I have a form, on isset function data is inserted in db also it returns a variable named id from db. i want to post this variable to next page.
here is code;
if (isset($_POST['preview'])){
echo $user = $_SESSION['ue'];
echo $title=$_POST['title'];
echo $dis=$_POST['dis'];
echo $a=$_POST['a'];
echo $b=$_POST['b'];
echo $c=$_POST['c'];
echo $d=$_POST['d'];
echo $timespan=$_POST['timespan'];
$sql="INSERT INTO survey (user, title, description, opta, optb,optc,optd,time) VALUES ('$user','$title', '$dis', '$a' , '$b', '$c', '$d','timespan')";
if (mysqli_query($con,$sql))
{
echo "Success";
}
else
{
echo "Error: " . mysql_error();
}
$id = mysqli_insert_id($con); //variable to send to next page
mysqli_close($con);
}
?>
Thanks in advance :)
You can send it not next page using GET attribute in url as:
header("Location: http://mydomain.com/myOtherPage.php?id=".$id);
In myOtherPage you can use:
if(isset($_GET['id']))
{
$idFromPreviousPage=$_GET['id'];
}
Related
I want to ask about checking a value in the database: when the value exist it will echo an alert message and when not it will do an insert function.
I have try a code like this but it seems not work.
<?php
include("../../Connections/koneksi.php");
if(isset($_POST['table-bordered'])){
$array=json_decode($_POST['table-bordered'],true);
foreach($array as $item) {
$sql = "SELECT * FROM wjm WHERE no_pol='".$item['no_pol']."',date='".$item['date']."',time='".$item['time']."'";
$query = mysqli_query($db,$sql);
if (mysqli_num_rows($query)> 0)
{
echo "Data sudah pernah di Input..!!";
}
else{
$sql = "INSERT INTO wjm (sloc,kode,nama,no_pol,id,date,time,netto,unit,uses,payroll) VALUES ('".$item['sloc']."', '".$item['kode']."', '".$item['nama']."', '".$item['no_pol']."', '".$item['id']."', '".$item['date']."', '".$item['time']."', '".$item['netto']."', '".$item['Unit']."', '".$item['uses']."', '".$item['payroll']."')";
if(mysqli_query($db, $sql)){
echo "Records inserted successfully.";
} else{
echo "Records inserted failed ";
}
}
}
}
?>
I want to check the value in my select function. when it exist in my table in will echo an alert in the page. I have try whit select count too but it not working
I'm new to php.I'm trying to build a signup webpage in which if email entered doesn't exist it should insert the values entered.The code works fine and it returns successful when a new mail is entered.But the problem is when I check my database the new values are not inserted.Is there any mistake in my code?
Thanks in advance.
<?php
session_start();
if(isset($_POST['signup'])){
include_once("db.php");
$email=strip_tags($_POST['emailid']);
$username=strip_tags($_POST['username']);
$password=strip_tags($_POST['password']);
if($email==NULL || $username== NULL || $password==NULL){
print "Missing one of the fields";
}
else{
$email=stripslashes($email);
$username=stripslashes($username);
$password=stripslashes($password);
$email=mysqli_real_escape_string($db,$email);
$username=mysqli_real_escape_string($db,$username);
$password=mysqli_real_escape_string($db,$password);
$query = "SELECT * FROM user WHERE email='$email'";
$result = mysqli_query($db,$query);
if($result && mysqli_num_rows($result) > 0 )
{
echo "Account already exists.Please login";
}
else{
$sql="INSERT INTO user (ID,email,username,password) VALUES
(NULL,'$email','$username','$password')";
if($sql)
{
echo "Account created successfully.";
}
else
{
echo "Error";
}
}
}
}
?>
You are not executing the insert query, it should look like:
$sql="INSERT INTO user (ID,email,username,password) VALUES
(NULL,'$email','$username','$password')";
$sql= mysqli_query($db,$sql); ///You are missing this
Change from:
$sql="INSERT INTO user (ID,email,username,password) VALUES
(NULL,'$email','$username','$password')";
if($sql)
{
echo "Account created successfully.";
}
To:
$sql="INSERT INTO user (ID,email,username,password) VALUES
(NULL,'$email','$username','$password')";
if(mysqli_query($db,$sql))
{
echo "Account created successfully.";
}
You need to execute the 2nd query ($sql)
$sql="INSERT INTO user (email,username,password) VALUES
('$email','$username','$password')";
if(mysqli_query($db,$sql))
{
echo "Account created successfully.";
}
Remove the null INSERT value it's not needed and should be auto generated if auto-incremental index.
execute the $sql statement a a MySQLi_query and then use the result of that in the IF statement.
