I would like to know how can i put the table from mysql in a new file.php.
I want the MySql table to be on the page.
This is my code that inserts data in MySql.
<?php
// Create connection
$con = mysqli_connect("host", "id_", "password", "xxxxxx");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$Task = $_POST['Task'];
$Date = $_POST['Date'];
$Desc = $_POST['Desc'];
$sql = "INSERT INTO tasklist (Task, Date, Description)
VALUES ('$Task', '$Date', '$Desc')";
if (!mysqli_query($con, $sql)) {
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
<html>
<body>
<form action="addtask.php" method="post">
Task: <input type="text" name="Task">
Date: <input type="text" id="datepicker" name="Date">
Decrption:<textarea type="text" name="Desc"></textarea>
<input type="submit" value="submit">
</form>
</body>
</html>
Try this code:
<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Persons");
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);
?>
also u can try w3schools sample code :
Display the Result in an HTML Table
The following example selects the same data as the example above, but will display the data in an HTML table:
<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Persons");
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);
?>
The output of the code above will be:
first code comes from "Jonnny" in this article
Related
When I run the code in localhost server it works properly but when we run the same code in our live server (Godaddy linux shared server) it gives sql connection failed error.
The database is quite big and in our local server the query takes approximately 2 minutes to return data, so in shared godaddy server it must take more time.
<?php
$conn = mysqli_connect("localhost","root","","doreme_eshop");
// $mysqli = new mysqli("localhost", "root", "", "doreme_eshop");
// if ($mysqli->connect_errno) {
// echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
// }
if(isset($_POST['submit'])){
$style=$_REQUEST['style_no'];
$result = mysqli_query($conn,"CALL style_report('".$style."')") or die("Query Failed: " . mysqli_error($conn));
//var_dump($result->fetch_assoc());
echo "<table border='1'>
<tr>
<th>total_set</th>
<th>style_no</th>
<th>size_id</th>
<th>size_description</th>
<th>Company_name</th>
<th>generate_no</th>
<th>order_Date</th>
<th>product_id</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['total_set'] . "</td>";
echo "<td>" . $row['style_no'] . "</td>";
echo "<td>" . $row['size_id'] . "</td>";
echo "<td>" . $row['size_description'] . "</td>";
echo "<td>" . $row['Company_name'] . "</td>";
echo "<td>" . $row['generate_no'] . "</td>";
echo "<td>" . $row['order_Date'] . "</td>";
echo "<td>" . $row['product_id'] . "</td>";
echo "</tr>";
}
echo "</table>";
}
?>
<html>
<head></head>
<body>
<form role="form" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" enctype="multipart/form-data">
<select name="style_no" >
<option value="">STYLE NO</option>
<?php
$style_result = mysqli_query($conn,"CALL get_all_style_no()") or die("Query Failed: " . mysqli_error($conn));
$row1 = mysqli_num_rows($style_result);
while ($row1 = mysqli_fetch_array($style_result)){
echo "<option value='". $row1['style_no'] ."'>" .$row1['style_no'] ."</option>" ;
}
?>
</select>
<input type="submit" value="Submit" name="submit"/>
</form>
</body>
</html>
You need to set time out on connection, secondly u can set the config file to be separated and require_once.
