I have a very simple use case where I am checking if a certain value is present in the table and it always seems to fail.This is my php code.
<?php
include "config.php";
$con = mysqli_connect($server_name,$mysql_user,$mysql_pass,$dbname);
if(!$con)
{
echo "Connection Error".mysqli_connect_error();
}
else{
//echo "";
}
$device_id = $_POST["device_id"];
$check = "SELECT magazine_id FROM registered_buyers WHERE device_id = $device_id";
$rs = mysqli_query($con,$check);
if(mysqli_num_rows($con,$rs) == 0)
{
$jsonarray = $_POST["jsonarray"];
echo "This will be inserted".$jsonarray;
}else
{
echo "User already registered";
}
?>
Can anyone please point out my mistake.Any help or suggestion is welcome.Thank you.
You can try to follow this code.
<?php
include "config.php";
$con = mysqli_connect($server_name,$mysql_user,$mysql_pass,$dbname);
if(!$con){
echo "Connection Error".mysqli_connect_error();
}
$device_id = $_POST["device_id"];
$check = "SELECT magazine_id FROM registered_buyers WHERE device_id = ".$device_id;
$rs = mysqli_query($con, $check);
if(mysqli_num_rows($rs) == 0){
$jsonarray = $_POST["jsonarray"];
echo "This will be inserted".$jsonarray;
}else{
echo "User already registered";
}
?>
since i dont have enough rep to add a comment, i will consider the device_id is string, if so try something like this:
"SELECT magazine_id FROM registered_buyers WHERE device_id = '$device_id'";
add '
Related
I have this php file and I want to echo one array that ticketnumber is located in it with json encode, in meaning I want to print only one array when isset POST the ticketnumber for this array, but I tried and every time I get error how to slove this problem in addition I used the loop (while) so is it correct for one array?
<?php
define('HOST', 'localhost');
define('USER', 'root');
define('PASS','');
define('DB', 'ala');
$con = mysqli_connect(HOST,USER,PASS,DB) or die ('unable to connect');
if ($_SERVER ['REQUEST_METHOD']=='POST') {
$ticketnumber = $_POST['ticketnumber'];
$sql = " SELECT * FROM contact WHERE ticketnumber = '$ticketnumber' ";
$res = mysqli_query($con, $sql);
$result = array();
while($get = mysqli_fetch_array($res))
{
array_push($result,array('ticketnumber' =>$get[0], 'subject' =>$get[1],'response' =>$get[2]));
}
if(isset($get)){
echo json_encode(array("responseticket"=>$result));
} else {
echo " error";
}
}
?>
You don't define variable $get .
I thinks there should be..
if(!empty($result)){
echo json_encode(array("responseticket"=>$result));
} else {
echo " error";
}
Try like this..
if(!empty($get)){
echo json_encode(array("responseticket"=>$result));
} else {
echo " error";
}
I have the following code to check if a row exists in MySQL:
<?php
if (!empty($_POST)) {
$code = $_POST['code'];
mysql_connect("$dbhost","$dbuser","$dbpass");
mysql_select_db("$dbname");
$result = mysql_query("SELECT 1 FROM files WHERE id='$code' LIMIT 1");
if (mysql_fetch_row($result)) {
echo 'Exists';
} else {
echo 'Does not exist';
}
}
?>
This works fine. But I need to change it a bit. I have the following fields:
id, title, url, type. When someone uses the code above ^ to check if a row exists, I need a variable to get the url from the same row, so I can redirect the user to there.
Do you have any idea how I can do that?
Thanks in advance! :)
Try this:
<?php
if (!empty($_POST)) {
$code = $_POST['code'];
mysql_connect("$dbhost","$dbuser","$dbpass");
mysql_select_db("$dbname");
$result = mysql_query("SELECT * FROM files WHERE id=" . $code . " LIMIT 1");
if (mysql_num_rows($result) > 0) {
while($rows = mysql_fetch_array($result)) {
echo 'Exists';
$url = $rows['url'];
}
} else {
echo 'Does not exist';
}
}
?>
It is quite simple. I think you don't show any effort to find the solution by yourself.
<?php
if (!empty($_POST)) {
$code = $_POST['code'];
mysql_connect("$dbhost","$dbuser","$dbpass");
mysql_select_db("$dbname");
$result = mysql_query("SELECT url FROM files WHERE id='$code' LIMIT 1");
if ($result) {
$url = mysql_fetch_row($resultado);
} else {
echo 'Does not exist';
}
}
<?php
$sql_query = "SELECT * FROM test WHERE userid ='$userid'";
$result1 =mysql_query($sql_query);
if(mysql_num_rows($result1)>0){
while($post = mysql_fetch_array($result1))
{
$url = $post['url'];
}
}
?>
If mysql_num_rows($result1)>0 it means row is existed fir the given user id
I'm trying to to test to see if an email address exists in my database by running a query check.
I can connect to the database fine.
However no matter what, even if the email exists it returns "doesn't exist".
