Im trying to display a database table in my html page but cant seem to get it to work. (code below):
<!DOCTYPE html>
<html>
<body>
<h1> Table </h1>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Library";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM `books` ORDER BY `Category` ASC ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<br> id: ". $row["ISBN"]. " Name: ". $row["firstname"]. " - Author: " . $row["BookAuthor"]. " Category: " . $row["Category"]. " - Quantity: " . $row["Quantity"]. " - Price: " . $row["Price"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
</body>
</html>
Im using WAMP server and the "books" table is located in a database called Library. However the table doesnt display and all i get is the php code being shown.
Any idea on whats gone wrong?
Thanks!
Is your WAMP running? (Check the system tray icon. It should be green)
Did you mess with PHP or Apache configs?
Is the file extension .php? It has to be.
Is other PHP displaying properly?
Related
So basically I am trying to simply just connect to my local mysql database with a football table to display on the site, but it doesn't seem to be able to connect. I feel like it's really obvious but I just can't see it.. I have checked the password is correct also, and it is.
<?php
$servername = "localhost";
$username = "root";
$password = "password"
$dbname = "import_excel_to_mysql";
/* Create connection */
$conn = new mysqli("localhost", "root", "password", "import_excel_to_mysql");
/* Check connection */
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error . mysql_error());
}
$sql = "SELECT position, team_name, played, wins, draws, losses, goals_scored, goals_against, goal_difference, points FROM football table";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "<table><tr><th>Pos</th><th>Team Name</th><th>P</th><th>W</th><th>D</th><th>L</th>(etc)</tr>";
/* output data of each row */
while($row = $result->fetch_assoc()) {
echo "<tr><td>" . $row["position"]. "</td><td>" . $row["team_name"]. " " . $row["played"]. " " . $row["wins"]. " " . $row["draws"]. (etc)"</td></tr>";
}
echo "</table>";
} else {
echo "0 results";
}
?>
It should display the table just on its own, but instead it displays all the php code starting at connect_error(. The rest of the site is unaffected.
I made a kind of plugin just for the database connection it works but if i activate my plugin the page wil go a little bit down so actually my site will become ugly
$servername = "xxxx"; // naam of ip van de machine ... 127.0.0.1
$username = "xxxx"; //naam van gebruiker
$password = "xxxx"; // wachtwoord
$dbname = "xxxx"; //db naam
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
if (isset($_GET["code"]) )
{
$code = $_GET['code'];
}
else
{
$code = '';
}
//$naam = '%DR%';
//$sql = "SELECT * FROM docenten WHERE naam=\"$naam\"";
$sql = "SELECT * FROM `cadeaubon` WHERE Kortingscode like \"$code\"";
//echo $sql; //debuggen
echo "<BR>";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
echo "<table>";
while($row = mysqli_fetch_assoc($result)) {
echo "- ID: " . $row["ID"]. " - Kortingscode: " . $row["Kortingscode"]. " - Aantal personen: " . $row["AantPersonen"] . " - Naam: " . $row["Naam"] . " - Achternaam: " . $row["Achternaam"] . " - Leeftijd: " . $row["Leeftijd"] . " - Datum: " . $row["Datum"]."<br>";
}
echo "</table>";
} else {
echo "";
}
mysqli_close($conn);
this is my code and i thought maybe it will help if the result will display at the bottom of the page but i dont know how
I guess you left there <br> from your debug echo:
//echo $sql; //debuggen
echo "<BR>";
However I don't know where in code this gets executed and how it integrates with WordPress. If the following table is shown inside content area of the page, the <br> shouldn't affect your site design other than pushing the table one line down.
It also seems to be a bit of overkill to create your own database connection - I would think WordPress already is connected and it would be preferable to use its own connection, unless it is a completely different database.
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.
I've tried many things. This is the simplest form I can try to see if all is working.
<?php
$mysql_host = "xxxx";
$mysql_username="xxxx";
$mysql_password="xxxxx";
$mysql_database="test2";
// Create connection
$conn = new mysqli($mysql_host, $mysql_username, $mysql_password, $mysql_database);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} echo "Connected successfully";
$sql = "INSERT INTO `test_table` (`id`,`firstname`) VALUES ('3','james')";
echo "-----------";
echo "-----------";
$sql = "SELECT * FROM `test_table`";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
The table already contains 2 dummy inputs that I can insert from phpmyadmin.
When I try the above I get:
Connected successfully----------------------id: 1 - Name: john
id: 2 - Name: ghost
I can't see where I am going wrong.
Could it be that I need some kind of permissions change for this in linux?
That would be odd, I don't remember doing this before.
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