PHP (mysqli) to find single value [duplicate] - php

This question already has answers here:
Single result from database using mysqli
(6 answers)
Closed 1 year ago.
I know how to search using mysqli to find if there is at least one match. How do I go about retrieving another value from a found row. It must just require changing 1 or 2 things.
Imagine I have the following DB:
ID | emailaddress | password
1 | dummy#email.com | DUMB1PASS
2 | second#email.com | DUMB2Pass
I can use the following code to verify if the the email address "dummy#email.com" exists. How would I look up the password associated with that row (i.e. the row containing dummy#email.com).
$email = "dummy#email.com";
$servername = "correct"; $username = "correct"; $DBpass = "correct"; $dbname = "correct";
$conn = new mysqli($servername, $username, $DBpass, $dbname);
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT emailaddress FROM registration WHERE emailaddress = '$email'";
$result = $conn->query($sql);
if ($result->num_rows > 0)
{
echo "we found a match";
}
else
{
echo "we did not find a match";
}
I imagine I can do something like:
$email = "dummy#email.com";
$servername = "correct"; $username = "correct"; $DBpass = "correct"; $dbname = "correct";
$conn = new mysqli($servername, $username, $DBpass, $dbname);
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT password FROM registration WHERE emailaddress = '$email'";
$result = $conn->query($sql);
if ($result->num_rows > 0)
{
echo "the value is '$result'";
}
else
{
echo "we did not find a match";
}
However, it produces this error:
Catchable fatal error: Object of class mysqli_result could not be converted to string in MYPAGE on line XX.
I think that the cause of this is likely that $result is an array or something. However, I don't know enough about sql/php to know if that is the problem, or how to pull the result from it if it is the case.
I'd really appreciate any help.

You need to fetch the row first, try this:
if ($result->num_rows) {
$row = $result->fetch_row();
echo 'the value is: ', $row[0];
}

Related

Retrieving value from database and checking it with PHP [duplicate]

This question already has answers here:
Single result from database using mysqli
(6 answers)
Closed 5 months ago.
I am trying to retrieve some value from a column in the database.
The idea is to retrieve the IP of the person, then check their country, after that check the database if the country is blacklisted or not. There seems to be something wrong in the code, may you please give an advise?
This is the code:
<?php
//Get IP and country name
$ip=$_SERVER['REMOTE_ADDR'];
$details = json_decode(file_get_contents("https://get.geojs.io/v1/ip/country/$ip.json"));
$country=$details->name;
$servername = "localhost";
$username = "My_Username";
$password = "My_password";
$dbname = "Database_name";
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT country_name FROM uni_country";
$result = $conn->query($sql);
if (strpos($result, $country) == true) {
echo $country ." " . "is banned";
} else {
echo "Your country is not banned";
}
$conn->close();
?>
When I run it, it shows "Your country is not banned" - when it should actually say that my country is banned (I added a country to the database to test this).
The issue seems to be with reading the data from the database.
If I do echo $country; --> It actually shows my country, so it is retrieving it correctly (from geojs.io). But the code is not pulling the data from the database and verifying it.
Update:
If I do echo $result; the page returns Error 500.
Update
tried this new code, now it's saying that the country is not in the database, although it is actually there.
<?php
$ip=$_SERVER['REMOTE_ADDR'];
$details = json_decode(file_get_contents("https://get.geojs.io/v1/ip/country/$ip.json"));
$country=$details->name;
$conn = mysqli_connect("localhost", "database", "password", "username");
$query = mysqli_query("SELECT * FROM `uni_country` WHERE country_name = '$country'");
if(mysqli_num_rows($query)>0) {
echo "Country Is in the database";
}
else {
echo "Country is not in the database";
}
?>
After trying for hours, this is the correct code:
<?php
$ip=$_SERVER['REMOTE_ADDR'];
$details = json_decode(file_get_contents("https://get.geojs.io/v1/ip/country/$ip.json"));
$country=$details->name;
// Create connection
$servername = "localhost";
$username = "MyUsername";
$password = "MyPassword";
$database = "MyDatabase";
$conn = new mysqli($servername, $username, $password, $database);
$sql = "SELECT * FROM uni_country WHERE country_name ='$country'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "You country is currently banned.";
}
} else {
echo "Your country is not banned!";
}
?>

SQL database connection in PHP successful, but I can't query it [duplicate]

This question already has answers here:
Reference - What does this error mean in PHP?
(38 answers)
Closed 6 years ago.
<?php
$servername = "localhost";
$username = "user";
$password = "pass";
$dbname = "test";
$tablename = "mapcoords";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
{
echo "Failure";
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$sql = "SELECT (lat, lng) FROM mapcoords";
$result = $conn->query($sql);
while($row = $result->fetch_assoc())
{
echo "ok";
}
$conn->close();
?>
Here is the code. So like I said, it can connect successfully, but the code won't successfully query. What's weird is that if I copy and paste the same code, which seems to be EXACTLY the same, it works. It makes no sense. I can't find a single difference between their code and my code besides the way they space things and the way I space things. Here is their code:
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT (lat, lng) FROM mapcoords";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo $row["lat"]. " " . $row["lng"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
The problem is that this query:
SELECT (lat, lng) FROM mapcoords
returns the folloowing error:
[21000][1241] Operand should contain 1 column(s)
You have to change the query to
SELECT lat, lng FROM mapcoords