Bonus: Use mysqli_error($db) to feed you back errors you will encounter, such as:
mysqli_query($db,$sql) or die("error: ".mysqli_error($db));
<?php
if(isset($_POST['sub']))
{
$mname=$_POST['sub'];
}
if(isset($_POST['pos']))
{
$pos=$_POST['pos'];
}
if(isset($_POST['rad1']))
{
$vis=$_POST['rad1'];
}
global $mname, $pos, $vis;
$q= "INSERT INTO subjects (menu_name, position, visible) VALUES ('$mname', '$pos', '$vis')";
$qs=mysql_query($q, $connection);
if($qs)
{
header("Location: content.php");
exit;
}
else{
echo mysql_error();
}
?>
I get sub, pos and rad1 from a form
my database's primary key is auto-incrementing but rows are empty
Where is the mistake?
It seems like your $_POST array is empty.
Do a
<?php
print_r($_POST);
exit();
?>
This way we can know if are you getting an empty $_POST.
Try this code it should be working properly
<?php
mysql_connect("localhost","root","");
mysql_select_db("test");
$mname='abc';
$pos='13';
$vis='true';
$q= "INSERT INTO subjects (`menu_name`, `position`, `visible`) VALUES ('".$mname."', '".$pos."', '".$vis."')";
$qs=mysql_query($q);
if($qs)
{
//header("Location: db.php");
exit;
}
else{
echo mysql_error();
}
?>
I'm trying to insert calculate values to Database.a and b from text field.this is my code.but it doesn't work.please help me
<?php
require('dbconnection.php');
$a=$_POST['a'];
$b=$_POST['b'];
class cal{
public function insert($a,$b){
if((($a)&&($b))!="")
{
$c=$a*$b;
$sql = "INSERT INTO cropplant (sbetweenQ ,sbetweenC ,sbetweenTC ) VALUES ('$a', '$b', '$c')";
$query = mysql_query($sql);
if(!$query)
{
echo '<script type="text/javascript">alert("DB Update error ! please Re-enter your details.");</script>';
}else
{
echo '<script type="text/javascript">alert("New value Added Successfuly.");</script>';
}
}else{
echo"please enter all";
}}}
?>
Your function basically does nothing. You fetch two post variables, and define a class. What you need to do is call the function in the class to process the two variables.
i have code for save 3 textbox in one field in databse
no problem when i am enter 3 textbox , but when i fill 1 textbox and press ok
save another textbox in database as blank
i want just take the textbox is fulled and ignore the textbox empty
this is my code
<?php
include("connect.php");
$expert_name = trim($_POST['expert_name']);
$expert_name2 = trim($_POST['expert_name2']);
$expert_name3 = trim($_POST['expert_name3']);
// this is for arabic language.
mysql_query("SET NAMES utf8");
// Insert data into mysql
$sql="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name')";
$sql2="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name2')";
$sql3="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name3')";
$result=mysql_query($sql);
$result2=mysql_query($sql2);
$result3=mysql_query($sql3);
// if successfully insert data into database, displays message "Successful".
if($result){
echo "Successful";
echo "<BR>";
// echo "<a href='formadd.php'>Back to main page</a>";
}
else {
echo "ERROR";
echo "<br>";
// this for print error in insert process
echo mysql_error();
echo "<a href='expert_add.php'><br>Please try again </a>";
}
//mysql_close($con);
?>
back to form add
Execute your sql query only the variable value not equal to empty.
try this,
$expert_name = trim($_POST['expert_name']);
$expert_name2 = trim($_POST['expert_name2']);
$expert_name3 = trim($_POST['expert_name3']);
// this is for arabic language.
mysql_query("SET NAMES utf8");
// Insert data into mysql
if ($expert_name != "") {
$sql = "INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name')";
$result = mysql_query($sql);
}
if ($expert_name2 != "") {
$sql2 = "INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name2')";
$result2 = mysql_query($sql2);
}
if ($expert_name != "") {
$sql3 = "INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name3')";
$result3 = mysql_query($sql3);
}
// if successfully insert data into database, displays message "Successful".
