I've made a like button on my site (which uses ExpressionEngine), and it works. However when I put almost identical code on another page, the like button's text is always what it should have been on the last page (the opposite to what it should be, though refreshing the page puts it right, so I can't just flip the values. Here's the code that works:
<?php
$DB1 = $this->EE->load->database('ext_db', TRUE);
$mapID = "{entry_id}";
$ipAddress = $_SERVER['REMOTE_ADDR'];
$thisquery = "SELECT * FROM mapLikes WHERE ipAddress = '$ipAddress' AND mapID = '$mapID'";
$q = $DB1->query($thisquery);
$results = $q->result_array();
foreach ($q->result() as $row)
{
$liked = $row->liked;
}
$buttontext = 'Like';
$buttonimage = "1";
if ($liked == "1") {
$buttontext = 'Unlike';
$buttonimage = "2";
}
if($_POST['like']) {
if ($liked == "1") {
$thisquery = "DELETE FROM mapLikes WHERE mapID = '$mapID' AND ipAddress = '$ipAddress'";
$DB1->query($thisquery);
} else {
$thisquery = "INSERT INTO mapLikes (mapLikeID, mapID, liked, ipAddress) VALUES ('null', '$mapID', '1', '$ipAddress')";
$DB1->query($thisquery);
}
}
?>
And the code that doesn't:
<?php
$DB1 = $this->EE->load->database('ext_db', TRUE);
$mapID = "{segment_3_category_id}";
$ipAddress = $_SERVER['REMOTE_ADDR'];
$thisquery = "SELECT * FROM categoryLikes WHERE ipAddress = '$ipAddress' AND mapID = '$mapID'";
$q = $DB1->query($thisquery);
$results = $q->result_array();
foreach ($q->result() as $row)
{
$liked = $row->liked;
}
$buttontext = 'Like';
$buttonimage = "1";
if ($liked == "1") {
$buttontext = 'Unlike';
$buttonimage = "2";
}
if($_POST['like']) {
if ($liked == "1") {
$thisquery = "DELETE FROM categoryLikes WHERE mapID = '$mapID' AND ipAddress = '$ipAddress'";
$DB1->query($thisquery);
} else {
$thisquery = "INSERT INTO categoryLikes (categoryLikeID, mapID, liked, ipAddress) VALUES ('null', '$mapID', '1', '$ipAddress')";
$DB1->query($thisquery);
}
}
?>
As you can see, the only difference between the two pieces of code are the $mapID and the queries. Yet for some reason one works and the other doesn't. Anyone have any idea what might be going on? I'm thinking this is more a php problem than an ExpressionEngine one.
The real question, I think, is why the first example does work...
You are setting the state of the button based on the current value of $liked, but then if $_POST['like'] is set, you're changing the state of the system and the button is now incorrect. If you switch your code around and test for $_POST['like'] first, then set the value of $like after you act on it, your button will show the correct state.
Figured it out now, and also realised the first bit of code had the same issues as the second. What I did was I made the POST script set the button state, like so:
if($_POST['like']) {
if ($liked == "1") {
$thisquery = "DELETE FROM mapLikes WHERE mapID = '$mapID' AND ipAddress = '$ipAddress'";
$DB1->query($thisquery);
$buttontext = 'Like';
$buttonimage = "1";
} else {
$thisquery = "INSERT INTO mapLikes (mapLikeID, mapID, liked, ipAddress) VALUES ('null', '$mapID', '1', '$ipAddress')";
$DB1->query($thisquery);
$buttontext = 'Unlike';
$buttonimage = "2";
}
}
Related
I want to send data to database, but if result = 1 status=plusone, result = 2 status=plustwo, etc..
It should work like that..
But no, it work like this: result = 2 status=plusone .
What did i missed? Help me..
I've tried this:
$item = '0';
$result = $item + $points;
and:
$result = $points + 0;
Here is rest (part of) code:
if($result = 1){
$Sql_Query ="INSERT INTO points SET unique_id = '$id', description = '$Description', points = '$points', status= 'plusone'";
if(mysqli_query($con,$Sql_Query)) {
echo 'Succcess!';
}
}elseif($result = 2){
$Sql_Query ="INSERT INTO points SET unique_id = '$id', description = ' $Description', points = '$points', status= 'plustwo'";
if(mysqli_query($con,$Sql_Query)) {
echo 'Succcess!';
}
}
= and == are different; = is for assignment, and == is to compare statement
This is my table in my database. It should only contain one row.
