I am trying to update a column of a database's table using this
$url = mysqli_connect($servername, $dbusername, $usrpassword, $dbname);
// Check connection
if (!$url) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT adv_val FROM current_advert";
if($result = mysqli_query($url, $sql)){
while($row = mysqli_fetch_assoc($result)){
$adv_val = $row['adv_val'];
}
}
echo "current advert is:" . $adv_val;
// Attempt select query execution
$sql = "SELECT advert_text FROM advertisements WHERE advert_id = $adv_val";
if($result = mysqli_query($url, $sql)){
while($row = mysqli_fetch_assoc($result)){
$advert = $row['advert_text'];
}
}
echo "<br>current advert text is:" . $advert;
if ($adv_val == 1 OR $adv_val == 2 OR $adv_val == 3) {
$adv_val = $adv_val + 1;
} else {
$adv_val = 1;
}
$sql = "UPDATE current_advert SET adv_val='$adv_val'";
// Close connection
echo "<br>next advert id is:" . $adv_val;
mysqli_close($url);
The connection to the database is ok since i'm able to read data from it in the beginning of my script. This is killing me!
$sql = "UPDATE current_advert SET adv_val='$adv_val'";
mysqli_query($url, $sql);
you are missing the 2nd line.
Related
Current update: I've cleaned up the code, and there are still some issues.
NOTE this code runs every 3 seconds. The outermost 'else' statement seems to run, setting the time to 0 in the database table, but then there is no activity.
After the initial time of running, the outermost 'else' statement should never run, and the time value stored under the user's alias should keep updating with the latest time stamp, but it just sits at '0'.
This is the JS that runs the php file:
//CHECK FOR NEW CHAT MESSAGES
setInterval(function()
{
$.post("chat_update.php", function(data) { $("#rect_comments_text").append(data);} );
}, 3000);
Code:
<?php
session_start();
$alias = $_SESSION['username'];
$host = 'localhost';
$user = '*';
$pass = '*';
$database = 'vethergen_db_accounts';
$table = 'table_messages';
$time_table = 'table_chat_sync';
$connection = mysqli_connect($host, $user, $pass) or die ("Unable to connect!");
mysqli_select_db($connection,$database) or die ("Unable to select database!");
$timestamp = time();
$last_time_query = "SELECT alias FROM $time_table";
$last_time_result = mysqli_query($connection,$last_time_query);
$last_time_rows = mysqli_fetch_array($last_time_result);
if ($last_time_rows['alias'] === $alias)
{
$last_time = $last_time_rows['time'];
$query = "SELECT * FROM $table WHERE time > $last_time ORDER BY text_id ASC"; //SELECT NEW MESSAGES
$result = mysqli_query($connection,$query);
//APPEND NEW MESSAGES
while($row = mysqli_fetch_array($result))
{
if ($row['alias'] === "Vether")
{
echo '<p id = "chat_text">'.'<b>'.$row['alias'].'</b>'.': '.$row['text']."</p>";
echo '<p id = "time_stamp">'.$row['time'].'</p>';
echo '<p id = "chat_number">'.$row['text_id'].'</p>';
}
else
{
echo '<p id = "chat_text">'.'<b class = "bold_green">'.$row['alias'].'</b>'.': '.$row['text']."</p>";
echo '<p id = "time_stamp">'.$row['time'].'</p>';
echo '<p id = "chat_number">'.$row['text_id'].'</p>';
}
echo '<hr class = "chat_line"></hr>';
}
//UPDATE LAST SYNC TIME
$update_query = "UPDATE $time_table SET time = '$timestamp' WHERE alias = '$alias'";
mysqli_query($connection,$update_query);
}
else
{
echo '<p> HERE </p>';
$update_query = "INSERT INTO $time_table (alias, time) VALUES('$alias','0')";
mysqli_query($connection,$update_query);
}
?>
You try this
$sql_update = "UPDATE time_table SET time= '$timestamp' WHERE alias = '$alias'";
if ($con->query($sql_update ) === TRUE) {
}
else{
echo "Error: " . $sql_update . "<br>" . $con->error;
}
You need to only check mysqli_num_rows to whether to insert or update data. You have to add ' around $alias in select query also. change your code as below:
//EITHER UPDATE THE EXISTING VALUE OR CREATE ONE FOR FIRST TIME VISITORS...
$last_time_query = "SELECT * FROM $time_table WHERE alias = '$alias'"; //change here add '
$last_time_result = mysqli_query($connection,$last_time_query);
if (mysqli_num_rows($last_time_result) == 0) //Only check number of rows
{
$update_query = "INSERT INTO $time_table (alias, time) VALUES('$alias','$timestamp')";
mysqli_query($connection,$update_query);
}
else
{
$update_query = "UPDATE $time_table SET time = '$timestamp' WHERE alias = '$alias'";
mysqli_query($connection,$update_query);
}
I'm trying to make something which will only display the name of the row which has ID 1 but I can't seem to get it to work. I can make it display all the names but I only want it to display the name of user ID 1. This is my current code but it doesn't work.
