How to split a multi value sql column into three single result - php

So I have a SQL column that stores data like this.
"[1338,0,8523]"
I was wondering if I can then garb each one individually and because the value is minutes, if I could times it by 60 to get the hours and then display it, I've used the below for my other results but I'm stuck on this one.
<?php
$servername = "localhost";
$username = "Time";
$password = "fakepassword";
$dbname = "time";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed to Timebase Highscores: " . $conn->connect_error);
}
$sql = "SELECT name, time FROM players ORDER BY time DESC LIMIT 5";
$result = $conn->query($sql);
echo "Top 5 Life Wasters". "<br>";
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo $row["name"]. " - $" . $row["time"] . "<br>";
}
} else {
echo "Error fetching any players from database.";
}
$conn->close();
?>

Related

want to fetch data from two tables in mysql

hello i want to display data from two tables
meaning
this table i need from it amount and created
and the user_id
i want to get the username
from other table name is users
withdraw table
users table
<?php
$servername = "localhost";
$username = "aa";
$password = "aat&IpnFt";
$dbname = "aa";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = " SELECT withdraw.amount, withdraw.account, withdraw.created
FROM withdraw
INNER JOIN users ON withdraw.user_id=users.username; ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["username"]. " Name: " . $row["amount"]. "<br> " . $row["account"]. " " . $row["created"]. "<br>";
}
} else {
echo "Nothing Found!";
}
$conn->close();
?>
You didn't select username of table2 and compare withdraw.user_id with users.username
Try:
$sql = " SELECT withdraw.amount, withdraw.account, withdraw.created, users.username
FROM withdraw
INNER JOIN users ON withdraw.user_id=users.id; ";

Displaying results of SQL query as a table?

I am trying to output the results of an SQL query as a table on a page on my website. I have found a few solutions online but I can't get any of them to work properly. Right now I copied and pasted a bit of code to just output the first two columns but I can't figure out how to get every column in a table. I am new to PHP and web development in general so any help would be appreciated.
My PHP:
<?php
SESSION_START() ;
$servername = "localhost";
$username = "MY USERNAME";
$password = "MY PASSSWORD";
$dbname = "MY DATABASE NAME";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
//$_session['userid'] = $userlogged;
$sql = "SELECT * FROM `climbs` WHERE `userlogged` = '" . $_SESSION['userid'] . "'";
$result = mysqli_query($conn,$sql);
if ($result->num_rows > 0) {
echo "<table><tr><th>ID</th><th>Name</th></tr>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>" . $row["climb-id"]. "</td><td>" . $row["climbname"]. " " . $row["cragname"]. "</td></tr>";
}
echo "</table>";
} else {
echo "0 results";
}
mysqli_close($conn);
?>
check with var_dump :
some like that:
$result = mysqli_query($conn,$sql);
var_dump($result);
if ($result->num_rows > 0) {
maybe the query it's wrong.

PHP select data from database

<?php
$servername = "localhost";
$username = "root";
$password = "william";
$dbname = "camping";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT fnavn,enavn,epost,tlf FROM knr ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["fnavn"]. " - Name: " . $row["enavn"]. " " . $row["epost"]. "ire: " . $row["tlf"]."<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
Tried it, and it doesnt work, anyone who knows why? It's something wrong with the second if statement I think
You would need to join the tables:
SELECT * FROM produkt
LEFT JOIN produkttype ON (produkt.ptype = producttype.ptype)
WHERE utleid="No"

CONCAT function in MySQL using PHP Undefined Variable

I need help on how to do this correctly. I need to execute this command:
SELECT concat(branchname, -->, itemtype, '(, quantity, ')') from monitoring
order by itemtype;
the syntax works in MySQL console. However, im having trouble with implementing it on php. I always get
"Undefined index: branchname"
"Undefined index: itemtype"
"Undefined index: quantity"
using this code:
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "dex_test";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT concat(branchname,itemtype,quantity) from monitoring order by itemtype";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo " " . $row["branchname"]. " " . $row["itemtype"]. " ".$row["quantity"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
The error says it's in this line
echo " " . $row["branchname"]. " " . $row["itemtype"]. " ".$row["quantity"]. "<br>";
Im confused because I basically ran the same code that worked that lets me see the itemtype in the table:
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "dex_test";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT itemtype FROM monitoring";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "itemtype: " . $row["itemtype"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
Help anyone?
It seems your query needs update
"SELECT concat(branchname,itemtype,quantity) from monitoring order by itemtype";
It should be
"SELECT branchname,itemtype,quantity from monitoring order by itemtype";
I have posted this answer in reference of how you were calling your fields in while loop
echo " " . $row["branchname"]. " " . $row["itemtype"]. " ".$row["quantity"]. "<br>";
and if you need to show the concat value within one field than it should be something like
$sql = "SELECT concat(branchname,' ',itemtype,' ',quantity) as branch from monitoring order by itemtype";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo $row["branch"]."<br>";
}
} else {
echo "0 results";
}
Just define the alias for the concatenated columns. Use this -
SELECT concat(branchname,itemtype,quantity) as branchname from monitoring order by itemtype
Or if you want them seperately then -
SELECT branchname, itemtype, quantityfrom monitoring order by itemtype

check if row exists else create one

Motive: I just want to crreate a row tih specified data if one with the same data does not exists.
Thing I have tried so far is -
CODE:
<?php
if(empty($_GET['a'])) {$xyz ="new";} else{$xyz=$_GET['a'];}
$servername = "mysql.soemwhere.com";
$username = "u130204422_acb";
$password = "YES-I-KNOW";
$dbname = "u130204422_acb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$output ="SELECT * FROM trailers
WHERE url='$xyz' LIMIT 1";
$result = mysqli_query($conn,$output);
while($row = mysqli_fetch_array($result)) {
$pid=$row["title"];
echo $pid;}
if (mysqli_num_rows($result) > 0) {echo 'yes';}
else{
$msql = "INSERT INTO `trailers`(`url`,`title`) VALUES ('$xyz','dekhlia')";
if ($conn->query($msql) === TRUE) {
echo "New record created successfully. Refrsh the page and it will echo dekhlia";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;}
echo "hehe";
$conn->close();}
?>
I am unable to figure out why this code is'nt working.
You are unnecessarily adding concatenation in the SQL string.
Its a part of the sting, so, remove it.
Change
$output ="SELECT * FROM trailers
WHERE url='.$xyz.' LIMIT 1";
To:
$output ="SELECT * FROM trailers
WHERE url='$xyz' LIMIT 1";

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