I have a php script made by me that can change a database info by taking db name,user,pass and host. Now I want to show a database column info to my php script.
Now how can i do that here is my main db.php script i didn't put the html from script cause i don't think it's necessary.
I have code this but it's showing a error like that
Parse error: syntax error, unexpected T_STRING in /home/abc/public_html/db.php on line 56
Here is my db.php
<style type="text/css">
body {
background-image: url('data:image/gif;base64,R0lGODlhAwADAPcAAAAAAABCAP///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////ywAAAAAAwADAAAICQADABhIcGBAAAA7');
font-family: Tahoma;text-align: center;color: green;
}
img{opacity:0.75; filter:alpha(opacity=75);}
.field_set{
border-color:#4AB825;
}
</style>
<?php
// escape received values
$dbusr = $_POST['usr'];
$dbpsw = $_POST['psw'];
$dbhost = $_POST['host'];
$dbname = $_POST['dbname'];
$admusr = $_POST['admusr'];
$prfx = $_POST['prfx'];
$admpsw = md5($_POST['admpsw']);
// Create connection
$conn = new mysqli($dbhost, $dbusr, $dbpsw);
// Check connection
if ($conn->connect_error) {
die("<br><br><br><br><br><br><br>Database Connection failed: " . $conn->connect_error);
}
echo "<br><br><br><br><br><br><br>Database Connected successfully";
mysqli_select_db($conn,"$dbname");
// use them in query
$sql = "UPDATE ".$prfx."_users SET user_login='".$admusr."',user_pass='".$admpsw."' WHERE id=1";
if ($conn->query($sql) === TRUE) {
echo "<br><br>Record updated successfully</br></br>Go to your login page <br><br>ex: www.site.com/wp-admin<br><br>and login with your given id and pass";
} else {
echo "<br>Error updating record:" . $conn->error;
}
$sql = "SELECT guid FROM".$prfx."_posts;
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "Site:" . $row["guid"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
$sql = "SELECT guid FROM".$prfx."_posts;
Should be
$sql = "SELECT guid FROM".$prfx."_posts";
You were missing the closing "
As Dagon said in his comment, In most editors code is shown with color coding, and you can usually see if a " or ' is missing it's closing tag because the color will be off.
UPDATE
Missing space between FROM and $prfx, use below:
$sql = "SELECT guid FROM ".$prfx."_posts";
You are missing a " at the end of line try replace this.
$sql = "SELECT guid FROM".$prfx."_posts";
Whenever you get this error it means that you are missing a closing double quote on that line which is mentioned on the error.
Change this,
$sql = "SELECT guid FROM".$prfx."_posts;
to this,
$sql = "SELECT guid FROM".$prfx."_posts";
Related
How to update a status from database if status is empty in using php? I have this condition in php. I have this if condition that decides if $getstatus is empty it will update from database to Avail. I tried refreshing the page after querying the database. But it will not update in database. Is there anyway to update this without using form submit in php?
<?php
session_start();
include "includes/connection.php";
// Display all parking slots
$sql = $connection->prepare('SELECT * FROM parkingslot where parkingslotid = 1');
$sql->execute(); // execute query
$result = $sql->get_result(); // fetch result
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$getstatus = $row["status"];
echo $getstatus;
}
}
if (empty($getstatus)) {
$sql = $connection->prepare("UPDATE parkingslot SET status = 'Avail' where parkingslotid = 1 ");
}
?>
Codes in connection for connecting to database
connection.php
<?php
$server = "localhost";
$username = "root";
$password = "";
// create connection
$connection = mysqli_connect($server,$username,$password);
// check connection
if(!$connection)
{
die("No connection found." . mysqli_connect_error());
}
else {
// select a database
$select_db = mysqli_select_db($connection,'smartparkingsystem');
if(!$select_db)
{
$sql = 'CREATE DATABASE sample';
// create database if no db found
if(mysqli_query($connection,$sql)) {
echo "Database Created";
}
else {
echo "Database not found" . mysqli_connect_error() . '\n';
}
}
else {
// Database already existed
// do nothing...
}
}
?>
If I understand your goal of: For row(s) whereparkingslotid=1 - Update status to 'Avail' but only if status is not currently set, this might help:
<?php
session_start();
include "includes/connection.php";
$connection->prepare("UPDATE `parkingslot` SET `status`=? WHERE `parkingslotid`=? AND (`status` IS NULL OR `status`=?)");
$connection->bind_param("sis", $status, $parkingslotid, $empty_str);
$status = 'Avail';
$parkingslotid = 1;
$empty_str = '';
$connection->execute();
echo $connection->affected_rows.' rows affected';
$connection->close();
?>
This saves a bit of processing by not checking with PHP first.
