I want to open a count in a php file in order to get the first register
<?php
include_once('dbConfig.php');
session_start();
$id_persona= $_SESSION["id"];
$query = "SELECT COUNT( * ) AS alias1 FROM `desayuno_ingreso` WHERE `desayuno_ingreso`.`id` =88757532";
echo $query;
$resultset = mysqli_query($conn,$query);
if(mysqli_num_rows($resultset) > 0){
$k=0;
while ($fila = mysqli_fetch_assoc($resultset)) {
$lat[$k]= $fila['alias1'];
$k=$k+1;
}
}
echo $lat[0];
?>
I tried in every single way I know (Every way worked for me before). However, it doesn't working right now.
I'm no sure what I'm doing wrong.
EIDTED
This is the dbConfig.php file:
<?php
$dbHost = 'localhost';
$dbUser = 'root';
$dbPass = 'mypassword';
$dbName = 'app';
$conn = mysqli_connect($dbHost,$dbUser,$dbPass,$dbName);
if(!$conn){
die("Database connection failed: " . mysqli_connect_error());
}
?>
Related
I'am setting up a new wampserver and created a new database with a table named 'users'
apache 2.4.23 & PHP 7.1.10 & mySQL 5.7.14
<?php
$server = 'localhost';
$serverUsername = 'root';
$serverPassword = '';
$dbName = 'test';
$connection = mysqli_connect($server,$serverUsername,$serverPassword);
msqli_select_db($dbName);
if(!$connection){
echo 'connection failed to database '.mysqli_connect_error();
}
$sql = "SELECT * FROM users";
$query = mysqli_query($sql);
while($row = mysqli_fetch_array($query)){
print_r($row);
}
?>
my value is really in th data base but nothing appears in the page after running code
Have a look at the comments mentioning the fix
$server = 'localhost';
$serverUsername = 'root';
$serverPassword = '';
$dbName = 'test';
// FIX 1
// You need to mention the database name as the last argument
$connection = mysqli_connect($server,$serverUsername,$serverPassword, $dbName);
if(!$connection){
echo 'connection failed to database '.mysqli_connect_error();
}
$sql = "SELECT * FROM users";
// FIX 2
// The first argument should be your mysqli connection
$query = mysqli_query($connection, $sql);
// Check for errors
if (!$query) {
printf("Error: %s\n", mysqli_error($connection));
exit();
}
while($row = mysqli_fetch_array($query)){
print_r($row);
}
?>
Reference for mysqli_connect: https://secure.php.net/manual/en/function.mysqli-connect.php
Reference for mysqli_query: https://secure.php.net/manual/en/mysqli.query.php
All i want to do is to store the name of the guest that is on the A1 seat in a PHP variable named "resulted".
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'root';
$dbname = 'test';
$con=mysqli_connect($dbhost,$dbuser,$dbpass,$dbname);
if ($con->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$resulted = mysqli_query("SELECT name FROM guests WHERE seat='A1');
echo $resulted;
?>
I know this is totally wrong but I don't know how i shoud do it....
There is some issue in your code. I have fetched associative array amd print gust name.
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'root';
$dbname = 'test';
$con=mysqli_connect($dbhost,$dbuser,$dbpass,$dbname);
if ($con->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$resulted = mysqli_query($conn, "SELECT name FROM guest WHERE seat='A1'");
if(mysqli_num_rows($resulted) > 0) {
$row = mysqli_fetch_assoc($resulted);
print_r($row['name']);
}
// Free result set
mysqli_free_result($resulted);
mysqli_close($con);
?>
$resulted is a mysqli resource. You need to fetch a row from this resource like this:
$row = mysqli_fetch_assoc($resulted);
echo $row['name'];
For more information, visit http://www.w3schools.com/php/func_mysqli_fetch_assoc.asp.
I am trying to connect my html page with MySQL database to show data from one specific table to my page, but I always get an error actually it just goes to die part of an SQL code. I am really new to PHP programming so please can someone help me, what am I doing wrong?
