I am trying trying to receive data by making a html form and then adding receiving data by $_GET method.Whenever I try to submit my data I get an error saying 'Your file was not found' in my chrome browser,so I tried opening my php page in chrome by typing "localhost/...." in my address bar and there it was displaying 'Database NOT Found'
here is my php code:-
<html>
<?php
$user_name = "root";
$password = "";
$database = "mysql"; //mysql is name of my database
$server = "localhost";
$db_handle = mysql_connect($server , $user_name, $password,"addresbook"); //adressbook is where all the tabels are
$db_found = mysql_select_db($database, $db_handle);
if (!$db_found) {
print "Database Found ";
$x=$_GET['fname'];
$y=$_GET['sname'];
$z=$_GET['address'];
$sql="INSERT INTO addresbook(First_Name,Surname,Address) VALUES('".$x."','".$y."','".$z."')";
mysql_query($sql,$db_handle);
}
else {
print "Database NOT Found ";
}
?>
</html>
here is mt html code:-
<form action="practic.php"method="get">
Firstname:<input type="text" name="fname"><br>
Lastname:<input type="text" name="sname"><br>
Address:<input type="text" name="address"><br>
<input type="submit">
</form>
btw i am using wamp server.Thanks in advance.
$db_found will be true on success, so your condition should be
if ($db_found) { // make DB changes etc.
and switch to MySQLi or PDO and use prepared statements as already mentioned, refer to the manual:
http://php.net/manual/en/mysqli.prepare.php
use mysqli_connect because mysql_connect is now deprecated.
$db_handle = mysqli_connect($server , $user_name, $password,$database);
refer here for connection
http://www.w3schools.com/php/func_mysqli_connect.asp
Hi change your code to use MysqLi or use PDO
<?php
$user_name = "root";
$password = "";
$database = "mysql"; //mysql is name of my database
$server = "localhost";
$con = mysqli_connect($server,$user_name,$password,'adresbook');//adresbook is where all the tabels are
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
exit();
}
// Change database to "adresbook"
$db_found = mysqli_select_db($con,'adresbook');
if ($db_found) {
print "Database Found ";
$x=$_GET['fname'];
$y=$_GET['sname'];
$z=$_GET['address'];
$sql="INSERT INTO addresbook(First_Name,Surname,Address) VALUES('".$x."','".$y."','".$z."')";
mysqli_query($con, $sql);
}
else {
print "Database NOT Found ";
}
?>
Your phpMyadmin should look like this with the database addresbook with a single s.
Related
<?php
$host = 'localhost';
$user = 'root'
$password = '';
$db ='members';
$connection = mysqli_connect("localhost", "user", "password") or die("Unable to connect to the server!");
mysqli_select_db("members", $connection) or die("Couldn't connect to the database!");
I have installed xampp and create database named "members". I tried to connect it to phpmyadmin but didn't work. I try to google all the answers since three days but in vain. Please help me...
<?php
$host = 'localhost';
$user = 'root';
$password = '';
$db ='members';
$connection = mysqli_connect($host,$user,$password,$db);// you can select db separately as you did already
if($connection){
// do all your stuff that you want
}else{
echo "db connection error because of".mysqli_connect_error();
}
Are your credentials for username and password correct?
By default, the localhost has username = root and password as blank.
Also, what's the issue? Is it showing "Unable to connect to the server!"?
You are missing a semicolon after $user = 'root' and you are using a mixture of mysql_ and mysqli_. Also, you could select a table by passing a fourth argument to mysqli_connect()
$host = 'localhost';
$user = 'root';
$password = '';
$db ='members';
$connection = mysqli_connect($host,$user,$password,$db);// you can select db separately as you did already
if($connection){ echo "Connected Successfully";}else{ echo "Error connecting: . mysqli_connect_error()"; }
Use mysqli_ to do queries:
mysqli_query($connection, "INSERT INTO user_login (uname,upassword,email) VALUES ('$uname','$upassword','$email')");
I recommend you to use prepared statements to avoid SQL injection.
So the above query would look like:
$stmt->prepare("INSERT INTO user_login (uname,upassword,email) VALUES (?,?,?)");
$stmt->bind_param('sss', $uname, $upassword, $email);
$stmt->execute();
I have been developing a CRUD application using PHP & MySQL database.
I was succeeded by creating, displaying, updation parts. But I stuck at the deletion part of a row from a database table.
I tried my best solving all the PHP shown errors but now in final it is now showing a message which I wrote to echo in case of failure.
I request someone to please help me with this problem.
Thankyou in advance.