You can use this query:
"UPDATE parkingslot SET status = 'Avail' where status IS NULL OR status = '' "
Edited:
#lumonald gave the right anwser in the comment. You're not executing your second SQL statement.
i just want to know how I can fix this. In my database : here i have a value of 123. My code is searching for it however it does not find it.
<html>
<head>
<?php
$userid = 123;
$con = mysqli_connect("","","*","");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}else{
echo "Connected";
}
echo "123";
// Perform queries
$sql_fetch_id = "SELECT * FROM AccessedIds WHERE 64id = '123";
$query_id = mysqli_query(sql2242953, $sql_fetch_id);
if(mysqli_num_rows($query_id) ==0) {
echo "empty";
}else{
echo "full";
}
mysqli_close($con);
?>
<title>PHP Test</title>
</head>
<body>
</body>
</html>
Data types
Your query is wrong, try:
$query_id = mysqli_query($con, $sql_fetch_id);
More information
Also as in the comments above, remove the single quote mark before 123.
Your Query and Query string is wrong. You are missing a Single Quote (') in your Query string. See the below statements:
$sql_fetch_id = "SELECT * FROM `AccessedIds` WHERE `64id` = '123'";
$query_id = mysqli_query($con, $sql_fetch_id);
Hope this helps.
I am trying to input data to MySQL using PHP. Don't know what's wrong. The connection succeeds, no errors but at the end there is not data being written to the database.
$dbhost = "localhost";
$dbname = "listings";
$un = $_POST["un"];
$pass = $_POST["pass"];
$name = $_POST["name"];
$des = $_POST["des"];
$quan = $_POST["quantity"];
$specs = $_POST["specs"];
$price = $_POST["price"];
$url1 = ".";
$url2 = ".";
$url3 = ".";
$url4 = ".";
$connection = mysqli_connect($dbhost,$un,$pass,$dbname);
if (!$connection) {
die("Error".mysqli_error);
} else {
echo "Database connection successfull ".$des;
}
$query = "INSERT INTO items
(name,description,quantity,specs,price,url1,url2,url3,url4) VALUES
'$name','$des','$quan','$specs','$price','$url1','$url2','$url3','$url4')
";
echo "Hellos";
$exeute_query = mysqli_query($query,$connection);
if(!execute_query){
die("error ".mysqli_error());
echo "query error";
} else {
echo "Query successfull";
}
mysqli_close($connection);
Any help?
There are several small mistakes in your code:
$query = "INSERT INTO items (name,description,quantity,specs,price,url1,url2,url3,url4) VALUES ('$name','$des','$quan','$specs','$price','$url1','$url2','$url3','$url4')";
echo "Hellos";
**$exeute_query** = mysqli_query($query,$connection); // $execute_query instead of $exeute_query
if(!**execute_query**){ //$execute_query instead of execute_query
die("error ".mysqli_error());
echo "query error";
}
else{echo "Query successfull";}
mysqli_close($connection);
?>
Your code breaks at the if statement because no fucntion with that name is found (if you do not use the dollarsign to show it is a variable, php will interpret it as a function. Also, when initiating your variable you forgot a 'c' so make sure to check if you have the correct variable name or php won't find your variable. Now your query will work or give an error message in case of wrong data formats or bad connection. Use code listed below to debug your php in the future.
error_reporting(E_ALL);
ini_set('display_errors', 'On');
I have XAMPP and I want to write a simple PHP page, that redirects me to the link that I specify, and also saves the link in an SQL database.
Let's say I want to visit www.google.com:
I'd visit something like:
localhost:80/redirect.php?url=https://google.com
And PHP would redirect me there and also save the www.google.com link in an SQL table.
Can you help me out?
Considering how you formed your question, it looks as if you had an idea an just want someone to give you the solution without you even making an effort (please correct me if I'm wrong but that's how it seams...)
The task you are trying to achieve is a simple one, and it's only fair to point you in the right direction. your "task" can be broken into several smaller ones:
Create database / table for storing data | PHP Create MySQL Tables
Get URL parameter in PHP
PHP Insert Data Into MySQL
How to make a redirect in PHP
Sorry if this is not the kind-a answer you are looking for, but I figure the point of this website is for people to learn something and not just copy+paste. The provided links can be used to solve your task problem.
This is what I came up with, after MySQLi Object-oriented did not validate this:
$sql = "SELECT * FROM logging WHERE link=$link";
if ($conn->query($sql) === TRUE) {}
It still increments the number of visits sometimes by +2. I don't know why.