I have this code, that checks if there is no data , if there is not.. The data is inserted, else it's updated. The problem is, that if i update only Brevet, the value of Baccalaureabt will disappear from my database. Any help?
Code:
if(isset($_POST['submit']))
{ $brevet = $_POST['Brevet'];
$baccalaureatbt = $_POST['Baccalaureatbt'];
$sql1="SELECT Brevet,Baccalaureatbt FROM university";
if ($result=mysqli_query($con,$sql1))
{
$rowcount=mysqli_num_rows($result);
}
if($rowcount==0)
{
$sql="INSERT INTO university(Brevet,Baccalaureatbt) VALUES('$brevet','$baccalaureatbt')";
$result = mysql_query($sql);
}
else
{
$sql2 = "UPDATE university SET Brevet = '$brevet' , Baccalaureatbt = '$baccalaureatbt'";
$result2 = mysql_query($sql2);
}
The issue is that you are setting both values, if however either $baccalaureatbt or $brevet is empty, it will be updated with a empty value, i.e the original value will be deleted.
There are multiple ways to avoid this, but one way to do it is like so:
$sql2 = "UPDATE university SET ";
if(!empty($brevet)){
$sql2 .= "Brevet = '$brevet'";
$first = 1;
}
if(!empty($baccalaureatbt)){
$sql2 .= isset($first) ? ", Baccalaureatbt = '$baccalaureatbt'" : "Baccalaureatbt = '$baccalaureatbt' " ;
}
$result2 = mysql_query($sql2);
To update the values dynamically, use the following code. Just add the values you want in the values array.
if(isset($_POST['submit']))
{
$brevet = $_POST['Brevet'];
$baccalaureatbt = $_POST['Baccalaureatbt'];
$sql1="SELECT Brevet,Baccalaureatbt FROM university";
$result=mysqli_query($con,$sql1);
if ($result)
{
$rowcount=mysqli_num_rows($result);
}
if($rowcount==0)
{
$sql="INSERT INTO university(Brevet,Baccalaureatbt) VALUES('$brevet','$baccalaureatbt')";
$result = mysql_query($sql);
}
else
{
$values = array("brevet","baccalaureatbt");
$sql2 = "UPDATE university ";
$n = 0;
foreach($_POST as $key => $val) {
if(in_array($key, $values)) {
$sql2 += ($n !== 0 ? ", " : 0);
$sql2 += "SET".$key." = ".$val;
$n++;
}
if($n > 0) {
$result2 = mysql_query($sql2);
}
}
}
}
I'm trying to create a voting system for artists played on my radio station. I'm using the source code from: http://dl.howcode.org/download/97ff383c7d4dc9939c65c9e6fab2a5dc
The problem I have found is that the votes update using the number from the first row in the database no matter which option is selected, thus if for instance the first row has 3 votes in and the user tries to vote on someone with 0 votes, it will change the votes for the correct artist to 4 instead of 1... I hope that makes sense?
The code I have is:
[EDIT] I have changed the queries to fetch assoc to make it easier to understand.