Code I wrote for deletion:
//include database connection
include 'db_connect.php';
//$mysqli->real_escape_string() function helps us prevent attacks such as SQL injection
$query = "DELETE
FROM `grocery`
WHERE `GrocerID` ='".$mysqli->real_escape_string($_GET['id'])."'
limit 0,1";
//execute query
if( $mysqli->query($query) ){
//if successful deletion
echo "User was deleted.";
}else{
//if there's a database problem
echo "Database Error: Unable to delete record.";
}
$mysqli->close();
?>
Code I wrote for delete link in display table:
//just preparing the delete link to delete the record
echo "<a href='delete.php?id={$GrocerID}'>Delete</a>";
Code I wrote for db config:
<?php
//set connection variables
$host = "localhost";
$username = "root";
$password = "secret";
$db_name = "crud"; //database name
//connect to mysql server
$mysqli = new mysqli($host, $username, $password, $db_name);
//check if any connection error was encountered
if(mysqli_connect_errno()) {
die("Connection failed: " . $conn->connect_error);
exit;
}
?>
I tried this and got working, can you update the code and see if this works?
$host = "localhost";
$username = "root";
$password = "secret";
$db_name = "crud"; //database name
//connect to mysql server
$mysqli = new mysqli($host, $username, $password, $db_name);
//check if any connection error was encountered
if(mysqli_connect_errno()) {
die("Connection failed: " . $conn->connect_error);
exit;
}
// Delete row
if ($mysqli->query (sprintf( "DELETE FROM grocery WHERE email = '".$mysqli->real_escape_string($_GET['id'])."' LIMIT 1") )) {
printf ( "Affected Rows %d rows.\n", $mysqli->affected_rows );
}
I hope this helps.
Provide a connection :
if( $mysqli->query($con, $query) ){
Im trying to create a login for my website and i need to store emails, usernames, passwords, ect in a database i have created already using phpMyAdmin. I have gone through article after article and nothing seems to be working. i have my connect.php like this:
<?
$hostname = "localhost";
$username = "username";
$password = "password";
$databaseName = "_mySiteUserDataBase";
mysql_connect($hostname, $username, $password) or die("Cannot connect to server");
mysql_select_db($databaseName) or die("Cannot select database");
?>
And my main.php like this:
<?
include("connect.php");
$tableName = "myUsers";
$sql = "SELECT * FROM $tableName";
$result = mysql_query($sql);
?>
And i have created a simple form in my html like this:
<html>
<head></head>
<body>
<form>
<input type = "submit" action = "main.php" method = "post" value = "Login">
</form>
</body>
</html>
After submitting the form it says cannot connect to server. I am new to php and mysql and i dont understand what each parameter in the mysql_connect is, and i dont know what they do therefore im not sure what im supposed to enter in but everyone i keep reading about seems to be inputing random values? I could use a brief explanation on that, because i am stuck at connecting and cant even get past this point sadly enough. Also i have been reading that mysql_connect is deprecated and isnt valid anymore but i dont understand what im supposed to use as an alternative. I know its mysqli but thats it and im unclear of the syntax.
mysqli:
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
echo "start<br/>";
try {
$mysqli= new mysqli('localhost', 'myusername', 'mypassword', 'dbname');
if ($mysqli->connect_error) {
die('Connect Error (' . $mysqli->connect_errno . ') '
. $mysqli->connect_error);
}
echo "I am connected and feel happy.<br/>";
$mysqli->close();
} catch (mysqli_sql_exception $e) {
throw $e;
}
?>
If you need to know how to create users, what the heck the hostname is, how to grant access (often useful after the connect :>), just ask.
Try this code in 'connect.php'
<?php
error_reporting(0);
$con=mysql_connect('localhost','root','');// here 'root' is your username and "" is password
if(!$con)
{
echo 'not connect';die;
}
mysql_select_db('dbname',$con);// here 'dbname' is your database name
?>
And also try following code to include sql connection in your other php file(main.php)
<?php
include 'connect.php';
$sql = "SELECT * FROM myUsers";
$result=mysql_query($sql);
?>
Let me convert it to mysqli for you and maybe that will fix the problem. Also, make sure the username, password, and database name are correct.
Try this code. At very least, it will provide a better error message for debugging.
<?
$hostname = "localhost";
$username = "username";
$password = "password";
$databaseName = "_mySiteUserDataBase";
$con = mysqli_connect($hostname, $username, $password, $databaseName) or die(mysqli_error($con));
?>
Main.php
<?
include("connect.php");
$tableName = "myUsers";
$sql = "SELECT * FROM $tableName";
$result = mysqli_query($con,$sql);
?>
I have created one HTML form which takes input from user ,Now I need to search user inputed name in Mysql database and print details related to that user inputed name which is stored in Mysql database.
