How to use comaprision in this case - php

if(isset($_POST["request"])){
$email = $_POST['email'];
$p_name = $_POST['p_name'];
$noc = $_POST['noc'];
$year = $_POST['year'];
$get_email = "select email from book where email = '$email' ";
$run_email = mysqli_query($con,$get_email);
$check = mysqli_num_rows($run_email);
if($check==1){
echo "<script>alert('You Have Already Booked') </script>";
exit();
}
$get_name = "SELECT capacity from party where type_party='$p_name' ";
$run_name = mysqli_query($con,$get_name);
$checkk = $run_name > '$noc';
if(checkk){
echo "<script>alert('Out Of Bound') </script>";
exit();
}
How to Compare capacity from party table with that the value of $noc??
Please help me out. Thank You!

You have missed the $ in last condition.
<?php
$email = $_POST['email'];
$p_name = $_POST['p_name'];
$noc = $_POST['noc'];
$year = $_POST['year'];
$get_email = "select email from book where email = '$email' ";
$run_email = mysqli_query($con,$get_email);
$check = mysqli_num_rows($run_email);
if($check==1){
echo "<script>alert('You Have Already Booked') </script>";
exit();
}
$get_name = "SELECT capacity from party where type_party='$p_name' ";
$run_name = mysqli_query($con,$get_name);
$checkname = mysqli_num_rows($run_name );
//$checkk = $run_name > '$noc';
if($checkname > $noc){
echo "<script>alert('Out Of Bound') </script>";
exit();
}
?>

Just change you code like below.
$get_name = "SELECT capacity from party where type_party='$p_name' ";
$run_name = mysqli_query($con,$get_name);
$row_cnt = mysqli_num_rows($run_name);
$rows = mysqli_fetch_assoc($run_name );
// please use any one condition which meet your requirement.
$checkk = $run_name > $noc;
// or
$checkk = $rows['capacity'] > $noc;
// if it's not working try below
$checkk = $rows[0]['capacity'] > $noc;
if($checkk){
echo "<script>alert('Out Of Bound') </script>";
exit();
}
I get the number of results found from the query using "mysqli_num_rows" and then compare with $noc for proper result. Or else use the second condition to compare $noc with "$run_name['capacity']".

Related

alert shows entry updated succesfully but entry does'nt update in database

I am creating a updating form page. it shows no error but there is no updation in my database.
I already check whole the database row names but that doesnt help.
that code is a little bit long but please help.
<?php
if (isset($_POST['update'])) {
$update_id = $_GET['edit_form'];
$bill = $_POST['b_no'];
$naam = $_POST['name'];
$mobile_no = $_POST['mobile'];
$addres = $_POST['add'];
$detail = $_POST['p_detail'];
$p_img_name = $_FILES['p_img']['name'];
$p_img_type = $_FILES['p_img']['type'];
$p_img_size = $_FILES['p_img']['size'];
$p_img_tmp = $_FILES['p_img']['tmp_name'];
$prc = $_POST['price'];
$deposite = $_POST['d_amt'];
$remaning = $_POST['r_amt'];
$b_img_name = $_FILES['b_img']['name'];
$b_img_type = $_FILES['b_img']['type'];
$b_img_size = $_FILES['b_img']['size'];
$b_img_tmp = $_FILES['b_img']['tmp_name'];
$p_date = date('y-m-d');
move_uploaded_file($p_img_tmp, "images/Product/$p_img_name");
move_uploaded_file($b_img_tmp, "images/Bill/$b_img_name");
$update_query = "UPDATE new_entry SET bill_no='$bill', name='$naam',
mobile_no='$mobile_no', address='$addres', product_detail='detail',
product_image='$p_img_name', price='$prc', deposite_amt='deposite',
remaining='$remaning', bill_image='$b_img_name',Product_date='$p_date'
WHERE s_no='$update_id' ";
if(mysqli_query($conn, $update_query)) {
echo "<script> alert('Entry Updated Successfully') </script>";
echo "<script> window.open('view_entry.php','_self') </script>";
} else {
echo "cant Update Entry.." .mysqli_error($conn);
}
}
?>

