I'm having issues getting this data from within a codeigniter Controller.
$q = $this->db->get('offers_orders');
$this->db->select('total');
$this->db->where('order_number', $orderid);
$orderdata = $q->result_array();
$orderamount = $orderdata[0]['total'];
Do you see anything wrong with this code ?.
Yes, Try :
$this->db->select('count(*) as total', false);
$this->db->where('order_number', $orderid);
$q = $this->db->get('offers_orders');
OR,
$q = $this->db->select('count(*) as total', false)->where('order_number', $orderid)->get('offers_orders');
You need first to defined your select and where method then you call the get() method which are used for get data.
Like this:
$this->db->select('total');
$this->db->where('order_number', $orderid);
$q = $this->db->get('offers_orders');
And as I see from your query, you need to fetch only one result and for it better solution is to use row() function because this function returns a single result row.
Like this:
$orderdata = $q->row();
$orderamount = $orderdata->total;
Also to learn more about it you can read this acticle.
<!DOCTYPE html>
<html>
<body>
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<br> id: ". $row["id"]. " - Name: ". $row["firstname"]. " " . $row["lastname"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
</body>
</html>
Related
I'm trying to echo the query made to my database. So far the code runs but nothing happens. I'm very new at coding so help or an explanation about what I'm doing wrong would be awesome! Thanks a lot!!!
<?php
$servername = "servername";
// REPLACE with your Database name
$dbname = "dbsname";
// REPLACE with Database user
$username = "username";
// REPLACE with Database user password
$password = "password";
$ID = 1;
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
echo "Error de conexion.";
}
$sql = "SELECT * FROM registro WHERE ID = '1' ";
$result = $conn->query($sql);
while ($data = $result->fetch_assoc()){
$sensor_data[] = $data;
}
echo $data;
$conn->close();
Thanks everyone who replied, #AbraCadaver was right, I can't echo an array so I used print_r() instead and printed the variable $data. Now the php file prints the array with the query results. Thank you all!
$sql = "SELECT * FROM registro WHERE ID = $ID";
while ($data = $result->fetch_assoc()){
$sensor_data[] = $data;
print_r($data);
}
I am trying to figure out a way to have one PHP page to display all of my blog post but have the URL decide what post is requested from that database. Something kind of like this: localhost/bolg/posts.php?pid=1 In my database I have it set up to where each post has an ID associated with it. So what I want is something that put the pid=1 and put it in the MySQL code. Here is the PHP code of the post.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, title, content, date FROM posts where id =3";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<h1> ". $row["title"]. "</h1>". $row["content"]. "" . $row["date"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
Assuming you enter example.com?pid=10 in the browser address bar, you can capture that variable pid using the $_GET (docs) array which PHP automatically fills for you when a page is called with a querystring.
Using your existing code as a start point you can
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";
if (isset($_GET['pid'])) {
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
$sql = "SELECT id, title, content, date FROM posts where id = ?";
$stmt = $conn->prepare($sql);
$stmt->bind_param('i', $_GET['pid']);
$stmt->execute();
$result = $stmt->get_result();
if ($result->num_rows > 0) {
// output data of each row
// while looop is not necessary, you are only returning one row
$row = $result->fetch_assoc();
echo "<h1> ". $row["title"]. "</h1>". $row["content"]. "" . $row["date"] . "<br>";
}
$conn->close();
} else {
echo "0 results";
}
Notice I took the liberty of amending your database access code to use prepared and parameterised query and binding the values to avoid SQL Injection Attack. You should always use this technique in the future
I got this work for me, but I'm sure there's a better way to get this done. But, I've searched many hours without finding the exact answer to what I'm looking to do. Basically getting the variable usrID from the URL, I need to search MySQL for the corresponding information to this user. Later I want to use the different fields on my page (better website) to personalize the experience.
<?php
$servername = "localhost";
$username = "authorized-user";
$password = "secret";
$dbname = "agentDB";
$usrID = "001";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM agentInfo WHERE usrID = '$usrID'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$Lname = $row["Lname"];
$Fname = $row["Fname"];
$tl = $row["tl"];
}
}
mysqli_close($conn);
?>
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Load MySQL Data into Corresponding PHP Variables</title>
</head>
<body>
here is the body<br>
My name is: <?php echo $Fname; ?> <?php echo $Lname; ?><?php echo $tl; ?>
</body>
</html>
You could create a variable to store a full name and then "tl" on it like this:
$user_info = $Lname . ", " . $Fname . ": " . $tl;
Then:
<?php echo $user_info; ?>
Wherever you need that information.