Try check if the DB has mysqli library,... just open connection only then close and check error
I had inserted the data from html and it is stored in mysql database, and the data is retrieved from database to html,now I had done how to delete the data that are displayed in html,my task is how to update the displayed.And my deleted code is:
Below is my HTML code:
<html>
<head>
<title>STUDENT_DATA</title>
</head>
<body>
<form action="tab.php" method="post" >
<center>
sname: <input type="text" name="sname" required><br></br>
sno:<input type="text" name="sno"><br></br>
marks:<input type="text" name="marks"><br></br>
class:<input type="text" name="class"><br></br>
phno:<input type="text" name="phno" onkeypress='return event.charCode >
= 48 && event.charCode <= 57'><br></br>
DOB:<input type="date" placeholder="DD-MM-YYYY"
required pattern="(0[1-9]|1[0-9]|2[0-9]|3[01]).(0[1-9]|1[012]).[0-9]{4}"
name="DOB"/><br></br>
<button>submit</button></br>
</center>
</form>
</body>
</html>
Below is my PHP code for displaying the data:
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<?php
$connection = mysql_connect('localhost', 'root','');
if (!$connection)
{
die("Database Connection Failed" . mysql_error());
}
$select_db = mysql_select_db( "student",$connection);
if (!$select_db)
{
die("Database Selection Failed" . mysql_error());
}
$sql = "SELECT * FROM hello1 ";
$result = mysql_query($sql) or die(mysql_error());
?>
<table border="2" style= " margin: 0 auto;" id="myTable">
<thead>
<tr>
<th>sname</th>
<th>sno</th>
<th>marks</th>
<th>class</th>
<th>phno</th>
<th>DOB</th>
</tr>
</thead>
<tbody>
<?php
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['sname'] . "</td>";
echo "<td>" . $row['sno'] . "</td>";
echo "<td>" . $row['marks'] . "</td>";
echo "<td>" . $row['class'] . "</td>";
echo "<td>" . $row['phno'] . "</td>";
echo "<td>" . $row['DOB'] . "</td>";
echo "<td><a href='delete.php?did=".$row['sname']."'>Delete</a></td>";
echo "</tr>";
}
?>
</tbody>
</table>
</body>
</html>
Below is my PHP code for deleting:
<?php
$connection = mysql_connect('localhost', 'root','');
if (!$connection)
{
die("Database Connection Failed" . mysql_error());
}
$select_db = mysql_select_db( "student",$connection);
if (!$select_db)
{
die("Database Selection Failed" . mysql_error());
}
?>
<?php
if(isset($_GET['did'])) {
$delete_id = $_GET['did'];
$sql = mysql_query("DELETE FROM hello1 WHERE sname = '".$delete_id."'");
if($sql) {
echo "<br/><br/><span>deleted successfully...!!</span>";
}
else
{
echo "ERROR";
}
}
?>
Below is what I have tried in update.php
<?php
$connection = mysql_connect('localhost', 'root','');
if (!$connection) { die("Database Connection Failed" . mysql_error());
}
$select_db = mysql_select_db( "emp",$connection);
if (!$select_db) { die("Database Selection Failed" . mysql_error()); } ?>
<?php if(isset($_GET['did']))
{
$update_id = $_GET['did']; $sql = mysql_query("UPDATE FROM venu WHERE id = '".$update_id."'");
if($sql) {
echo "<br/><br/><span>updated successfully...!!</span>";
} else {
echo "ERROR";
}
} ?>
I am new to PHP and I made a simple program where you can apply your name and age, it will take the data to the database and the table will be added with a new row.
I want to add a new column where you can click "change", only the data from that particular row will show up in a few textboxes and can be changed. when pressing submit I want to use the UPDATE function to update the records.
example/plot:
Mike Towards 23 Change
Tyler Frankenstein 24
Change Sophie Baker 22
Change
I want to change the age of Sophie Baker to 24 so I press Change on that row.
Now I only want to get the data from that row and make some changes.
The code I have this far:
Drawing the table above the input fields and the input:
$result = mysqli_query($con,"SELECT * FROM Persons");
echo "<table border='2'> <tr> <th>Voornaam</th> <th>Achternaam</th> <th>Leeftijd</th></tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "<td>" . $row['Age'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
<html>
<body>
<br />
<form action="insert.php" method="post"><br />
<input type="text" name="firstname"> Firstname <br />
<input type="text" name="lastname"> Lastname <br />
<input type="text" name="age"> Age
<p><input type="submit"></p>
</form>
</body>
</html>
Parser:
<?php
$con = mysqli_connect("localhost", "user" , "", "personInfo");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="INSERT INTO Persons (FirstName, LastName, Age) VALUES ('$_POST[firstname]','$_POST[lastname]','$_POST[age]')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added to the database";
echo "<p><a href=sql2.php>Back to form</a></p>";
mysqli_close($con);
?>
I have tried a few things, but I cant figure out how to show the content on the row I want to select.