<a style="font-size: 17px; color: #ff0000;"><?php
$q = "SELECT * FROM `Team` WHERE id =1";
$result=mysqli_query($q);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
if ($row != FALSE) {
echo '<br />$row is not false.';
$name = $row['name'];
echo $name;
} else{echo "it's false :(";};
?></a>
It returns:
it's false :(
you may need the while() check on there.
Try something like:
Your database connection:
$servername = "YOUR_HOST";
$username = "YOUR_USER";
$password = "YOUR_PASSWORD";
$dbname = "YOUR_DATABASE";
$mysqli = new mysqli($servername, $username, $password, $dbname);
if ($mysqli->connect_error) {
echo "There was a slight problem, please contact your webmaster before continuing.";
exit();
}
Then your main file with displaying the row you want:
// create query
$q = "SELECT * FROM Team WHERE id = 1";
// get the records from the database
if ($result = $mysqli->query($q))
{
// display records if there are records to display
if ($result->num_rows > 0)
{
// fetch the results
while ($row = $result->fetch_object())
{
$name = $row->name;
echo $name;
}
}
else
{
echo "No results to display!<br><hr><br>";
}
}
else
{ // show an error if there is an issue with the database query
echo "Error: " . $mysqli->error;
}
// close database connection
$mysqli->close();
mysqli_query requires first parameter should be connection string and second is the query
mysqli_query($link, "your query");
Ref: http://php.net/manual/en/mysqli.query.php
You need to add the Connection-Parameter!
$result=mysqli_query($db, $q);
instead of
$result=mysqli_query($q);
I'm making a form for some project. The problem is when the user enter their data using the field and then submit, the input not keep in the database.
Also when clicking submit the direct page show blank empty page even the connection test also not showing.
I'm using almost similar code for other project and it work except for this.
below is my code:
<?php
//check connection
require 'config.php';
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
//asas (table name)
$id = $_POST["Sid"]; $ic = $_POST["Sic"];
$name = $_POST["Snp"]; $jant = $_POST["J1"];
$trum = $_POST["Chr"];$tbim = $_POST["Chp"];
$mel = $_POST["Sem"]; $arum = $_POST["Ar"];
$asum = $_POST["As"];
//institusi
$thp = $_POST["T1"]; $uni = $_POST["Sis"];
$bid = $_POST["tpe"];$Aint = $_POST["Ai"];
//industri
$bip = $_POST["bid"];$bik = $_POST["B1"];
$tem = $_POST["te"];$mula = $_POST["tm"];
$tamm = $_POST["tt"]; $res = $_POST["fileToUpload1"];
$tran = $_POST["fileToUpload2"];$keb = $_POST["fileToUpload3"];
$link = mysqli_connect($h,$u,$p,$db);
if('id' != '$Sid'){
$asas = "insert into asas Values ('$id','$ic','$name','$jant','$trum','$tbim','$mel','$arum','$asum')";
$inst = "insert into institusi Values ('$thp','$uni','$bid','$Aint')";
$indr = "insert into industri Values ('$bip','$bik','$tem','$mula','$tamm','$res','$tran','$keb')";
mysqli_query($link,$asas);
mysqli_query($link,$inst);
mysqli_query($link,$indr);
mysqli_close($link);
}
else
{
echo "failed"
}
?>
<b>Register complete</b>
Can anybody tell me what the error or maybe some solution. Thanks
I think you are having problem in insert query, please check this:
$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('John', 'Doe', 'john#example.com')";
Write this kind.