if ($result || $result2 || $result3) {
echo "Successful";
echo "<BR>";
// echo "<a href='formadd.php'>Back to main page</a>";
} else {
echo "ERROR";
echo "<br>";
// this for print error in insert process
echo mysql_error();
echo "<a href='expert_add.php'><br>Please try again </a>";
}
//mysql_close($con);
?>
back to form add
You should also check $result2 and $result3. I added that in this answer
try this
if ( !empty($_POST['expert_name']) ){
$sql="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name')";
$result=mysql_query($sql);
}
if ( !empty($_POST['expert_name2']) ){
$sql2="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name2')";
$result2=mysql_query($sql2);
}
if ( !empty($_POST['expert_name3']) ){
$sql3 ="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name3')";
$result3 =mysql_query($sql3 );
}
Then you might want to check if the variable is empty().
<?php
include("connect.php");
$expert_name = trim($_POST['expert_name']);
$expert_name2 = trim($_POST['expert_name2']);
$expert_name3 = trim($_POST['expert_name3']);
// this is for arabic language.
mysql_query("SET NAMES utf8");
// Insert data into mysql
if(!empty($expert_name)) {
$sql="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name')";
$result=mysql_query($sql);
}
if(!empty($expert_name2)) {
$sql2="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name2')";
$result2=mysql_query($sql2);
}
if(!empty($expert_name3)) {
$sql3="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name3')";
$result3=mysql_query($sql3);
}
// if successfully insert data into database, displays message "Successful".
if($result){
echo "Successful";
echo "<BR>";
// echo "<a href='formadd.php'>Back to main page</a>";
}
else {
echo "ERROR";
echo "<br>";
// this for print error in insert process
echo mysql_error();
echo "<a href='expert_add.php'><br>Please try again </a>";
}
Also note: You only check if $result is okay. If you only fill textbox 2 and leave 1 empty, the value of 2 it will get inserted but an error is shown.
I'd say your code need general review, but as it is for now you will have to do something like this each query:
if (!empty($expert_name2){
$result2=mysql_query($sql2)
}
But you should try to loop your queries in foreach rather than manually write every on query. And by the way:
if($result){
echo "Successful";
echo "<BR>";
// echo "<a href='formadd.php'>Back to main page</a>";
}
This code only return succes when 1st wuery success because you use $result which is set in 1st query only
The ID is probably NOT NULL AUTO_INCREMENT, so that won't accept NULL as value.
try sending blank value, such as:
$sql="INSERT INTO experts(id,expert_name) VALUES ('', '$expert_name')";
Also, build bulk insert, rather than multiple.
I will explain why, when you insert single insert into the database, the values being inserted, then, the DB engine flushes indexes (they written to disk), unless you have set delay_key_write=ALL in you my.cnf. Index flushing directly affects your db performance.
Please, check the reworked code out. The code adjusted for bulk insert, sql string escaping for security purposes and additional verification for post keys existence.
<?php
include("connect.php");
// this is for arabic language.
mysql_query("SET NAMES utf8");
$values = array();
$skipInsert = true;
$fields = array('expert_name', 'expert_name2', 'expert_name3');
$insert = "INSERT INTO experts(id,expert_name) VALUES ";
// Loop through predefined fields, and prepare values.
foreach($fields AS $field) {
if(isset($_POST[$field]) && !empty($_POST[$field])) {
$values[] = "('', '".mysql_real_escape_string(trim($_POST[$field]))."')";
}
}
if(0 < sizeof($values)) {
$skipInsert = false;
$values = implode(',', $values);
$insert .= $values;
}
if(false === $skipInsert) {
mysql_query($insert);
}
// if successfully insert data into database, displays message "Successful".
if($result){
echo "Successful","<BR>";
// echo "<a href='formadd.php'>Back to main page</a>";
} else {
echo "ERROR","<br>",mysql_error(),"<a href='expert_add.php'><br>Please try again </a>";
}
HTH,
VR
if(!empty($textbox1_value)) {
//DO SQL
}
You can repeat this for multiple boxes however you wish, the empty operator checks if its empty, so if its not empty the "//DO SQL" area will get run.