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; ";
Related
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();
?>
i have 4 tables in my db
(0)users
user id,user name,password
(1)messages
user id,user name(fk),message,location,time
(2) images
user name(fk),image id,image path,time
(3)videos
user name(fk),video id,video path,location,time
i want to show data from multiple tables
what will be query syntax to do it
i want to show like
"rana " has upload this image and this video and this messages"
image should be select from images table video from video tables and message from message table
thanks
sorry if question already asked
thanks please help
Your expected mysql query is :
select *from users a
join messages b on a.user_name = b.user_name
join images c on a.user_name = c.user_name
join videos d on a.user_name = d.user_name
Example (MySQLi Procedural) & PHP
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "myDB";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "select *from users a
join messages b on a.user_name = b.user_name
join images c on a.user_name = c.user_name
join videos d on a.user_name = d.user_name";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "User Name: " . $row["user_name"]. "<br />";
echo "Message: " . $row["message"]. "<br />";
echo "Image Path: " . $row["image_path"]. "<br />";
echo "Video Path: " . $row["video_path"]. "<br />";
//you get more data as your wise................
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
<?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"
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
I Currently have this code that displays all entries rows from a mysql table and displays them on a web page;
<?php
$servername = "localhost";
$username = "appuser1";
$password = "*****";
$dbname = "acmefg_app";
$row = mysql_fetch_array(mysql_query("SELECT jobnumber FROM appdata WHERE id = '5608' LIMIT 1"));
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, jobnumber, assetnumber FROM appdata";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<br> id: ". $row["id"]. "<br>";
echo "<br> Job number: ". $row["jobnumber"]. "<br>";
echo "<br> Asset number: " . $row["assetnumber"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
How can I change this so it only displays one result? Lets say in this case I wanted to display the row with an id of 5611? Many Thanks
If you want to show only one row then no need to use while() loop. Remove loop
if ($result->num_rows > 0) {
$row = $result->fetch_assoc();
echo "<br> id: ". $row["id"]. "<br>";
echo "<br> Job number: ". $row["jobnumber"]. "<br>";
echo "<br> Asset number: " . $row["assetnumber"] . "<br>";
}
Or can set LIMIT 1 in your query string.
SELECT id, jobnumber, assetnumber FROM appdata LIMIT 1