depending on condition show error message in php

i have code like this
<?php
require('../config.php');
require_once($CFG->dirroot . '/user/editlib.php');
$errorMessage = '';
$successMessage = '';
if(isset($_SESSION['successMessage']))
{
$successMessage = $_SESSION['successMessage'];
unset($_SESSION['successMessage']);
}
if (isset($_POST['register'])) {
if(!preg_match("/^(?=.*[0-9])(?=.*[a-z])(\S+)$/i", $_POST['password']))
{
$errorMessage="don't allow spaces";
}
$errors = array();
$data = array();
$chk_sql = "SELECT * FROM {user} u where username = ?";
if (!empty($chk_sql) ) {
$errorMessage='Username already taken';
}
if(!$chk_username = $DB->get_record_sql($chk_sql, array($_POST['username'])) )
{
$secret = $_POST['secret'];
$access_code_sql = "SELECT * FROM {accesscode} WHERE random_no= ? and status=1";
if($chk_secret = $DB->get_record_sql($access_code_sql, array($secret)) )
{
if ( $chk_secret->used >= $chk_secret->number ) {
$errorMessage = "your access code limit completed..";
}
else
{
$cadminid = $chk_secret->cadmin_id;
$clientid = $chk_secret->clientid;
$DB->execute("UPDATE {accesscode} SET used = used+1 WHERE random_no = '$secret'");
$insert_record = new stdClass();
$insert_record->firstname = $_POST['firstname'];
$insert_record->lastname = $_POST['lastname'];
$insert_record->username = $_POST['username'];
$insert_record->secret = $secret;
$insert_record->password = password_hash($_POST['password'], PASSWORD_DEFAULT);
$insert_record->timecreated = time();
$insert_record->maildigest = $cadminid;
$insert_record->maildisplay = $clientid;
$insert_record->idnumber = 1;
$insert_record->mnethostid = 1;
$insert_record->confirmed = 1;
$insert_record->email = $_POST['email'];
if($result = $DB->insert_record('user', $insert_record))
{
$_SESSION['successMessage'] = "record created successfully";
header('Location: register.php');
}
else
$errorMessage = "error! can you please try again";
}
}
else
$errorMessage = "your access code is wrong..";
}
}
?>
so i want to write condition like another if condition
if ( $chk_secret->status='0' ) {
$errorMessage = "your access code deactivated..";
}
if not they can register
i tried..but i didn't get idea where i have to add that if..
before i have condition like if number>used it will show some error message like your accesscode limit completed
can anyone help me..
thanks in advance..
= is for value assignment
== is compare two operands
so you need to change
if ( $chk_secret->status='0' ) {
to
if ( $chk_secret->status=='0' ) {
UPDATE:
your query SELECT * FROM {accesscode} WHERE random_no= ? and status=1
which means it going to return only status == 1
you can check with number of rows returned is ZERO then through status zero error message.
Or else
Get rows only based on random_no exists and then check status key