<?php
//----------------------------------------------------------------------------------//
//Setup
require_once('SB_Constants.php');
//----------------------------------------------------------------------------------//
//Connect to the database
//----------------------------------------------------------------------------------//
$connection = mysqli_connect(DATABASE_HOST, SAVE_USERNAME, SAVE_PASSWORD, DATABASE_NAME);
// check the connection was successful
if (mysqli_connect_errno($connection)) {
header('HTTP/1.0 500 Internal Server Error', true, 500);
die(FailedToAccessDatabase . ". Failed to connect to Database");
} else {
echo "Connection Success!";
}
//Query Check
$assessorEmail = mysqli_query($connection, "SELECT email_address FROM assessorID WHERE email_address = 'ryan#ablah.com'");
if (mysqli_num_rows($query_identifier) == 0) {
die(UnregisteredAssessor . ". Doesn't Exist");
} else {
// Exists
echo "Exists getting ace id.";
//Get the assessor ID
$result = mysqli_query($connection, "SELECT ace_id FROM assessorID WHERE email_address = 'ryan#blah.com'");
echo $result;
}
/* close connection */
mysqli_close($connection);
?>
Any ideas of the problem? :)
Various mistakes. Fix:
$assessorEmail = mysqli_query($connection, "SELECT ace_id,email_address FROM assessorID WHERE email_address = 'ryan#ablah.com'");
if (mysqli_num_rows($assessorEmail) == 0) {
die(UnregisteredAssessor . ". Doesn't Exist");
} else {
// Exists
echo "Exists getting ace id.";
//Get the assessor ID
$result = mysqli_fetch_assoc($assessorEmail);
echo $result['ace_id'];
}
Your problem is mysqli_num_rows($query_identifier) is accessing an undefined variable instead of $assessorEmail.
Additionally, you only need one query if you just want the ace_id:
$assessorEmail = mysqli_query($connection, "SELECT ace_id FROM assessorID WHERE email_address = 'ryan#ablah.com'");
If mysqli_num_rows($assessorEmail) returns a row, than the email exists and you already have the ace_id
while(mysqli_fetch_assoc($assessorEmail) = $row) {
echo $result['ace_id'];
}
<?php
include 'config.php'; //connect to db
if(isset($_REQUEST["pwd"]) && isset($_REQUEST["name"])) {
$password = $_REQUEST['pwd']; //pass from previous page
$name = $_REQUEST['name']; //pass from previous page
$checkUserPass = mysql_query("SELECT * FROM validPersonnel WHERE Passkey = '$password' and Name = '$name'", $conn); //check if the user exist
if(mysql_num_rows($checkUserPass) == 1) {
$personnelId = mysql_query("SELECT PersonnelID FROM validPersonnel WHERE Passkey = '$password' and Name = '$name'", $conn); //query user id
while($row = mysql_fetch_assoc($personnelId)) {
echo $row['PersonnelD']; // print user id
}
mysql_close($conn);
//echo "<br/><br/>";
//echo "<script>alert('Logged In.')</script>";
//header("Refresh: 1; url=profile/profile.php?id="'.$id.');
//header('Refresh: 1; url=test.php?id=$personnelId');
} else {
echo "<br/><br/>";
echo "<script>alert('Wrong Password.')</script>";
header('Refresh: 1; url=personnelselect.php');
}
}
?>
i cannot echo the $row['PersonnelD'] the page shows blank. i cannot understand where did i go wrong. this page quesion have been solved
Looks like you have mistake in code:
echo $row['PersonnelD'];
shouldn't it be following?
echo $row['PersonnelID'];
check the mysql_fetch_assoc() function may be its parameter is empty so it can't enter the while loop
Try to debug and check the values came in the variables using var_dump() function. Ex: var_dump($row); in while loop.
In both your querys, you have
"SELECT * FROM validPersonnel WHERE Passkey = '$password' and Name = '$name'"
It should be:
"SELECT * FROM validPersonnel WHERE Passkey = '".$password."' and Name = '".$name."';"
PHP doesn't recognize the $var unless you close the quotes. The period adds the $var to the string.
I am trying to check if the session username matches the record in my database and if it does, I want to include a file.
This is my code
<?php
$username = $_SESSION['username'];
echo $username;
include('connect.php');
mysqli_select_db($connect,"persons");
$sql = "SELECT * FROM users WHERE sessionusername='$username'";
$r = mysqli_query($connect,$sql) or die(mysqli_error($connect));
$geez = mysqli_fetch_array($r);
if($geez)
{
include('check.php');
}
else
{
echo "error";
}
?>
The session username does not match the record in my database, yet the file is being included. Why?
OH, I FOUND THE ISSUE. IT IS CONSIDERING MY USERNAME TO BE ROOT...BUT WHEN I SAY ECHO $_SESSION['USERNAME'] IT IS CRAIG#CRAIG.COM..WHY SO>
<?php
$username = $_SESSION['username'];
echo $username;
include('connect.php');
mysqli_select_db($connect,"persons");
$sql = "SELECT sessionusername FROM users WHERE sessionusername='$username'";
$r = mysqli_query($connect,$sql) or die(mysqli_error($connect));
$geez = mysqli_fetch_array($r);
if($geez["sessionusername"]==$username)
{
include('check.php');
}
else
{
echo "error";
}
?>
You are simply testing whether the array $geez is empty or not. If the array has anything in it, you if($geez) will return true. To stop this behaviour, please see ceteras' answer, particularly this part:
if($geez["sessionusername"]==$username)
{
include('check.php');
}
I believe that's the only part that has changed.
Thanks,
James