<?php
$servername = " ";
$username = " ";
$password = " ";
$dbname = " ";
$datetime = date_create()->format('Y-m-d H:i:s');
$datetime = "'".$datetime."'";
$link_clean = $_GET['link'];
$link = "'".$link_clean."'";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM logging WHERE link=$link";
if ($result = mysqli_query($conn, $sql))
{
if(mysqli_num_rows($result)>0)
{
$sql="UPDATE logging SET last_visit_date = $datetime, visit_count = visit_count + 1 WHERE link=$link";
if (mysqli_query($conn, $sql)) {
$conn->close();
header("Location: https://$link_clean");
exit;
} else {
echo "1Error: " . $sql . "<br>" . mysqli_error($conn);
$conn->close();
exit;
}
}
else
{
$sql="INSERT INTO logging (link, last_visit_date, visit_count) VALUES ($link , $datetime , 1)";
if (mysqli_query($conn, $sql)) {
mysqli_close($conn);
header("Location: https://$link_clean");
exit;
} else {
echo "2Error: " . $sql . "<br>" . mysqli_error($conn);
mysqli_close($conn);
exit;
}
}
}
else
{
echo "3Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
after I managed to connect my website form to my database, I decided to try to transfer over my files to my work computer.
Initially I only had one error: mysqli_fetch_row() expects parameter 1 to be mysqli_result, boolean given in...
However now I get an extra mysqli_fetch_row() error the same as above but the error is on a different line.
Additionally I also get the error: Undefined index: fill which I never got before. Are there any mistakes in my code? The form still works and can connect to my database.
<center><form action="fill.php" method="post">
Fill
<input type="text" id="fill"" name="fill">
<input type="submit" id ="submit" name="submit" value="Submit here!">
</form></center>
</div>
<?php
$val1 = $_POST['fill'];
$conn = mysqli_connect('localhost', 'root', '')or
die("Could not connect");
mysqli_select_db($conn, 'rfid');
$val2 = "SELECT * FROM card_refill WHERE refill = $val1";
$result1= $conn->query($val2);
$row = mysqli_fetch_row($result1);
$refill1 = $row[2];
$value = "SELECT *FROM card_credit ORDER BY id DESC LIMIT 1:";
$result = $conn->query($value);
$row = mysqli_fetch_row($result);
$refill = $row[2];
$money= $refill+$refill1;
echo $money;
$sql = "UPDATE card_credit SET value = '$money'";
if ($conn->query($sql) === TRUE) {
echo "Success";
}
else {
echo "Warning: " . $sql . "<br>" . $conn->error;
}
mysqli_close($conn);
?>
</body>
</html>
You're getting that error because you use $_POST['fill'] without checking whether it's set first. It will only be set when the form is submitted, not when the form is first displayed. You need to put all the code that processes the form input into:
if (isset($_POST['submit'])) {
...
}
BTW, you can do that entire update in a single query.
UPDATE card_credit AS cc
CROSS JOIN card_refill AS cr
CROSS JOIN (SELECT * FROM card_credit ORDER BY id DESC LIMIT 1) AS cc1
SET cc.value = cr.col2 + cc1.col2
WHERE cr.refill = '$val1'
Like GolezTrol said from his comment. You're mixing object and functional notation.
Although this might not work exactly how you need it to because I don't have all the information. I have written you something I think is close to what you're looking for.
<?php
// Define the below connections via $username = ""; EXTRA....
// This is best done in a separate file.
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$val1 = $_POST['fill'];
$result1 = $conn->query("SELECT * FROM card_refill WHERE refill = '$val1' ");
$result2 = $conn->query("SELECT * FROM card_credit ORDER BY id DESC LIMIT 1:");
$refill1 = array(); // Pass Results1 Into Array
while($row = $result1->fetch_assoc()) {
$refill1[] = $row[2];
}
$refill = array(); // Pass Results2 Into Array
while($row = $result2->fetch_assoc()) {
$refill[] = $row[2];
}
/* Without an example of what data you are getting from your tables you will have to figure out what data you want from the arrays.
$money= $refill+$refill1;
echo "DEBUG: $money";
*/
// This code will not be functional until your populate the $money value.
$sql = "UPDATE card_credit SET value = '$money' ";
if ($conn->query($sql) === TRUE) {
echo nl2br("Record updated successfully"); // DEBUG
print_r(array_values($refill1)); // DEBUG
print_r(array_values($refill)); // DEBUG
echo nl2br("\n"); // DEBUG
} else { // DEBUG
echo "Error updating record: " . $conn->error; // DEBUG
echo nl2br("\n"); // DEBUG
}
$conn->close();
?>