Passing php variables through pages / sql

i have the following information displayed
<?php
$my_query="SELECT * FROM games";
$result= mysqli_query($connection, $my_query);
if (mysqli_num_rows($result) > 0)
while ($myrow = mysqli_fetch_array($result))
{
$description = $myrow["game_description"];
$image = $myrow["gamepic"];
$game_id = $myrow["game_id"];
$gamename = $myrow["game_name"];
echo "<div class='cover'>
</div>";
}
?>
as you can see i have created a game_details page which will display that specific Game_id when the image is clicked
im having trouble understanding how to pull the data out from that game_id in sql on the other page.
here is my attempt on the game_details page
<?php
if (!isset($_GET['$game_id']) || empty($_GET['game_id']))
{
echo "Invalid category ID.";
exit();
}
$game_id = mysqli_real_escape_string($connection, $_GET['game_id']);
$sql1 = "SELECT * games WHERE game_id={$game_id}'";
$res4 = mysqli_query($connection, $sql1);
if(!$res4 || mysqli_num_rows($res4) <= 0)
{
while ($row = mysqli_fetch_assoc($res4))
{
$gameid = $row['$game_id'];
$title = $row['game_name'];
$descrip = $row['game_description'];
$genre = $row['genretype'];
echo "<p> {$title} </p>";
}
}
?>
This attempt is giving me the "invalid category ID" error
Would appreciate help
There are a few issues with your code.
Let's start from the top.
['$game_id'] you need to remove the dollar sign from it in $_GET['$game_id']
Then, $row['$game_id'] same thing; remove the dollar sign.
Then, game_id={$game_id}' will throw a syntax error.
In your first body of code; you should also use proper bracing for all your conditional statements.
This one has none if (mysqli_num_rows($result) > 0) and will cause potential havoc.
Rewrites:
<?php
$my_query="SELECT * FROM games";
$result= mysqli_query($connection, $my_query);
if (mysqli_num_rows($result) > 0){
while ($myrow = mysqli_fetch_array($result))
{
$description = $myrow["game_description"];
$image = $myrow["gamepic"];
$game_id = $myrow["game_id"];
$gamename = $myrow["game_name"];
echo "<div class='cover'>
</div>";
}
}
?>
Sidenote for WHERE game_id='{$game_id}' in below. If that doesn't work, remove the quotes from it.
WHERE game_id={$game_id}
2nd body:
<?php
if (!isset($_GET['game_id']) || empty($_GET['game_id']))
{
echo "Invalid category ID.";
exit();
}
$game_id = mysqli_real_escape_string($connection, $_GET['game_id']);
$sql1 = "SELECT * games WHERE game_id='{$game_id}'";
$res4 = mysqli_query($connection, $sql1);
if(!$res4 || mysqli_num_rows($res4) <= 0)
{
while ($row = mysqli_fetch_assoc($res4))
{
$gameid = $row['game_id'];
$title = $row['game_name'];
$descrip = $row['game_description'];
$genre = $row['genretype'];
echo "<p> {$title} </p>";
}
}
?>
Use error checking tools at your disposal during testing:
http://php.net/manual/en/mysqli.error.php
http://php.net/manual/en/function.error-reporting.php
You want to be using $_GET['gameid'] as that's the parameter you passed.
You are calling for game_id when the link to go to game_details.php has the variable gameid. Either change the parameter in the link to game_id or call for gameid in your $_GET['$game_id'].
Also, as Fred -ii- said, take out the dollar sign in $_GET['$game_id']

