I have a such select:
$sql = "SELECT milestone_id, project_id, sum(estimated_hours) as value_sum
FROM project_has_tasks GROUP BY milestone_id";
Getting a multiple results.
Is this possible to create a query to INSERT these results to multiple rows in DB?
Thanks to myself :) Is so simple as ....
<?php
$servername = "localhost";
$username = "db";
$password = "password";
$dbname = "db";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO `burndown_snap` (`project_id`,`milestone_id`,`estimated_sum`)
SELECT `project_id`, `milestone_id`, sum(estimated_hours) as value_sum FROM project_has_tasks WHERE status !='done' GROUP BY milestone_id";
$result = $conn->query($sql);
$conn->close();
?>
Related
This question already has answers here:
How to include a PHP variable inside a MySQL statement
(5 answers)
Closed 2 years ago.
I want to select an email from the DB which needs to return the related Account_ID
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "connected accounts details";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT acc_id FROM users
WHERE
email = $user_email";
$result = $conn->query($sql);
echo "$result";
$conn->close();
This code returns nothing, just blank.
While using the same connection/db data is being inserted in db successfully.
Try this code
// Php Sql Select From DB....
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "connected accounts details";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT `acc_id` FROM `users` WHERE `email` = ?";
$result->$conn->prepare($sql);
$result->bind_param('s',$user_email);
$result->execute();
$data = $result->get_result();
// Fetch all
$data->fetch_all(MYSQLI_ASSOC);
print_r($data);
$mysqli -> close();
// #link http://php.net/manual/en/mysqli-result.fetch-all.php
$sql = "SELECT acc_id FROM users
WHERE
email LIKE '%$user_email%' ";
$result = mysqli_query($conn, $sql);
$row = $result->fetch_assoc();
$id = $row['acc_id'];
echo "$id";
$conn->close();
Got the wanted Record, trying this way.
Thanks everyone answering on this post.
I am trying to select data from a database. I do have a successful connection, but it seems like the query doesn't work even though I know for sure that the query is right. What am I doing wrong?
If I execute the code below, the result I get is: "Connected successfullyBad query". The 'Bad query' should mean that the query is wrong, but I checked it and it isn't wrong...
<?php
$servername = "localhost";
$username = "root";
$password = "usbw";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$sql ="SELECT * FROM `producten`";
$result = mysqli_query($conn, $sql) or die("Bad query");
$conn->close();
?>
I expect to only see "connected successfully"
You are missing your database name. You can do it two ways, or in the connect statement:
$conn = new mysqli($servername, $username, $password,$database);
Or you can do it in your select statement:
$sql ="SELECT * FROM `yourdatabase`.`producten`";
If you don´t set your database your query is wrong
Your query is right just write your database name in mysqli constructor.
$conn = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);
Visit: https://www.php.net/manual/en/mysqli.construct.php
Please give database name also, check below code.
<?php
$servername = "localhost";
$username = "root";
$password = "usbw";
$dbname = ""; //Enter database name
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$sql ="SELECT * FROM `producten`";
$result = mysqli_query($conn, $sql) or die("Bad query");
$conn->close();
?>
I want to fetch time column from mysql database and to compare with current time.
It should execute only if it matches with the current time.
Please tell me a MySql query or php code to do this.
I am new to php.
Try This:
<?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);
}
$today = date("h:i a")
$sql = "SELECT * FROM Table_name WHERE SUBSTRING_INDEX(time, ' ', 2) = '".$today."'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// execute you code
} else {
}
$conn->close();
?>
Try this: Updated Code
<?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);
}
$today = date("h:i a")
$sql = "SELECT * FROM Table_name WHERE time = $today";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// execute you code
} else {
}
$conn->close();
?>
This program is meant to delete a record when given the id.
php:
if ($_GET['type']=="file"){
$servername = "localhost";
$username = "****";
$password = "****";
$dbname = "****";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (mysqli_connect_error($conn)) {
die("Connection failed: " . mysqli_connect_error($conn));
}
$sql = "SELECT id,user, FROM CreationsAndFiles WHERE id =".$_GET['id']." LIMIT 1";
$result = mysqli_query($conn,$sql);
$row = mysqli_fetch_assoc($result);
if ($row['user'] == $login_session){
$sql = "DELETE FROM CreationsAndFiles WHERE id=".$_GET['id'];
if(mysqli_query($conn, $sql)){echo "deleted";}
}
mysqli_close($conn);
//header("location: index.php?page=CreationsAndFiles");
}
the header is type=file&id=9
there is a record where id=9
It for no apparent reason will not work.
Your SQL syntax is wrong;
SELECT id,user, FROM CreationsAndFiles...
^ extra comma
should be simply
SELECT id,user FROM CreationsAndFiles...
You may want to sanitize your input though, for example simply entering type=file&id=id will most likely do bad things.
I have one number in a mysql database
This is my code:
$servername = "xxxxx";
$username = "xxxxxx";
$password = "xxxxxx";
$dbname = "xxxxxx";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql1 = "SELECT date FROM test";
$result1 = $conn->query($sql1);
$res1 = mysqli_fetch_assoc($result1);
echo $res1['date'];
I want to load that number into variable and count with that variable. After that i will store in same the db and table on the same place
$numvar = $res1['date'];
There yah go.