<?php
$voteID = $_GET['voteID'];
$connect = mysqli_connect('xxx', 'xxx', 'xxx', 'xxx');
$query = "SELECT * FROM listenervotes WHERE voteID='$voteID'" ;
$q = mysqli_query($connect, $query);
while($row = mysqli_fetch_assoc($q)){
$id = $row["id"];
$voteTitle = $row["voteTitle"];
$voteID = $row["voteID"];
$ipaddress = $row["ipAddress"];
echo "<h3>$voteTitle</h3>";
?>
<table>
<form action="" method="POST">
<?php
$artists = "SELECT * FROM artists WHERE voteID='$voteID'" ;
$q2 = mysqli_query($connect, $artists);
while($r = mysqli_fetch_assoc($q2)){
$artist = $r["artistName"];
$votes = $r["votes"];
$genre = $r["genre"];
$ip = $_SERVER['REMOTE_ADDR'];
$newIpAddress = $ipaddress."$ip, ";
$newVotes = $votes + 1;
if (isset($_POST['vote'])) {
$voteOption = $_POST['voteOption'];
if ($voteOption == ""){
die("You haven't selected anyone!");
}else{
$ipaddressE = explode(",", $ipaddress);
if(in_array($ip, $ipaddressE)){
die("You have already voted!");
}else{
mysqli_query($connect, "UPDATE artists SET votes='$newVotes' WHERE voteID='$voteID' AND artistName='$voteOption'");
mysqli_query($connect, "UPDATE listenervotes SET ipaddress='$newIpAddress' WHERE voteID='$voteID'");
die('You voted successfully!<br><tr><td>'.$artist.'</td><td>'.$genre.'</td><td>'.$votes.' Votes</td></tr>');
}
}
}
echo '<tr><td>'.$artist.'</td><td>'.$genre.'</td><td><input type="radio" name="voteOption" value="'.$artist.'"</td></tr>';
}
}
?>
I could be missing something obvious, in my mind I'm thinking that I somehow need to iterate through the rows before setting the new value, if so, how and where?
It looks like you are always looping over all rows and updating the relevant row with the first value found. Adding a check on the ID should do:
<?php
$voteID = $_GET['voteID'];
$connect = mysqli_connect('xxx', 'xxx', 'xxx', 'xxx');
$query = "SELECT * FROM listenervotes WHERE voteID='$voteID'" ;
$q = mysqli_query($connect, $query);
while($row = mysqli_fetch_assoc($q)){
$id = $row["id"];
$voteTitle = $row["voteTitle"];
$voteID = $row["voteID"];
$ipaddress = $row["ipAddress"];
echo "<h3>$voteTitle</h3>";
?>
<table>
<form action="" method="POST">
<?php
$artists = "SELECT * FROM artists WHERE voteID='$voteID'" ;
$q2 = mysqli_query($connect, $artists);
while($r = mysqli_fetch_assoc($q2)){
$artist = $r["artistName"];
$votes = $r["votes"];
$genre = $r["genre"];
$ip = $_SERVER['REMOTE_ADDR'];
$newIpAddress = $ipaddress."$ip, ";
$newVotes = $votes + 1;
if (isset($_POST['vote'])) {
$voteOption = $_POST['voteOption'];
if ($voteOption == ""){
die("You haven't selected anyone!");
}else{
$ipaddressE = explode(",", $ipaddress);
if(in_array($ip, $ipaddressE)){
die("You have already voted!");
}elseif ($voteOption === $artist) { // Don't run UPDATE when we're on the wrong row.
mysqli_query($connect, "UPDATE artists SET votes='$newVotes' WHERE voteID='$voteID' AND artistName='$voteOption'");
mysqli_query($connect, "UPDATE listenervotes SET ipaddress='$newIpAddress' WHERE voteID='$voteID'");
die('You voted successfully!<br><tr><td>'.$artist.'</td><td>'.$genre.'</td><td>'.$votes.' Votes</td></tr>');
}
}
}
echo '<tr><td>'.$artist.'</td><td>'.$genre.'</td><td><input type="radio" name="voteOption" value="'.$artist.'"</td></tr>';
}
}
?>
I'm using PHP/MySql and I'm trying to team users into pairs based on a skill that they have. I have an existing table for users, and an existing table for the teams.
For example, I executed a query which returned 6 members which needs to be paired up into a team.
SELECT * FROM users WHERE skill = 'Office'
users
id/name/skill
1/Bob/Office
2/Ted/Office
3/Tim/Office
4/Bill/Office
5/Shawn/Office
6/Gab/Office
These results must be then paired up, and the expected output should be:
teams
name/member
Office1/Bob
Office1/Ted
Office2/Tim
Office2/Bill
Office3/Shawn
Office3/Gab
Once 2 members are placed in a team, the team name should increment by one.
Any help will be greatly appreciated Thanks.