Resource id #6 error, Not sure how to fix it

I keep getting a 'Resource id # 6' failure when submitting a script on my website. The code I'm using is the same type of code I use for registering for the website and that works but this script doesn't work at all. What my code does is send a booking request with the fields as shown to the database. I keep getting a Resource id#6 error , and I've googled what that is but I can't seem to figure out whats wrong. I am a beginner at php , so any tips on whats to look for to avoid a resource id # 6 error would be a lot of help
<?php
//$pattern="/^.+#.+/.com/";
//error_reporting(0);
if(isset($_POST["submit"])){
$Name_of_Person = $_POST['Name_of_Person'];
$Name_of_Group = $_POST['Name_of_Group'];
$room = $_POST['room'];
$How_Many_People = $_POST['How_Many_People'];
$Date_of_Booking = $_POST['Date_of_Booking'];
$End_time = $_POST['End_time'];
$Purpose = $_POST['Purpose'];
$Contact_Number = $_POST['Contact_Number'];
$Contact_Email = $_POST['Contact_Email'];
$Alcohol = $_POST['Alcohol'];
$Security = $_POST['Security'];
$Projector = $_POST['Projector'];
$Extra_Chairs = $_POST['Extra_Chairs'];
$Extra_Info = $_POST['Extra_Info'];
$Activated = '0';
$con = mysql_connect('localhost','root','test123') or die("couldn't connect");
mysql_select_db('bookerdb') or die("couldn't connect to DB");
//if(filter_var($email, FILTER_VALIDATE_EMAIL)){//(preg_match($pattern, $_POST['Contact_Email'])){
$query = mysql_query("SELECT * FROM `booking_table` WHERE Date_of_Booking='".$Date_of_Booking."' AND room='".$room."'");
$numrows = mysql_num_rows($query);
echo $query;
if($numrows==0){
$sql="INSERT INTO `booking_table` (Name_of_Person,Name_of_Group,room,How_Many_People,Date_of_Booking,End_time,Purpose,Contact_Number,Contact_Email,Alcohol,Security,Projector,Extra_Chairs,Extra_Info, Activated) VALUES ('$Name_of_Person','$Name_of_Group','$room','$How_Many_People','$Date_of_Booking','$End_time','$Purpose','$Contact_Number','$Alcohol','$Security','$Projector','$Extra_Chairs','$Extra_Info',$Activated)";
$result = mysql_query($sql);
if($result){
echo "Sent to be approved";
$redirect_page = '../ASC.php';
$redirect = true;
if($redirect==true){
header('Location: ' .$redirect_page);
}
}else{
echo "Failed";
}
}else{
echo"There is already a requested booking on that date & time";
$redirect_page = '../EAR.php';
$redirect = true;
if($redirect==true){
header('Location: ' .$redirect_page);
}
}
/*}else{
echo "error";
$redirect_page = '../EWF.php';
$redirect = true;
if($redirect==true){
header('Location: ' .$redirect_page);
}
}*/
}
?>
You have error in your second SQL query. You try to insert 14 values into 15 columns (in values you forgot $Contact_Email).
$sql="INSERT INTO `booking_table` (Name_of_Person,Name_of_Group,room,How_Many_People,Date_of_Booking,End_time,Purpose,Contact_Number,Contact_Email,Alcohol,Security,Projector,Extra_Chairs,Extra_Info, Activated) VALUES ('$Name_of_Person','$Name_of_Group','$room','$How_Many_People','$Date_of_Booking','$End_time','$Purpose','$Contact_Number','$Contact_Email','$Alcohol','$Security','$Projector','$Extra_Chairs','$Extra_Info',$Activated)";
Than remove echo $query from your code, line 30.
In $query isn't query, but mysql result object. You can't work with that by this way, you can't echo it.

how to check the exisitng data from database before inserting

This is my code and it does not work, it's not inserting into the database. Please help me fix this problem.
$orgexist = $_POST['orgName1'];
$_SESSION['id'] = $_POST['id'];
$orgid = $_POST['id'];
$orgnme = $_POST['orgName1'];
$orgdesc = $_POST['orgDesc'];
$orgcat = $_POST['cat'];
$orgdept = $_POST['coldept'];
$orgvis = $_POST['vision'];
$orgmis = $_POST['mision'];
//get the value of category from database
//echo $orgdept;
$dept = "SELECT `col_id`, `col_description` FROM `college` WHERE `col_description` = '$orgdept'";
$deptresult = mysql_query($dept);
while ($rows = mysql_fetch_array($deptresult)) {
$getcol = $rows['col_id'];
//echo $getcol;
}
$sqlorg = mysql_query("SELECT * FROM `organization`");
while ($orgrows = mysql_fetch_array($sqlorg)) {
//$dborgid = $orgrows['org_id'];
$dborgnme = $orgrows['org_name'];
}
if ($dborgnme == $orgexist) {
echo "<script type='text/javascript'>
alert('Organization Name Already Used by other Organization');
history.back();
</script>";
} else {
$orginsrt = mysql_query("INSERT INTO `organization`(`org_id`,`org_name`,`org_desc`,`category`,`vision`,`mission`,`col_id`,`image`) VALUES ('$orgid','$orgexist','$orgdesc','$orgcat','$orgvis','$orgmis','$getcol','$image')");
echo "<script type='text/javascript'>
alert('Proceed to next Step');</script>";
//require ('orgsignup.php');
header('Location:orgsignup2.php');
//echo "Not in the Record";
}
}
You're using deprected functions, replace mysql_query by mysqli_query.
For more references see http://php.net/manual/en/function.mysql-query.php

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