I want to insert foreign key value in table.I have two tales employee(employee_id) and attendance. Here employee_id is foreign key in attendance table.
I try a lot but value is not inserted.
Here is my code
if(isset($_POST['submit']))
{
$date = date('Y-m-d',strtotime($_POST['daily_date']));
$in = $_POST['daily_in'];
$l_out = $_POST['lunch_out'];
$l_in = $_POST['lunch_in'];
$out = $_POST['daily_out'];
$emp_remarks = $_POST['remarks'];
$sql = "INSERT INTO attendance (atten_id,daily_date,daily_in,lunch_out,lunch_in,daily_out,remarks,employee_id)
VALUES('NULL','$date','$in','$l_out','$l_in','$out','$emp_remarks','".$_REQUEST['employee_id']."')";
$res = mysql_query($sql);
if ($res > 0) {
echo "inserted";
}
If I run below code then
if(isset($_POST['submit']))
{
$date = $_POST['daily_date'];
$in = $_POST['daily_in'];
$l_out = $_POST['lunch_out'];
$l_in = $_POST['lunch_in'];
$out = $_POST['daily_out'];
$emp_remarks = $_POST['remarks'];
if(isset($_REQUEST['employee_id']))
{
echo "Employee Id" .$_REQUEST['employee_id'];
}
else {
echo "Smoething went wrong";
}
$sql = "INSERT INTO attendance (atten_id,daily_date,daily_in,lunch_out,lunch_in,daily_out,remarks,employee_id)
VALUES
('NULL','$date','$in','$l_out','$l_in','$out','$emp_remarks','".$_REQUEST['employee_id']."')";
its gives
Smoething went wrong not inserted error
Before inserting try this , try this to check whether value was passed or not.
isset($_REQUEST['employee_id'])
{ echo "Employee Id" .$_REQUEST['employee_id'];
} else {
echo "Smoething went wrong";
}
Related
<?php include('dbcon.php');
include('header.php');
//variable
$clientID='';
$billAmount='';
$arrear='';
$monthlyBill='';
$surcharge='';
//data add
$clientID=$_POST['clientID'];
$sqll = "SELECT `client`.`clientId` , (
`arrear` + `surcharge` + `monthlyBill`) AS 'billamount', (
FROM `billifno`
JOIN `client` ON `billifno`.`clientID` = `client`.`clientID` WHERE `billifno`.`clientID`='".$_POST["clientID"]."'";
$result = mysqli_query($con,$sqll);
// if ($result->num_rows > 0)
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
$clientId = $row['clientId'];
$arrear = $row['arrear'];
$monthlyBill= $row['monthlyBill'];
$billAmount = $row['billAmount'];
$surcharge = $row['surcharge'];
}
//echo $billAmount;
$sql = "INSERT INTO `billIfno`
(`clientID`, `billAmount`,`arrear`, `monthlyBill`,`surcharge`) VALUES
('$clientID','$billAmount','$arrear','$monthlyBill','$surcharge')";
if ($con->query($sql)=== true)
{
echo "Recorded!!!";
}
else
{
echo "Not Recorded !!!";
}
$con->close();
?>
Hare I am taking some value and process it using query then again insert into billInfo table but it not inserting the new values like
If I echo billAmonu it print the correct value that is calculated by query bt that value is not inserting into database.
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
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
What am i doing wrong here?
(Know its mysql)
this else is not working it look like. user can insert same modul id many times.
$besvarelse = $_GET['besvarelse'];
$modulid = $_GET['modulid'];
$username = $_GET['username'];
$tilkobling = kobleTil();
if (!empty($besvarelse) ) {
$insert = mysql_query("INSERT INTO oppgave (besvarelse, modulid, username) VALUES ('$besvarelse', '$modulid','$username')");
if($insert) {
echo "<h1>Levering OK</h1><br>";
}
}
else {
die("<h1>Du har lever før</h1>");
}
?>
Try with this and post the errors that PHP shows, please:
//Assuming that connection to mysql server its done somewhere with mysql_connect()
error_reporting(E_ALL); ini_set('display_errors', 1);
explode($_GET);
echo "<pre>";
print_r($_GET); //Just to see what's on the variables...
