Use id or name to differentiate records
$sql = "SELECT * FROM contatos WHERE nome = ".$_SESSION['user'];
$sql = "SELECT * FROM contatos WHERE id = ".$_SESSION['id'];
You must have an autoincremented primary key which wil be your id.
I finally get reviews to work, which will allow user to post their review about the product and display them in a review page with product details, however I can't get it to work perfectly. When a user posts their review, it will update in the table but only the second post will show up.
The following image is from the test i was running,
Before posting a review
After posting the first review
After posting the second review
As the images display, the first review will never show up, only starting from the second and above,
Here my code for review page updated with full code
<?php
if (!isset($_SESSION)) {session_start();} //start session
if (!isset($_SESSION['client_ID'])) {
//echo "<script>alert('not logged in');</script>";
header("Location: index.html" );
}
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<meta name="keywords" content="Games, Gaming, PS4, PS3, XBOX, Video games">
<meta name="description" content="Games 4 You">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<title>Games 4 You</title>
<link rel="stylesheet" type="text/css" href="Styles/ProductsStyle.css">
<!-- javascript/jQuery -->
<script src="http://code.jquery.com/jquery-1.11.0.min.js"></script>
</head>
<body>
<!--Add the following script at the bottom of the web page (before </body></html>)
Support system using MyLiveChat.com
-->
<script type="text/javascript" async="async" defer="defer" data-cfasync="false" src="https://mylivechat.com/chatinline.aspx?hccid=42206151"></script>
<script>// disable zoom to keep image fit and always in position
document.firstElementChild.style.zoom = "reset";
// the above script will disable zoom in and out
</script>
<script type="text/javascript">
// this will auto change the background image to the following 7 images which are in the root Images/
// this is set to change every five second
// declare list of backgrounds
var images = ['bg-01.jpg', 'bg-02.jpg', 'bg-03.jpg', 'bg-04.jpg', 'bg-05.jpg', 'bg-06.jpg', 'bg-07.jpg'];
// declare function that changes the background
function setRandomBackground() {
// choose random background
var randomBackground = images[Math.floor(Math.random() * images.length)];
// set background with jQuery
$('body').css('background-image', 'url("Images/' + randomBackground + '")');
}
// declare function that sets the initial background, and starts the loop.
function startLoop() {
// Set initial background.
setRandomBackground();
// Tell browser to execute the setRandomBackground every 5 seconds.
setInterval(setRandomBackground, 5 * 1000);
}
// One the page has finished loading, execute the startLoop function
$(document).ready(startLoop);
</script>
<header id="header">
<div class="container">
<center><img src="Images/Title.png" alt="Title"></div>
</center>
</header>
<center>
<nav>
<?php
echo "<p> Welcome ".$_SESSION['client_name']."</p>";
//create connection
$con = new mysqli("localhost", "student", "student", "cib4003_h00233671_at");
if ($con->connect_errno) { //failed
echo "Failed to connect to MySQL: (" . $con->connect_errno . ") " . $con->connect_error;
}?>
<div class="wrapper">
<ul id="category" >
<li>Home</li>
<li>Products</li>
<li>View Cart</li>
<li>About</li>
<li>Settings</li>
<li>Logoff</li>
</ul>
</nav>
</div>
</center>
<main>
<h3>Available Products</h3>
<?php
$product = $_GET["RID"];
$_SESSION["product_name_RID"] = $_GET["RID"];
$sql="SELECT * FROM products,reviews WHERE products.Product_Name = '$product' AND reviews.Product_Name = '$product'";
//$sql="SELECT * FROM reviews WHERE Product_Name = '$product'";
