Not able to insert the data from url to database in php - php

final.php
Here I am trying to get the data from the url using GET method and trying to insert into the database. I was able to insert the data for first few rows after that the data is not inserted. Can anyone help me regarding this?
when I try to run the url: www.myii.com/app/final.php?name=123&glucose=3232...
the data is not inserting.
<?php
include("query_connect.php");
$name = $_GET['name'];
$glucose = $_GET['glucose'];
$temp = $_GET['temp'];
$battery = $_GET['battery'];
$tgs_a = $_GET['tgs_a'];
$tgs_g = $_GET['tgs_g'];
$heartrate = $_GET['heartrate'];
$spo2 = $_GET['spo2'];
$rr = $_GET['rr'];
$hb = $_GET['hb'];
$ina22 = $_GET['ina22'];
$accucheck = $_GET['accucheck'];
$isactive = $_GET['isactive'];
$address = $_GET['address'];
$deviceno = $_GET['deviceno'];
$sql_insert = "insert into query (name,glucose,temp,battery,tgs_a,tgs_g,heartrate,spo2,rr,hb,ina22,accucheck,isactive,address,deviceno) values ('$name','$glucose','$temp',$battery','$tgs_a','$tgs_g','$heartrate','$spo2','$rr','$hb','$ina22','$accucheck','$isactive','$address','$deviceno')";
mysqli_query($sql_insert);
if($sql_insert)
{
echo "Saving succeed";
//echo $date_time;
}
else{
echo "Error occured";
}
?>
Query_connect.php
This is my database config php file.
<?php
$user = "m33root";
$password = "me3i434";
$host = "localhost";
$connection = mysqli_connect($host,$user,$password);
$select = mysqli_select_db('miiyy',$connection);
if($connection)
{
echo "connection succesfull<br>";
}
else {
echo "Error";
}
?>

Make sure that all columns can contain NULL so that not filled fields will stay NULL instead of throwing an error.

Try below to see mysql error:
mysql_query($sql_insert);
echo mysql_error()

Try these
In SQL
Change the table name of query to some other name. Because "query" is reserved in SQL
In code
if (!mysqli_query($con,$sql_insert ))
{
echo("Error description: " . mysqli_error($con));
}
else
{
echo "Success";
}
Use mysqli_error() function

Related

How to update status in database if status is empty without submitting a form in php?

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.

MySQL INSERT in PHP no error feedback

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');

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

PHP certain part of code is not executing

Hi I am facing a unique issue with my PHP script.Here is my code. After writeToDB() is executed I dont see the echo ("<script> top.location.href=www.facebook.com</script>");
Can someone let me know why my script stops executing after writing to db?
<?php
function writeToDB($access_token,$uid,$username,$birthday,$gender,$age)
{
/* Database Connection */
$user = "xxxx";
$password = "xxxx";
$host = "xxxxxxxxxxxxxxxxxx";
//connect to database, where tsnames.ora is setup
$connect_obj = oci_connect($user, $password, $host);
if ($connect_obj) {
error_log("connected okay");
} else {
$err = OCIError();
echo "Oracle connection error " . $err[text];
return;
}
$select_query = "SELECT USER_ID FROM FBTABLE WHERE USER_ID= '$uid'";
$select_sql_stmt = oci_parse($connect_obj, $select_query);
//execute statement
try {
$r = oci_execute($select_sql_stmt, OCI_DEFAULT);
if (!$r) {
$p = oci_error($select_sql_stmt);
echo "Oci Execute error";
}
} catch (Exception $e) {
echo "<br>Failed to get database info" . $e->getMessage();
}
$user_id_in_db = null;
while (oci_fetch($select_sql_stmt)) {
$user_id_in_db = oci_result($select_sql_stmt, 'USER_ID');
}
// User already exists in db so update instead of insert
if ($user_id_in_db != null) {
$query ="UPDATE FBTABLE SET ACCESS_TOKEN='$access_token' WHERE USER_ID='$uid'";
} else {
$query = "INSERT INTO FBTABLE(ACCESS_TOKEN, USER_ID,USER_NAME,BIRTHDAY,GENDER,AGE)
VALUES
('$access_token','$uid','$username','$birthday','$gender','$age')";
}
//create sql statement
$sql_statement = oci_parse($connect_obj, $query);
//execute statement
try {
$r = oci_execute($sql_statement, OCI_DEFAULT);
if (!$r) {
$p = oci_error($sql_statement);
echo "Oci Execute error";
}
} catch (Exception $e) {
echo "<br>Failed to get database info" . $e->getMessage();
}
//Commit transaction
$committed = oci_commit($connect_obj);
//Test whether commit was successful. If error occurred, return error message
if (!$committed) {
$error = oci_error($conn);
echo 'Commit failed. Oracle reports: ' . $error['message'];
}
//close the connection
$oci_free_statement($sql_statement);
if (oci_close($connect_obj)) {
echo " oci connection not closed!!!";
}
//close the connection
$oci_free_statement($sql_statement);
}
?>
<html>
<body>
<?php
$access_token = $_GET['access_token'];
$uid = $_GET['uid'];
$username = $_GET['username'];
$birthday = $_GET['birthday'];
$gender = $_GET['gender'];
$age = $_GET['age'];
echo $username;
writeToDB($access_token,$uid,$username,$birthday,$gender,$age);
echo ("<script> top.location.href=www.facebook.com</script>");
?>
</body>
</html>
i think error is in $oci_free_statement($sql_statement); must be oci_free_statement($sql_statement); extra $ before oci_free_statement
http://php.net/manual/en/function.oci-free-statement.php
no any error show because of error_display is off
Your JavaScript code should be
echo ("<script> top.location.href='http://www.facebook.com';</script>");
It's happen because writeToDB() causes error. You don't see this error because error_display is off or error_reporting = 0
Also maybe you didn't install OCI8. So when you call oci_connect it will cause error.
Thanks.
you are not using quotes around the string:
www.facebook.com should be 'www.facebook.com'

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