If you want to minimize the amount of variables being assigned you could wrap it in a function and return the desired data field:
function fetchUserData(userData) {
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM agentInfo WHERE usrID = '$usrID'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$userData = $row[userData];
}
}
return $userData;
}
mysqli_close($conn);
You can the get the specified data like this:
<?php echo fetchUserData("Fname"); ?>
MySQL database Start = "2017-03-29 01:30:00"
The problem is:
I print the $SQL and search the record in MySQL database, it can search the record. However, I need to debug and test if there is no result, it will be expected to echo "No". But, it always to echo "Yes" no matter I can get the record in MySQL or not.
How can I fixed it.
Main purpose: Get the record if there are and Echo "No" if there don't have record
<?php
$serverName = "localhost";
$username = "root";
$password = "";
$dbName = "fyp";
$tbName = "events";
$String_start = '2017-03-29 01:28:00';
$String_end = '2017-03-29 01:32:00';
$conn = new mysqli($serverName, $username, $password, $dbName);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//$staffID = $_SESSION['userID'];
$staff_ID = '15207800';
$sql = "SELECT *
FROM `$tbName`
WHERE `start` BETWEEN ('$String_start') AND ('$String_end')";
echo $sql;
$result=mysqli_query($conn,$sql);
if($result)
{
echo "Yes";
}
else{
echo "no";
}
?>
Mysql query only return fails if there is an issue, for successful execution it returns TRUE even if there is no record. You can achieve with follwing way
<?php
$serverName = "localhost";
$username = "root";
$password = "";
$dbName = "fyp";
$tbName = "events";
$String_start = '2017-03-29 01:28:00';
$String_end = '2017-03-29 01:32:00';
$conn = new mysqli($serverName, $username, $password, $dbName);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//$staffID = $_SESSION['userID'];
$staff_ID = '15207800';
$sql = "SELECT *
FROM `$tbName`
WHERE `start` BETWEEN ('$String_start') AND ('$String_end')";
echo $sql;
$result=mysqli_query($conn,$sql);
$num_rows = mysqli_num_rows($result);
if($num_rows > 0)
{
echo "Yes";
}
else{
echo "no";
}
?>
Use mysqli_num_rows that will return total results returned by the query. If total > 0 then results found else no results.
To fetch rows from MySQL result set, use mysqli_fetch_assoc
$result = mysqli_query($conn,$sql);
// $result contains result set and will return `TRUE` if query was successfully executed
// and will return `FALSE` only in case of error
$total = mysqli_num_rows($result);
if($total > 0)
{
echo "Yes";
// Fetch rows from mysql result set
while($row=mysqli_fetch_assoc($result))
{
print_r($row);
}
}
else
{
echo "no";
}
This first field is where a web visitor will enter in the 'cardname' hit submit and be directed to another page (dashboard2.php) where only his or her content will appear.
Enter your cardname to access your content<br>
<form action='dashboard2.php'>
<input type='text' name='cardname'/><input type='submit' value='retrieve card'/>
</form>
</body>
The page below is the page that is directed after the user enters in the 'cardname' from the first input field. However, I only want this second page to show the information based on the cardname that was entered. Right now, it shows every single cardname, questionone, answerone from that table.
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "flashcards";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT cardname, questionone, answerone FROM cards";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<br> ". $row["cardname"]. " ". $row["questionone"]. " " . $row["answerone"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
You have to modify the query to accept a WHERE clause. For instance, WHERE cardname = mysqli_real_escape_string($conn, $_GET['cardname']) (The default method for any form is GET unless you specify method="post".).
You should learn about prepared statements for MySQLi and perhaps consider using PDO, it's really not hard.
It seems that you want to perform a search and not a display all the records.
Usually a search returns records that match a certain field, unless a specific ID or unique value was entered in the search. I'm not sure this is the case.
I put this together a little quick but hopefully it helps...
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "flashcards";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// escape the string to avoid SQL injections
$searchEscaped = $conn->real_escape_string($_POST['cardname']);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT cardname, questionone, answerone FROM cards WHERE cardname = '$searchEscaped' ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
if($result->num_rows == 1){
// only one result found, show just that
$row = $result->fetch_assoc()
echo "<br> ". $row["cardname"]. " ". $row["questionone"]. " " . $row["answerone"] . "<br>";
}else{
// multiple rows found, show them all
while($row = $result->fetch_assoc()) {
echo "<br> ". $row["cardname"]. " ". $row["questionone"]. " " . $row["answerone"] . "<br>";
}
}
} else {
echo "0 results";
}
$conn->close();
?>