// $sql="SELECT * FROM pizza,pizzacart WHERE pizza.Pizza_ID=pizzacart.Pizza_ID AND pizzacart.client_ID=".$_SESSION['client_ID'];
//echo "connected to DB";
//run SQL query
$result = mysqli_query($con,$sql);
//output result
if(mysqli_num_rows($result)==0) //no records found
{
$sql="SELECT * FROM products WHERE Product_Name = '$product'";
$result = mysqli_query($con,$sql);
// echo "<p>no records in DB".mysqli_num_rows($result)."</p>";
// echo "<p><a href=products.php></a>";
// link has been disable because i am using the <a for something else so i can't force the image to be in the center when using <a
// so the result will only be image that tell the customers no products found click all or search with different data
?>
<table class="table-style-one">
<tr>
<th>Product Image</th>
<th>Product Name</th>
<th>Description</th>
<th>Product Type</th>
<th>Console Type</th>
</tr>
<?php
while($row = mysqli_fetch_array($result)) { //loops through records
echo "<tr>";
echo "<td><img src='".$row['picture']."'/>";
echo "<td>".$row['Product_Name']."</td>";
echo "<td>".$row['Description']." <center><b><br>".$row['Trailer']."<br></b></center></td>";
echo "<td>".$row['Product_Type']."</td>";
echo "<td>".$row['Console_Type']."</td>";
echo "</tr>";
}
//end of loop
echo "</table>";
echo "<p>No Reviews available for this product.<br> To post a review of this product, fill up the below form.</p>";
//end of else
}
else
{
?>
<table class="table-style-one">
<tr>
<th>Product Image</th>
<th>Product Name</th>
<th>Description</th>
<th>Product Type</th>
<th>Console Type</th>
</tr>
<?php
while($row = mysqli_fetch_array($result)) { //loops through records
echo "<tr>";
echo "<td><img src='".$row['picture']."'/>";
echo "<td>".$row['Product_Name']."</td>";
echo "<td>".$row['Description']." <center><b><br>".$row['Trailer']."<br></b></center></td>";
echo "<td>".$row['Product_Type']."</td>";
echo "<td>".$row['Console_Type']."</td>";
echo "</tr>";
echo "<br>";
?>
<?php
while($row = mysqli_fetch_array($result)) {
echo "<table class=table-style-one align=center>";
// echo "<tr><th>Review ID</th><td>".$row['Review_ID']."</td></tr>";
echo "<tr><th>Review By:</th><td>".$_SESSION['client_name']."</td></tr>";
echo "<tr><th>Review Title</th><td>".$row['Review_Title']."</td></tr>";
echo "<tr><th>Rate:</th><td>".$row['Review_Rate']."/5</td></tr>";
echo "<tr><th>Review</th><td colspan=2>".$row['Review']."</td></tr>";
echo "<tr><th>Submitted On</th><td>".$row['Review_Date']."</td></tr>";
echo "<br>";
?>
<?php
}
//end of loop
echo "</table>";
//end of else
}
}
?>
<table class="table-style-one" align="center">
<tr>
<form method="POST" action="submitreview.php">
<!--<th>Product Name:</th><td> <input type="text" size="30" id="ReviewTitle" name="ReviewTitle" pattern=".{5,}" required title="5 characters minimum" placeholder="Review Title"></td> -->
<th>Review Title:</th><td> <input type="text" required size="30" id="ReviewTitle" name="ReviewTitle" pattern=".{5,}" required title="5 characters minimum" placeholder="Review Title"> </td>
<th>Rate: </th>
<td>
<select name="Review_Rate" required>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
</select>
</td>
<tr>
<td colspan="4">
<textarea name="WriteReview" id="WriteReview" required rows="10" cols="50" wrap="physical" placeholder="Write your Review here" style="margin: 0px; width: 437px; height: 150px;"></textarea>
</td>
</tr>
<td align="center" colspan="2"><input type="submit" value="submit"></td>
<td colspan="2"><input type="Reset"></td>
</form> </tr>
</table>
</h2>
<br>
<br>
<br>
</main>
</body>
<footer>
<p>Made by Humaid Al Ali - H00233671</p>
<div id="google_translate_element"></div><script type="text/javascript">
function googleTranslateElementInit() {
new google.translate.TranslateElement({pageLanguage: 'en', layout: google.translate.TranslateElement.InlineLayout.HORIZONTAL, multilanguagePage: true}, 'google_translate_element');