Here is my code:
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>AJDE</title>
</head>
<?php
$servername = "localhost";
$username = "******";
$password = "*****";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$query = "select * from PERCENTILE";
$result = mysqli_query($conn,$query);
if(!$result) {
die ("Umro!");
}
/* close connection */
$mysqli->close();
?>
<body>
</body>
</html>
Thank you!
Do you want to show the table as a HTML table or just an array?
The following is what I did to display my table as a HTML table:
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '*****';
$dbname = 'dbname';
$selectedTable = 'whateverTableYouWant';
$conn = mysqli_connect($dbhost, $dbuser, $dbpass,$dbname);
if (!$conn){
die('cannot connect to mysql');
}
$query = "SELECT * FROM $selectedTable";
if ($result = mysqli_query($conn , $query)) {
echo("<div class = 'data_wrapper'>");
// Display Header of the table
$fieldcount=mysqli_num_fields($result); //value = number of columns
$row = mysqli_fetch_assoc($result); //Fetch a result row as an associative array:
//array to string conversion
echo("<table id='example' class='table table-striped table-bordered' cellspacing='0' width='100%'>");
echo("<thead> <tr>");
foreach($row as $item){
echo "<th>" .$item. "</th>";
}
echo("</tr> </thead>");
//Footer
echo("<tfoot> <tr>");
foreach($row as $item){
echo "<th>" .$item. "</th>";
}
echo("</tr> </tfoot>");
//Display Data within the table
echo("<tbody>");
while ($row = mysqli_fetch_assoc($result)){
echo "<tr>";
foreach ($row as $item){
echo "<td contenteditable = 'true'>" . $item . "</td>"; //Change contenteditable later
//Editable data should be constricted, int = numbers only, string = words, date = date
}
echo "</tr>";
}
echo("<tbody>");
echo "</table>";
echo("</div>");
}
The following is just displaying as an array:
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '*****';
$dbname = 'dbname';
$selectedTable = 'whateverTableYouWant';
$conn = mysqli_connect($dbhost, $dbuser, $dbpass,$dbname);
if (!$conn){
die('cannot connect to mysql');
}
$query = "SELECT * FROM $selectedTable";
if ($result = mysqli_query($conn , $query)) {
while ($row = mysqli_fetch_array($result)){
print_r($row);
}
}
Please change the connection string like mentioned below.
<?php
$con = mysqli_connect("localhost","username","password","dbname");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
Please let me know if you've any queries.
I Think you need to select a database where you will run the query:
Try with this code:
<?php
$username = "your_name";
$password = "your_password";
$hostname = "localhost";
//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
echo "Connected to MySQL<br>";
//select a database to work with (so replace the database name examples)
$selected = mysql_select_db("examples",$dbhandle)
or die("Could not select examples");
//execute the SQL query and return records
$result = mysql_query("SELECT * FROM PERCENTILE");
//fetch the data from the database and display results
//replace id,name,year with the columns you have on your table PERCENTILE and you want to show
while ($row = mysql_fetch_array($result)) {
echo "ID:".$row{'id'}." Name:".$row{'name'}."Year: ". $row{'year'}."<br>";
}
//close the connection
mysql_close($dbhandle);
?>
Hope it helps,
Vince.
How do I choose a tbl_uploads table in the database?
PHPMYADMİN IMG
<?php
$dbhost = "";
$dbuser = "";
$dbpass = "";
$dbname = "";
mysql_connect($dbhost,$dbuser,$dbpass) or die('cannot connect to the server');
mysql_select_db($dbname) or die('database selection problem');
?>
You don't need to choose any table just fire the Query on the database in which the table exists, and you're done. For example,
<?php
$dbhost = "";
$dbuser = "";
$dbpass = "";
$dbname = "";
mysql_connect($dbhost,$dbuser,$dbpass) or die('cannot connect to the server');
mysql_select_db($dbname) or die('database selection problem');
$query = "SELECT * FROM tbl_uploads;"
//executing the query and printing it's results
$results= mysql_query($query);
print_r($results);
?>
This will print the results of query in the variable $results.