Change the actual data with the update function won't be the problem, so I only need help to get the actual data from the correct row.
you'd need to select with the primary key of that table if any exists. if not you should create one. I assume you have a primary key named PersonID:
$query = "SELECT * FROM Persons WHERE PersonID = '" . ($_GET['PersonID']) . "'";
to add the edit button:
echo "<table border='2'> <tr> <th>Voornaam</th> <th>Achternaam</th> <th>Leeftijd</th><th>Action</th></tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "<td>" . $row['Age'] . "</td>";
echo "<td><a href = '?PersonID=" . $row['PersonID'] . "'>Edit</a></td>";
echo "</tr>";
}
echo "</table>";
I assume you have a column named "id".
you can do the following:
<?php
$con = mysqli_connect("localhost", "user" , "", "personInfo");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// when you are in "edit mode" just display the row you will edit row
if (isset($_GET['id'])
$result = mysqli_query($con,"SELECT * FROM Persons where id = ".(int)$_GET['id']);
else
$result = mysqli_query($con,"SELECT * FROM Persons");
echo "<table border='2'> <tr> <th>Voornaam</th> <th>Achternaam</th> <th>Leeftijd</th></tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "<td>" . $row['Age'] . "</td>";
echo "<td><a href='?id=" . $row['id'] . "'>change</a></td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
<html>
<body>
<br />
<form action="update.php" method="post"><br />
<input type="hidden" name="id" value="<?php echo isset($_GET['id']?$_GET['id']:'') ?>" />
<input type="text" name="firstname" value="<?php echo isset($row['FirstName'])?$row['FirstName']:'' ?>"/> Firstname <br />
<input type="text" name="lastname" value="<?php echo isset($row['LastName'])?$row['LastName']:'' ?>"/> Lastname <br />
<input type="text" name="age" value="<?php echo isset($row['Age'])?$row['Age']:'' ?>"/> Age
<p><input type="submit"></p>
</form>
</body>
</html>
update.php (handle both insertion and update):
<?php
$con = mysqli_connect("localhost", "user" , "", "personInfo");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if (isset($_POST['id'])
$sql="UPDATE Persons set FirstName = ?, LastName = ?, Age = ?
WHERE id = ".(int)$_POST['id'];
else
$sql="INSERT INTO Persons (FirstName, LastName, Age) VALUES (?, ?, ?)";
$sth = mysqli_prepare($con, $sql);
$sth->bind_param($_POST[firstname],$_POST[lastname],$_POST[age]);
if (!$sth->execute())
{
die('Error: ' . mysqli_error($con));
}
echo "1 record ".(isset($_POST['id']?'modified':'added')." to the database";
echo "<p><a href=sql2.php>Back to form</a></p>";
I am trying to get a form to input data into my "mysql database" however i am getting an error message and also it is inputting a blank data everytime the page loads.