}
</script><script type="text/javascript" src="//translate.google.com/translate_a/element.js?cb=googleTranslateElementInit"></script>
</footer>
</html>
and here the php file where I insert the review into the table
<?php
if (!isset($_SESSION)) {session_start();} //start session
if (!isset($_SESSION['client_ID'])) {
//echo "<script>alert('not logged in');</script>";
header("Location: index.html" );
}
?>
<?php
//new connection
$con = new mysqli("localhost", "student", "student", "cib4003_h00233671_at");
if ($con->connect_errno) { //failed
echo "Failed to connect to MySQL: (" . $con->connect_errno . ") " . $con->connect_error;
}
//success
//if ($_SERVER['REQUEST_METHOD'] === 'POST') {
// run sql
$sql ="INSERT INTO `cib4003_h00233671_at`.`reviews`(`Review_ID`, `Product_Name`, `client_ID`, `Review_Title`, `Review_Rate`, `Review`) VALUES (NULL, '".$_SESSION['product_name_RID']."', '".$_SESSION['client_ID']."', '".$_POST["ReviewTitle"]."', '".$_POST['Review_Rate']."', '".$_POST['WriteReview']."');";
if ($con->query($sql) === TRUE) {echo "<h3> New record created successfully</h3>";
header('Location: '. $_SERVER['HTTP_REFERER'] );
} else {
echo "Error : " . $sql . "<br>" . $con->error;
}
$con->close();
?>
Problem:
As the images display, the first review will never show up, only starting from the second and above
That's because you're fetching the first row which includes the first review but you don't display it, rather you starting fetching next reviews and display them. Look at the two conditions in while block inside the else block.
Solution:
To display reviews, you should do something like this:
<?php
// your code
if(mysqli_num_rows($result)==0){
// your code
}else{
echo "<table class='table-style-one'>";
echo "<tr>";
echo "<th>Product Image</th>";
echo "<th>Product Name</th>";
echo "<th>Description</th>";
echo "<th>Product Type</th>";
echo "<th>Console Type</th>";
echo "</tr>";
// fetch first row, which also contains first review
$row = mysqli_fetch_array($result);
echo "<tr>";
echo "<td><img src='" . $row['picture'] . "'/>";
echo "<td>" . $row['Product_Name'] . "</td>";
echo "<td>" . $row['Description'] . "<center><b><br>" . $row['Trailer'] . "<br></b></center></td>";
echo "<td>" . $row['Product_Type'] . "</td>";
echo "<td>" . $row['Console_Type'] . "</td>";
echo "</tr>";
echo "</table>";
echo "<table class=table-style-one align=center>";
echo "<tr>";
echo "<th>Review ID</th>";
echo "<th>Review By:</th>";
echo "<th>Review Title</th>";
echo "<th>Rate:</th>";
echo "<th>Review</th>";
echo "<th>Submitted On</th>";
echo "</tr>";
do{
// display first review and then fetch rest of the reviews
echo "<tr>";
echo "<td>" . $row['Review_ID'] . "</td>";
echo "<td>" . $_SESSION['client_name'] . "</td>";
echo "<td>" . $row['Review_Title'] . "</td>";
echo "<td>" . $row['Review_Rate'] . "/5</td>";
echo "<td colspan=2>" . $row['Review'] . "</td>";
echo "<td>" . $row['Review_Date'] . "</td>";
echo "</tr>";
}while($row = mysqli_fetch_array($result));
echo "</table>";
?>
</table>
<?php
}
// your code
?>
The problem looks to me to be in your SQL query. It looks as though your SQL query should actually be doing a join on the related column to bring together data across two different tables
$sql="SELECT *
FROM products AS p,
JOIN reviews AS r
ON p.Product_Name = r.Product_Name
WHERE products.Product_Name = '$product'";
But I think the easiest solution would be to run another query within the table output loop where you get the reviews for the product name you are currently outputting and loop over those.
I have an HTML table being displayed from php data and I'm trying to move data from one table to another based on what is clicked. I cannot get it to move from table to another but I"m not getting any code error.