echo "</pre>;
// $tilkobling = kobleTil(); Need to explain this line
if (!empty($besvarelse) ) {
$sql = "INSERT INTO oppgave (besvarelse, modulid, username) VALUES ('$besvarelse', '$modulid','$username')";
if($insert = mysql_query($sql))
{
echo "<h1>Levering OK</h1><br>";
}
else {
die("<h1>Du har lever før</h1>");
}
} // you were missmatching the IF closing bracket
?>
try to check for exist rows like
$besvarelse = $_GET['besvarelse'];
$modulid = $_GET['modulid'];
$username = $_GET['username'];
$tilkobling = kobleTil();
if (!empty($besvarelse) ) {
// check exist moduleid
$check = mysql_query("select * from oppgave where modulid = '$modulid'");
$conut_rows= mysql_num_rows($check);
if($conut_rows > 0) {
echo "module id already exist";
}
else {
$insert = mysql_query("INSERT INTO oppgave (besvarelse, modulid, username) VALUES ('$besvarelse', '$modulid','$username')");
if($insert) {
echo "<h1>Levering OK</h1><br>";
}
else {
die("<h1>Du har lever før</h1>");
}
}
}
I'm working on a project where a user can click on an item. If the user clicked at it before , then when he tries to click at it again it shouldn't work or INSERT value on the DB. When I click the first item(I'm displaying the items straight from database by id) it inserts into DB and then when I click at it again it works(gives me the error code) doesn't insert into DB. All other items when I click at them , even if I click for the second, third, fourth time all of it inserts into DB. Please help guys. Thanks
<?php
session_start();
$date = date("Y-m-d H:i:s");
include("php/connect.php");
$query = "SELECT * FROM test ORDER BY `id` ASC LIMIT 3";
$result = mysql_query($query);
if (isset($_SESSION['username'])) {
$username = $_SESSION['username'];
$submit = mysql_real_escape_string($_POST["submit"]);
$tests = $_POST["test"];
// If the user submitted the form.
// Do the updating on the database.
if (!empty($submit)) {
if (count($tests) > 0) {
foreach ($tests as $test_id => $test_value) {
$match = "SELECT user_id, match_id FROM match_select";
$row1 = mysql_query($match)or die(mysql_error());
while ($row2 = mysql_fetch_assoc($row1)) {
$user_match = $row2["user_id"];
$match = $row2['match_id'];
}
if ($match == $test_id) {
echo "You have already bet.";
} else {
switch ($test_value) {
case 1:
mysql_query("UPDATE test SET win = win + 1 WHERE id = '$test_id'");
mysql_query("INSERT INTO match_select (user_id, match_id) VALUES ('1','$test_id')");
break;
case 'X':
mysql_query("UPDATE test SET draw = draw + 1 WHERE id = '$test_id'");
mysql_query("INSERT INTO match_select (user_id, match_id) VALUES ('1','$test_id')");
break;
case 2:
mysql_query("UPDATE test SET lose = lose + 1 WHERE id = '$test_id'");
mysql_query("INSERT INTO match_select (user_id, match_id) VALUES ('1','$test_id')");
break;
default:
}
}
}
}
}
echo "<h2>Seria A</h2><hr/>
<br/>Welcome,".$username."! <a href='php/logout.php'><b>LogOut</b></a><br/>";
while ($row = mysql_fetch_array($result)) {
$id = $row['id'];
$home = $row['home'];
$away = $row['away'];
$win = $row['win'];
$draw = $row['draw'];
$lose = $row['lose'];
echo "<br/>",$id,") " ,$home, " - ", $away;
echo "
<form action='seria.php' method='post'>
<select name='test[$id]'>
<option value=\"\">Parashiko</option>
<option value='1'>1</option>
<option value='X'>X</option>
<option value='2'>2</option>
</select>
<input type='submit' name='submit' value='Submit'/>
<br/>
</form>
<br/>";
echo "Totali ", $sum = $win+$lose+$draw, "<br/><hr/>";
}
} else {
$error = "<div id='hello'>Duhet te besh Log In qe te vendosesh parashikime ndeshjesh<br/><a href='php/login.php'>Kycu Ketu</a></div>";
}
?>
Your problem is here :
$match = "SELECT user_id, match_id FROM match_select";
$row1 = mysql_query($match)or die(mysql_error());
while ($row2 = mysql_fetch_assoc($row1)) {
$user_match = $row2["user_id"];
$match = $row2['match_id'];
}
You are not checking it correctly. You have to check if the entry in match_select exists for the user_id and the match_id concerned. Otherwise, $match would always be equal to the match_id field of the last inserted row in your database :
$match = "SELECT *
FROM `match_select`
WHERE `user_id` = '<your_id>'
AND `match_id` = '$test_id'";
$matchResult = mysql_query($match)or die(mysql_error());
if(mysql_num_rows($matchResult)) {
echo "You have already bet.";
}
By the way, consider using PDO or mysqli for manipulating database. mysql_ functions are deprecated :
http://www.php.net/manual/fr/function.mysql-query.php
validate insertion of record by looking up on the table if the data already exists.
Simplest way for example is to
$query = "SELECT * FROM match_select WHERE user_id = '$user_id'";
$result = mysql_query($query);
if(mysql_num_rows($result) > 0)
{
// do not insert
}
else
{
// do something here..
}
In your form you have <select name='test[$id]'> (one for each item), then when you submit the form you are getting $tests = $_POST["test"]; You don't need to specify the index in the form and can simply do <select name='test[]'>, you can eventually add a hidden field with the id with <input type="hidden" value="$id"/>. The second part is the verification wich is not good at the moment; you can simply check if the itemalready exist in the database with a query