Here is my code:
<form action="insert.php" method="post">
Name: <input type="text" name="name">
<input type="submit" value="Submit">
</form>
<?php
// This is the connection to my database
$con = mysql_connect('127.0.0.1', 'shane', 'diamond89');
if (!$con){
die('Could not Connect: ' . mysql_error());
}
// This creates my table layout
echo "<table border='1'>
<tr>
<th>ID</th>
<th>Name</th>
<th>Delete</th>
</tr>";
// This selects which database i want to connect to
$selected = mysql_select_db("shane",$con);
if (!$con){
die("Could not select examples");
}
// This inserts new information to the Database
$query = "INSERT INTO test1 VALUES('id', '$name')";
$result = mysql_query($query);
if ($result){
echo("Input data is Successful");
}else{
echo("Input data failed");
}
// This chooses which results i want to select from
$result = mysql_query("SELECT `id`, `name` FROM `test1` WHERE 1");
// This outputs the information into my table
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . "[D]" . "</td>";
echo "</tr>";
}
echo "</table>";
// This closes my connection
mysql_close($con);
?>
Here is the error message:
( ! ) SCREAM: Error suppression ignored for
( ! ) Notice: Undefined variable: name in C:\wamp\www\sql_table.php on line 36
Call Stack
Time Memory Function Location
1 0.0006 250360 {main}( ) ..\sql_table.php:0
You are trying to access to the POST data, so you should do something like that :
EDIT: be careful about the data you put into your database. You should use a modern database API, or, at least, escape your data (cf bellow code)
<form action="insert.php" method="post">
Name: <input type="text" name="name">
<input type="submit" value="Submit">
</form>
<?php
// Following code will be called if you submit your form
if (!empty($_POST['name'])) :
// This is the connection to my database
$con = mysql_connect('127.0.0.1', 'shane', 'diamond89');
if (!$con){
die('Could not Connect: ' . mysql_error());
}
// This creates my table layout
echo "<table border='1'>
<tr>
<th>ID</th>
<th>Name</th>
<th>Delete</th>
</tr>";
// This selects which database i want to connect to
$selected = mysql_select_db("shane",$con);
if (!$con){
die("Could not select examples");
}
// This inserts new information to the Database
$query = "INSERT INTO test1 VALUES('id', \'".mysql_real_escape_string($_POST['name'])."\')";
$result = mysql_query($query);
if ($result){
echo("Input data is Successful");
}else{
echo("Input data failed");
}
// This chooses which results i want to select from
$result = mysql_query("SELECT `id`, `name` FROM `test1` WHERE 1");
// This outputs the information into my table
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . "[D]" . "</td>";
echo "</tr>";
}
echo "</table>";
// This closes my connection
mysql_close($con);
endif;
?>
I am just unable to understand why the rows are not getting deleted!
please note that i am getting the login values of corrected check boxes in the php page.
from my point of view, most probably error should be in php page where i am using
'DELETE FROM' query.
<?php
session_start();
?>
<html>
<head>
<form id="delete_customers" action="deletecustomers.php" method="post">
<?php
$con = mysql_connect("localhost","root","");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("car_rental_system", $con);
$result = mysql_query("SELECT * FROM customer");
echo "<table border='1' align='center'>
<tr>
<th>first_name</th>
<th>Last_name</th>
<th>login</th>
</tr>";
while($row = mysql_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['first_name'] . "</td>";
echo "<td>" . $row['login'] . "</td>";
echo "<td>"."<input type='checkbox' name='deletingcustomers[]'
value=$row['login']}"."</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>
<p class='submit'>
<button type='submit' name='dscustomer'>Delete selected</button>
</p>
</head>
</html>
//NOW deletecustomers.php
<?php
session_start();
$_SESSION['deletingcustomers'] = $_POST['deletingcustomers'];
$N = count($_SESSION['deletingcustomers']);
$con = mysql_connect("localhost","root","");
if (!$con) die('Could not connect: ' . mysql_error());
mysql_select_db("car_rental_system", $con);
if(empty($_SESSION['deletingcustomers'])) {
echo("No customers selected");
} else {
for ($i=0; $i<$N; $i++) {
$sql1="delete from `customer`
where login='{$_SESSION[deletingcustomers][$i]}'";
if(mysql_query($sql1,$con))
echo 'executed';
}
}
?>
NO! No! Why do people keep using mysql_query().....(head desk)
Please look up PDO. http://php.net/manual/en/book.pdo.php it helps prevent sql injections and gives you a better understanding of how to harness oop's power.
Your $_SESSION[deletingcustomers][$i] needs to be $_SESSION['deletingcustomers'][$i]
Example on its way
$tempVar = $_SESSION['deletingcustomers'][$i];
$dbConnection = new PDO("mysql:host=".$hostName.";dbname=".$dbName, $username, $password);
$sql = "delete from `customer` where login='$tempVar'";
$stmt = $newObj->prepare($sql);
$stmt->execute();
Replace
echo "<td>"."<input type='checkbox' name='deletingcustomers[]' value=$row['login']}"."</td>";
To
echo "<td><input type='checkbox' name='deletingcustomers[]' value='".$row['login']."'</td>";
and try