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($db));
Related
I want to ask about checking a value in the database: when the value exist it will echo an alert message and when not it will do an insert function.
I have try a code like this but it seems not work.
<?php
include("../../Connections/koneksi.php");
if(isset($_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
How to return a success/failure message after running the query and how to check for exist data in product_code data array before saving the data?
$query = mysql_query("INSERT INTO `produk2`
(product_code,product_name,product_desc,product_type,product_price,product_img,product_img_name)
VALUES ('$kod','$namaproduk','$spesifikasi','$jenis','$harga','$image','$name')");
}
I tried to run this code and doesn't seems to work. Any ideas? I've gone wrong?
$result=mysql_query($query);
if($result)
{
echo "<br/>Data uploaded.";
}
else
{
echo "<br/>Data not uploaded.";
}
You should use mysql_error() or mysql_errno().
See the manual
manual
Add if else condition to the return result of query.
$query = mysql_query("INSERT INTO `produk2`
(product_code,product_name,product_desc,product_type,product_price,product_img,product_img_name)
VALUES ('$kod','$namaproduk','$spesifikasi','$jenis','$harga','$image','$name')");
if($query)
{
echo "Data inserted";
}
else
{
echo "Sorry! Data not inserted";
}
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.
The following is my code that I have written for not inserting same data
I would like if the record exist in mysql then it should show me error message that the record already exist the else part should insert record to database but it not working
can any one help me plz
the help would be highly appreciated
function addcontact()
{
if(isset($_POST['addContact']))
{
$officeName = strip_tags($_POST['office_name']);
$contactName = strip_tags($_POST['contactName']);
$contactNo = strip_tags($_POST['contactNo']);
$digitalNo = strip_tags($_POST['digitalNo']);
$mobileNo = strip_tags($_POST['mobileNo']);
$check="SELECT * FROM contacts WHERE office_name = '$officeName'";
if(mysql_num_rows($check) != 0)
{
echo "Already in Exists<br/>";
}else
{
$sql = mysql_query("INSERT INTO contacts (office_name, contact_no,
digital_no, mobile_no) VALUES
('$contactName','$contactNo','$digitalNo','$mobileNo')") or die(mysql_error());
if($sql)
{
header("Location: index.php?admin&done"); exit;
}
else
{
header("Location: index.php?admin&failed"); exit;
}
}
}
}
you did mistake here.
$check="SELECT * FROM contacts WHERE office_name = '$officeName'";
if(mysql_num_rows($check) != 0)
{
echo "Already in Exists<br/>";
}
just add mysql_query like
$check=mysql_query("SELECT * FROM contacts WHERE office_name = '$officeName'");
if(mysql_num_rows($check) != 0)
{
echo "Already in Exists<br/>";
}
or you can also use like
$name=$_POST['username'];
$q="select * from login where name='$name' ";
$rs=mysql_query($q);
if(mysql_fetch_row($rs)>0)
{
echo "already exist";
}
else
{
$msg="done";
}
Add the ON Duplicate KEY Update. This way you don't need to check if the record already exists, which means you don't need an extra select query just to check. If it exists, nothing happens.
INSERT INTO contacts (office_name, contact_no, digital_no, mobile_no)
VALUES ('$contactName','$contactNo','$digitalNo','$mobileNo')
ON DUPLICATE KEY UPDATE office_name = office_name
And set the office_name to be the primary key or a unique index.
There is missing one step, your first query is not executed, please try this:-
function addcontact()
{
if(isset($_POST['addContact']))
{
$officeName = strip_tags($_POST['office_name']);
$contactName = strip_tags($_POST['contactName']);
$contactNo = strip_tags($_POST['contactNo']);
$digitalNo = strip_tags($_POST['digitalNo']);
$mobileNo = strip_tags($_POST['mobileNo']);
$check= mysql_query("SELECT * FROM contacts WHERE office_name = '{$officeName}'");
if(mysql_num_rows($check) != 0)
{
echo "Already in Exists<br/>";
}else
{
$sql = mysql_query("INSERT INTO contacts (office_name, contact_no,
digital_no, mobile_no) VALUES
('$contactName','$contactNo','$digitalNo','$mobileNo')") or die(mysql_error());
if($sql)
{
header("Location: index.php?admin&done"); exit;
}
else
{
header("Location: index.php?admin&failed"); exit;
}
}
}
}
you can handle it from database side. write a stored procedure such a way that first check weather the record is in database or not if exist then ignore it and get back the text "Record already exist", if not exist then insert it to table. use conditional statements in mysql.
when the info is successfully inserted, it's displaying the error message and saying that it's a duplicate entry for a primary key...I can't figure out why!
<?
$email=$_POST['email'];
$pw=$_POST['pw'];
mysql_connect('***','***','***');
#mysql_select_db('***') or die('Unable to select database');
$query = "INSERT INTO test_table VALUES ('','$email','$pw')";
mysql_query($query) or die(mysql_error());
if(mysql_query($query))
{
echo 'success';
}
else
{
echo 'failure' .mysql_error();
}
mysql_close();
?>
You are executing the query twice: first, in mysql_query($query) or die(mysql_error()); and second, in if(mysql_query($query)). So the second time the query executes the record is already there and thus the insertion fails.
You are executing same query twice.
$query_result = mysql_query($query) or die(mysql_error());
if ($query_result) {
echo 'success';
} else {
echo 'failure' . mysql_error();
}
Write this way, hope it will work.
Just delete this code from your php script and it will be fine.
if(mysql_query($query))
{
echo 'success';
}
else
{
echo 'failure' .mysql_error();
}
You make it running error twice in a time. You can also use mysql_affected_rows() to make sure the data is executed in database server. Return a string type value.
<?
$email=$_POST['email'];
$pw=$_POST['pw'];
mysql_connect('***','***','***');
#mysql_select_db('***') or die('Unable to select database');
$query = "INSERT INTO test_table VALUES ('','$email','$pw')";
if(mysql_query($query))
{
echo 'Data executed : '.mysql_affected_rows();
}
else
{
echo 'failure' .mysql_error();
}
mysql_close();
?>
Good luck and let me know the result.
$email=$_POST['email'];
$pw=$_POST['pw'];
$alerts = array();
if (trim($_POST['email']) == '') {
$alerts[] = "<div class='alert alert-danger' role='alert'> Enter your Email! </div>"; }
if (trim($_POST['pw']) == '') {
$alerts[] = "<div class='alert alert-danger' role='alert'> Enter your PW! </div>"; }
if (!count($alerts)) {
$query = "INSERT INTO test_table (email, pw) VALUES ('".$email."', '".$pw."')";
mysqli_query($this->conn, $query) or die (mysqli_connect_error());
return ['success' => true];
} else {
return ['success' => false, 'alert_m' => implode($alerts)."<br>"];
}