Edit: I tried this:
$results = mysql_query("SELECT * FROM users WHERE skill = 'Office'");
$numrows = mysql_num_rows($results); $name ="";
if($numrows!=0) {
while($row = mysql_fetch_assoc($results)) {
$name = $row['userName'];
}
}
//For incrementing the team name
$namectr=0;
for($ctr=0;$ctr<$results_num;$ctr++) {
if($ctr%2==0) {
$query = mysql_query("INSERT INTO teams VALUES ('Office$namectr','$name')");
$ctr++; if($ctr%2==1) {
$query = mysql_query("INSERT INTO teams VALUES (Office$namectr','$name')");
$namectr++;
}
}
}
why not try:
$result = mysql_query("SELECT * FROM users WHERE users.skill = 'office'");
$count = 0;
$sqlcount = 0;
$offcount = 1;
while ($source = mysql_fetch_array($result)) {
if ($count < 1) {
$office = $source['skill'] . $offcount;
$name = $source['name'];
$result[$sqlcount] = mysql_query("INSERT INTO teams ('name', 'member') VALUES ('$office', '$name')");
$sqlcount++;
} else if ($count >= 1) {
$offcount++;
$count = 0;
$office = $source['skill'] . $offcount;
$name = $source['name'];
$result[$sqlcount] = mysql_query("INSERT INTO teams ('name', 'member') VALUES ('$office', '$name')");
$sqlcount++;
} else {
echo "ERROR" . mysql_error();
}
}//end while
$sqlcount = 0;
while ($result[$sqlcount] != "") {
if ($source = $result) {
} else {
echo "ERROR! " . mysql_error();
}
}//end while
Couldn't you do your database query, then do something like the below:
$count=0;
$team=1;
$teams = array();
foreach($result as $output){
if($count % 2 == 0){
// if even number, reset count and increment position
$count=0;
$team++;
}
$teams[] = $output['skill']." ".$team." - ".$output['member'];
$count++;
}
Something like the above, but it is untested, but should work in theory.
I am trying to write to my database if a variable is equal to 0. The problem is that it still writes to the database even when the variables equals 1. What is wrong??
echo $new_user;
if ($new_user == 0) {
//SENT NEW USER WELCOME MESSAGE
$adminid = '9';
$welcomemessagetitle = 'Welcome to The site';
$welcomemessagecontent = 'Hello and welcome';
$addmessages = "INSERT into `user_messages`(`to_user`,`from_user`,`title`,`content`)
VALUES ('$userid','$adminid','$welcomemessagetitle','$welcomemessagecontent');";
$query = mysql_query($addmessages) or die(mysql_error());
//SET USER AS NOT NEW USER
$newuservalue = '1';
$notnewuser = "UPDATE users SET new_user = $newuservalue WHERE id = $userid" ;
$query2 = mysql_query($notnewuser) or die(mysql_error());
} elseif ($new_user == 1) {};
UPDATE FULL CODE::
<?php
session_start();
include "../includes/db_connect.php";
///profile/index.php
if($_SESSION['id'])
{
$username = $_SESSION['username'];
$userid = $_SESSION['id'];
//WRITE FIRST TIME LOGIN INFORMATION TO DATABASE
$sql="SELECT new_user from `users` WHERE `id`= $userid ";
$res=mysql_query($sql) or die(mysql_error());
while($row=mysql_fetch_assoc($res)) $new_user = $row['new_user'] ;
echo $new_user;
if ($new_user == 0) {
//SENT NEW USER WELCOME MESSAGE
$adminid = '9';
$welcomemessagetitle = 'Welcome to Escorvee';
$welcomemessagecontent = 'Hello and welcome';
$addmessages = "INSERT into `user_messages`(`to_user`,`from_user`,`title`,`content`)
VALUES ('$userid','$adminid','$welcomemessagetitle','$welcomemessagecontent');";
$query = mysql_query($addmessages) or die(mysql_error());
//SET USER AS NOT NEW USER
$newuservalue = '1';
$notnewuser = "UPDATE users SET new_user = $newuservalue WHERE id = $userid" ;
$query2 = mysql_query($notnewuser) or die(mysql_error());
} elseif ($new_user == 1) {};
}
?>
It the variable $new_user is the value 1 then your code won't be executed so I would guess one of the following applies:
$new_user isn't 1.
The database is being modified from another location.
Your script is being called twice.
To work out which you will have to provide more information in your question.