<html>
<head>
<title>View Requests</title>
</head>
<body>
<div class="breadcrumbs">
<center>
Home · Requests
</center>
</div>
<?php
require_once '../../scripts/app_config.php';
require_once '../../scripts/database_connection.php';
// get results from database
$result = mysql_query("SELECT * FROM temp")
or die(mysql_error());
echo "<table border='1' cellpadding='10'>";
echo "<tr> <th>ID</th> <th>First Name</th> <th>Last Name</th> <th>Username</th> <th></th> <th></th></tr>";
// loop through results of database query, displaying them in the table
while($row = mysql_fetch_array( $result )) {
// echo out the contents of each row into a table
echo "<tr>";
echo '<td>' . $row['User_id'] . '</td>';
echo '<td>' . $row['First_name'] . '</td>';
echo '<td>' . $row['Last_name'] . '</td>';
echo '<td>' . $row['Username'] . '</td>';
echo '<td>Approve</td>';
echo '<td>Delete</td>';
echo "</tr>";
}
// close table>
echo "</table>";
?>
</body>
</html>
PHP EDIT SCRIPT:
<html>
<head>
<title>View Requests</title>
</head>
<body>
<div class="breadcrumbs">
<center>
Home · Requests · All Users
</center>
<?php
require_once '../../scripts/app_config.php';
require_once '../../scripts/database_connection.php';
$id = $_GET['id'];
$sql = ("INSERT INTO users
SELECT * FROM temp WHERE User_id = $id ");
mysql_query($sql)
or die(mysql_error());
?>
</body>
</html>
I needed to use
$id = $_get['id];
and in the insert statement i had to replace id with User_id
<html>
<head>
<title>View Requests</title>
</head>
<body>
<div class="breadcrumbs">
<center>
Home · Requests · All Users
</center>
<?php
require_once '../../scripts/app_config.php';
require_once '../../scripts/database_connection.php';
$id = $_GET['id'];
$sql = ("INSERT INTO users
SELECT * FROM temp WHERE User_id = $id ");
mysql_query($sql)
or die(mysql_error());
?>
</body>
</html>
here is my code. actually i am displaying some data from mysql on the page and creating dynamic link.i want started a session with session_start() in the very begining of code before starting any code. i want to store the value of the link that is to be display on other pagepage..
page1.php
<a style="color:#F00; font-family:Arial, Helvetica, sans-serif; margin-left:33px; font-weight:bold">
No. of registered students:
</a>
<table border='1' align="center" style="font-size:14px" width="95%" cellspacing="3" class="db_table">
<tr class="db_table_tr" >
<th class="db_table_th" name="submit">USN</th>
</tr>
<?php
include('includes/login_connection.php');
$query = "select p.usn, p.name from personal_details p, course_codes c where p.usn = c.usn order by p.usn";
$run = mysql_query($query) or die($query."<br/><br/>".mysql_error());
$num = mysql_numrows($run);
echo $num;
while($row = mysql_fetch_assoc($run)){
echo "<tr>";
echo "<td>" . $row['usn'] . "" . "</td>";
echo "<td>" . $row['name'] . " </td>";
if(isset($_GET['submit'])){
$_SESSION['session_usn'] = $_GET['usn'];
}
}
echo "</tr>";
mysql_close($bd);
?>
</table>
page2.php
<?php
session_start();
if(isset($_SESSION['session_usn']))
{
$_POST['usn'] = $_SESSION['session_usn'];
echo $_POST['usn'];
}
?>
You need to provide a fall-back, in case the URL provided does not contain the proper variables in the $_GET section.
You have:
if(isset($_GET['submit'])){
$_SESSION['session_usn'] = $_GET['usn'];
}
You should do something else if $_GET['submit'] isn't set:
if(isset($_GET['submit'])){
$_SESSION['session_usn'] = $_GET['usn'];
} else {
$_SESSION['session_usn'] = "unset";
// or set a warning flag like "unset"
}
You should be feeding your php file a url like:
http://yoururl.com/page1.php?usn='333'
Where 333 is the value you want to store.