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, $sql)){
echo "Records inserted successfully.";
} else{
echo "Records inserted failed ";
}
}
}
}
?>
I want to check the value in my select function. when it exist in my table in will echo an alert in the page. I have try whit select count too but it not working
Related
I'm new to php.I'm trying to build a signup webpage in which if email entered doesn't exist it should insert the values entered.The code works fine and it returns successful when a new mail is entered.But the problem is when I check my database the new values are not inserted.Is there any mistake in my code?
Thanks in advance.
<?php
session_start();
if(isset($_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));
this is my php code. hire is a problem i use check box selection and insert data in my database. but it insert data by foreach loop. so when press ok if data successfully insert then for 3 selection it say
successfully register
successfully register
successfully register
but i want it say only once successfully register
<?php
$db=require "script_database/connect.php";
$query = "SELECT * FROM course";
$query1="select * from selection where student_id='1229CSE00241' and semester='FALL2015' ";
$key=mysql_query($query1);
if(mysql_num_rows($key)>0)
{
echo "you already selected courses for registration";
}
else if($_POST['buy']==''){
echo "<h2><center>You didn't select any courses</h2></center>";
}
else{
foreach($_POST['buy'] as $item) {
$query = "SELECT * FROM course WHERE id = $item
";
if ($r = mysql_query($query)) {
while ($row = mysql_fetch_array($r)) {
$student_id="1229CSE00241";
$id=$item;
$course_id=$row['course_id'];
$course_title=$row['course_title'];
$course_credits=$row['course_credits'];
$course_status=$row['course_status'];
$semester="FALL2015";
}
} else {
print '<p style="color: blue">Error!</p>';
}
{
$insert_query="insert into selection(student_id,semester,course_id,course_title,course_credits,course_status,date_time) values ('$student_id','$semester','$course_id','$course_title','$course_credits','$course_status',NOW())";
}
//here is my problem
//it repeat every time when insert data but i want to make it only once
if(mysql_query($insert_query))
{
echo "successfully register";
}
else
echo "problem show";
}
}?>
Before you start your foreach loop set a flag like:
$error = false;
Then in your loop
if(!mysql_query($insert_query))
{
$error =true;
}
And after the loop has closed
if($error){
echo "problem show";
}else{
echo "successfully register";
}
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'];
}
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.
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>"];
}