thank you
there are few issues with the code like variable id was used without $
and need to use die method with mysqli_query() function to check for
errors, please check below improved codes, it may help you -
<?php
//check connection
require 'config.php';
if (isset($_POST)) {
//asas (table name)
$id = $_POST["Sid"];
$ic = $_POST["Sic"];
$name = $_POST["Snp"];
$jant = $_POST["J1"];
$trum = $_POST["Chr"];
$tbim = $_POST["Chp"];
$mel = $_POST["Sem"];
$arum = $_POST["Ar"];
$asum = $_POST["As"];
//institusi
$thp = $_POST["T1"];
$uni = $_POST["Sis"];
$bid = $_POST["tpe"];
$Aint = $_POST["Ai"];
//industri
$bip = $_POST["bid"];
$bik = $_POST["B1"];
$tem = $_POST["te"];
$mula = $_POST["tm"];
$tamm = $_POST["tt"];
$res = $_POST["fileToUpload1"];
$tran = $_POST["fileToUpload2"];
$keb = $_POST["fileToUpload3"];
}
$link = mysqli_connect($h, $u, $p, $db);
if (!$link) {
die("Connection failed: " . mysqli_connect_error());
}
// if('id' != '$Sid'){
if ($id != '$Sid') {
$asas = "insert into asas Values
('$id','$ic','$name','$jant','$trum','$tbim','$mel','$arum','$asum')";
$inst = "insert into institusi Values ('$thp','$uni','$bid','$Aint')";
$indr = "insert into industri Values
('$bip','$bik','$tem','$mula','$tamm','$res','$tran','$keb')";
if (mysqli_query($link, $asas)) {
echo "records inserted";
} else {
echo "failed".mysqli_error($link) ;
}
if (mysqli_query($link, $inst)) {
echo "records inserted";
} else {
echo "failed".mysqli_error($link) ;
}
if (mysqli_query($link, $indr)) {
echo "records inserted";
} else {
echo "failed".mysqli_error($link) ;
}
}
mysqli_close($link);
?>
<b>Register complete</b>
just use or die after mysqli_query
mysqli_query($link,$asas)or die ('Unable to execute query. '. mysqli_error($link));
you will get to know what is the actual problem
My mySQLi query is failing but the error is blank. Can anybody tell me what the problem is, or how I can output info on the error?
Heres my PHP code:
// Create connection
$conn = new mysqli($hostname, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['in_wordarray']))
{
$words = $_POST['in_wordarray'];
$sql = "SELECT *, (";
$i = 0;
foreach ($words as $value) {
if($i == 0) {
$sql .= "(`words` LIKE '%$value%')";
} else {
$sql .= " + (`words` LIKE '%$value%')";
}
$i++;
}
$sql .= ") AS `numMatches` FROM `mytable` HAVING `numMatches` >= 3 ORDER BY `numMatches` DESC";
//echo $sql;
$result = $conn->query($sql);
if ($result === TRUE) {
$text_result_array = array();
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
$text_result_array[] = $row;
}
// Encode the response
echo json_encode($text_result_array);
}
} else {
echo $conn->error;
}
} else {
echo "Bad Input";
}
And the resulting SQL query looks like this:
SELECT *, ((`words` LIKE '%word1%') + (`words` LIKE '%word2%') + (`words` LIKE '%word3%')) AS `numMatches` FROM `words`HAVING `numMatches` >= 3 ORDER BY `numMatches` DESC
Sorry I'm not so experienced with php alone, but are you sure this returns a Boolean?
$result = $conn->query($sql);
If not can you remove that if statement?
I wrote it as an answer because sadly I can't comment yet.
$sql = "UPDATE reservations SET status = '$this->status',remaining_time ='$this->remain',cost = '$this->cost' WHERE id = '$this->id'";
This code is not working although it's correct
I am using object oriented php.
$this->id is a variable passed by link from another page.
When I run the code it tells me it was successful but that there are zero affected rows.
The one line above is part of the following code:
<?php
class edit {
private $status;
private $remain;
private $cost;
private $id;
public function edit_data() {
$this->status = strtoupper(strip_tags($_POST['status']));
$this->remain = strip_tags($_POST['remain']);
$this->cost = strip_tags($_POST['cost']);
$submit = $_POST['submit'];
$this->id = $_GET['edit'];
$con = mysql_connect("localhost","root","")
or die("Failed to connect to the server: " . mysql_error());
mysql_select_db("Users")
or die("Failed to connect to the database: " . mysql_error());
if($submit) {
if($this->status and $this->remain and $this->cost) {
$sql = "UPDATE reservations SET status = '".$this->status."',remaining_time ='".$this->remain."',cost = '".$this->cost."' WHERE id = '".$this->id."'";
$query = mysql_query($sql,$con);
if(!$query) {
echo("Could not update data: " . mysql_error());
}
echo "<h4>Customer reservation data has been updated successfully.</h4>";
echo "Number of affected rows: " . mysql_affected_rows();
}
else {
echo "Please fill in all fields.";
}
}
mysql_close($con);
}
}
$edit = new edit();
echo $edit->edit_data();
?>
Are you sure about your concatenation?
$sql = "UPDATE reservations SET status = '$this->status',remaining_time ='$this->remain',cost = '$this->cost' WHERE id = '$this->id'";
Print $sql to see the value.
If your database is already updated, you will receive 0 affected lines.
I am not totally sure but try this,
"UPDATE reservations SET status = '".$this->status."',remaining_time ='".$this->remain."',cost = '".$this->cost."' WHERE id = '".$this->id."'";
It seems that your table doesn't contain a value which satisfies where condition.
You can check this by executing a simple query.
$sql = "select * from reservations where id='$this->id'";