Note :
This extension was deprecated in PHP 5.5.0, and it was removed in PHP 7.0.0. 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_query()
PDO::query()
First, connect to your database (keinotis_iletisim). For example:
$dbhost = "localhost"; // or any other address
$dbuser = ""; // the user you created for the database in phpmyadmin
$dbpass = ""; // the password you created for the database in phpmyadmin
$dbname = "keinotis_iletisim";
Then, in a query you can mention a table, for example:
SELECT * FROM tbl_uploads;
Also see this link on W3Schools.
You can select your table using sql statement like this
$sql = "SELECT * FROM MyGuests";
$result = mysql_query($conn, $sql);
Now you can fetch the all records by using
while($row=mysql_fetch_array($result))
{
echo $row['column_name'];
}
This one will work for you
<?php
//the connction
$hostname_localhost = "localhost";
$database_localhost = "keinotis_iletisim";
$username_localhost = "root";
$password_localhost = "";
$localhost = mysql_pconnect($hostname_localhost, $username_localhost, $password_localhost) or trigger_error(mysql_error(),E_USER_ERROR);
?>
<?php
mysql_select_db($database_localhost, $localhost);
$query_record = "SELECT * FROM tbl_uploads";
$record = mysql_query($query_record, $localhost) or die(mysql_error());
$row_record = mysql_fetch_assoc($record);
$totalRows_record = mysql_num_rows($record);
//echo $row_record['tbl_uploads_table_colum'];
?>
<!--And if you want to display the list-->
<?php do { ?>
<?php echo $row_record['tbl_uploads_table_colum']; ?>
<?php } while ($row_record = mysql_fetch_assoc($record)); ?>
I was trying to make a backup of my MySQL db called "backup" using a PHP script below, but for some reason, it doesnt work. Any ideas what is wrong? I wanted to create a file called test.sql in the same folder that would contain the data (because the db is quite big I only selected values of Temp>35, but I could change that later). Right now when I run it, I get the echo, but no file created.
<?php
$dbhost = '...';
$dbuser = '...';
$dbpass = '...';
$dbname = 'backup';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbname);
$tableName = 'backup';
$backupFile = 'test.sql';
$query = "SELECT * WHERE Temp>35 INTO OUTFILE '$backupFile' FROM $tableName";
$result = mysql_query($query);
echo "Backed up";
?>
Just Try.
$query = "SELECT * INTO OUTFILE '$backupFile' FROM $tableName WHERE Temp>35";
$result = mysql_query($query) or die(mysql_error());
Try this and let me know:
<?php
$dbhost = '...';
$dbuser = '...';
$dbpass = '...';
$dbname = 'backup';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbname);
$databasename = 'backup';
$backupFile = 'test.sql';
$query = "SELECT * INTO OUTFILE '$backupFile' FROM $databasename WHERE Temp>35";
$result = mysql_query($query);
echo "Backed up";
?>
<?php
$source_db='source_db';
$target_db='target_db';
$server='127.0.0.1';
$user='root';
$password='';
mysql_connect($server,$user,$password);
mysql_select_db($source_db);
// Get names of all tables in source database
$result=mysql_query("show tables");
while($row=mysql_fetch_array($result)){
$name=$row[0];
$this_result=mysql_query("show create table $name");
$this_row=mysql_fetch_array($this_result);
$tables[]=array('name'=>$name,'query'=>$this_row[1]);
}
// Connect target database to create and populate tables
mysql_select_db($target_db);
$total=count($tables);
for($i=0;$i < $total;$i++){
$name=$tables[$i]['name'];
$q=$tables[$i]['query'];
mysql_query($q);
mysql_query("insert into $name select * from $source_db.$name");
}
?>