Below script is creating HTML form to take user input, Saved as "ProcessTracking.html".
<form action="details.php" method="get"/>
<h3 align="center"><FONT color=#CCFF66>ENTER SO NUMBER</h3>
<p align="center">
<input type="text" id="SO_Number" name="SO_Number"/>
</p>
<div style="text-align:center">
<button type="submit" value="SEARCH">
<img alt="ok" src=
"http://www.blueprintcss.org/blueprint/plugins/buttons/icons/tick.png"/>
SEARCH
</button>
</form>
Below PHP script named as "details.php"
<?php
$userinput = $_GET['SO_Number'];
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ProcessTrackingSystem";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_errno) {
printf("Connect failed: %s\n", $conn->connect_error);
exit();
}
$result = mysqli_query($conn, "SELECT * FROM ProcessTrackingSystem.ProcessDetails WHERE SO_Number = '$userinput'") or die(mysqli_error($conn));
$row = mysqli_fetch_assoc($result);
#printf ("SO_Number: %s \n",$row["SO_Number"])
#print_r($row);
printf ("SO_Number:");
printf($row["SO_Number"]);
printf ('--||--');
printf ("Name:");
printf($row["Name"]);
printf ('--||--');
$conn->close();
?>
Firstlly you are not using the $_GET['SO_Number'] parameter in a WHERE of SQL statement. Secondlly you are using both mysql and mysqli which are totaly diffrent and don't work together. For usage see mysqli_fetch_row() and mysqli_query(). Also use print_r($row);.
Here is the corrected code:
<?php
$userinput = $_GET['SO_Number'];
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ProcessTrackingSystem";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_errno) {
printf("Connect failed: %s\n", $conn->connect_error);
exit();
}
$result = mysqli_query($conn, "SELECT * FROM ProcessTrackingSystem.ProcessDetails WHERE SO_Number = '$userinput'") or die(mysqli_error($conn));
$row = mysqli_fetch_row($result);
print_r($row);
$conn->close();
?>
EDIT: Added code example.
You have mixed mysql and mysqli api together. Try using either one.
Note: mysql api is deprectaed as of php 5.5.0
1st Error
As saty says it is because of the syntax error you have in this line
print_r"$row";
which should be as print_r($row)
2nd Error
You're mixing mysql & mysqli
I am not sure about the table that you have.
Recommendation :
I would recommend you to turn on the error_reporting if not those errors will be in your errors_log file
Also for debugging your sql, you can first construct your sql query, run in the phpmyadmin or related tools for your query, then fire the query and make this done.
Note :
If you are using these code in online then the error_log will be in the directory where you execute this page. (But it may change according to your hosting)
If you are running in local machine the error log may locate according to the server you use...
You can find by printing the php's configuration by phpinfo and find for
error_log
May this thing will fix your issue
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ProcessTrackingSystem";
$so = $_POST['SO_Number'];
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$stmt = $conn->prepare("SELECT * FROM ProcessDetails WHERE SO_Number=$so");
$stmt->execute();
$result = $stmt->get_result();
$row = $result->fetch_assoc();
print_r($row[SO_Number]);
$conn->close();
?>
I'm getting the error: No database selected
Here is the code:
<?php
$host = "localhost";
$user = "root";
$password = "";
$database_name = "Student";
mysql_connect($host, $user, $password);
mysql_select_db($database_name);
?>
This code is for php-connect.php class and for the form:
<!DOCTYPE html>
<html>
<body>
<?php
//load database connection
include("php-connect.php");
$id = "1";
$name = "Elena";
$city = "Lahore";
//Command to insert into table
$query = "INSERT INTO studentdata (id,name,city) VALUES ('$id','$name','$city')";
//run the query to insert the person.
$result = mysql_query($query) OR die(mysql_error());
//let them know the person has been added.
echo "Data successfully inserted into the database table ... ";
?>
</body>
</html>
This is the code for the form.. I've tried a lot of things to fix this error but it does not work. Is there any problem with my database?
Try this...
<?php
$host = "localhost";
$user = "root";
$password = "";
$database_name = "Student";
$mLink = mysql_connect($host, $user, $password) or die(mysql_error());
mysql_select_db($database_name , $mLink);
another file
<?php
require 'php-connect.php';
//...... etc..
but, read this first:
This extension is deprecated as of PHP 5.5.0, and will be removed in
the future. Instead, the MySQLi or PDO_MySQL extension should be used.
See also MySQL: choosing an API guide and related FAQ for more
information. Alternatives to this function include:
mysqli_select_db()
PDO::__construct() (part of dsn)
http://br.php.net/mysql_select_db