If variable is in db then stop- if variable is not- then enter it

i have been trying to get this script done for a while now - im kind of new to php and mysql but i have been trying to get this to check the db for the username and then if the username exists - stop checking the db and if it doesn't exists add it to the db.
here is my code:
//input from application
$test = "wheelsmanx";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT mainusername FROM CCCpro_test";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
if ($row["mainusername"] === $test) {
echo "User Name Already In Use.";
}if($row["mainusername"] !== $test){
echo "this statement";
[code that inserts into db i can do this part myself]
}
}
$conn->close();
} else {
echo "0 results";
}
$conn->close();
The problem with your code is that you do the INSERT of the new name inside an if statement that has confirmed the existence of that user already. In addition I think you messed up your SELECT statement by selecting all the users.
Look into INSERT ON DUPLICATE for a better way to do it, or revise your code as below.
$sql = "SELECT mainusername FROM CCCpro_test WHERE mainusername = $test";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "User Name Already In Use.";
}
else{ //no rows selected therefore the user doesn't exist
[code that inserts into db i can do this part myself]
}
$conn->close();
PLEASE READ I have somewhere to go so I am being lazy so I did not bind the $test variable therefore DO NOT copy and paste this code without updating it to bind the $test variable. Please read this post about PDO and variable binding to prevent SQL injection.
here is my full working code if anyone needs it - it uses the post method - from an html form .... in case some one needs to hack it to pieces for something else
well guys i appreciate all of your help :D but i have found an answer or a way around it i suppose- i thought of it all night and day on how i could make it work and i came up with this
$servername = "127.0.0.1";
$username = "TESTUSER";
$password = "TESTPASS";
$dbname = "TESTDB";
$testusername = $_POST['mainusername'];
$testpassword = $_POST['mainpassword'];
//input from application
$test = $_POST['mainusername'];
$test2 = "0";
//Count switch
$countswitch = "0";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql1 = "INSERT INTO CCCpro_test ( mainusername, mainpassword ) VALUES ('$testusername','$testpassword' )";
$sql = "SELECT mainusername FROM CCCpro_test";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
if ($row["mainusername"] === $test) {
echo "Im Sorry Username Already In Use";
$countswitch ++;
}
}
if($countswitch == $test2){
echo "User Name Registered";
$db_handle = mysql_connect($servername, $username, $password);
$db_found = mysql_select_db($dbname, $db_handle);
if ($db_found) {
$result1 = mysql_query($sql1);
mysql_close($db_handle);
}
}
if ($countswitch == 3){
echo "this";
}
} else {
echo "0 results";
}
$conn->close();

"mysql_num_rows expects parameter 1 to be resource" error when querying [duplicate]

This question already has answers here:
Can I mix MySQL APIs in PHP?
(4 answers)
Closed 7 years ago.
When I execute the following code, I gets this error:
Warning: mysql_num_rows() expects parameter 1 to be resource, object given in C:\wamp\www\my\myWork.php on line 53
Whats the wrong with this code?
<?php
if(isset($_POST['login']))
{
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Begin";
$uName = $_POST["txtUsername"];
$uPwd = $_POST["txtPwd"];
echo "Your Username is: ".$uName."<br>";
echo "Your password is: ".$uPwd."<br>";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if(!$conn)
die("Connection faild: " . mysqli_connect_error());
$sql = "SELECT firstname from myguests WHERE firstname = '$uName'";
$result = $conn->query($sql);
if(mysql_num_rows($result) > 0)
{
echo "You have a login";
$_SESSION['uname'] = $uName;
}
else
echo "You don't have a login";
}
?>
I believe that because you are using mysqli_connect that you need to use the mysqli_query($conn,$sql) notation.
Try this:
$result = mysqli_query($conn,$sql)
if(mysqli_num_rows($result) > 0)

mysql login username password [duplicate]

This question already has answers here:
Can I mix MySQL APIs in PHP?
(4 answers)
Closed 1 year ago.
$servername = "localhost";
$username = "csc4370FA14_18";
$password = "1db23";
$dbname = "csc4370FA14_18";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$username_login = $_POST["username"];
$password_login = $_POST["pw"];
$query2 = mysql_query("SELECT * FROM users WHERE name='$username_login'");
$numrow = mysql_num_rows($query2);
if ($numrow != 0) {
while ($row = mysql_fetch_assoc($query2)) {
$dbusername = $row['name'];
$dbpassword = $row['password'];
}
// Check to see if username and password match
if ($username_login==$dbusername && $password_login==$dbpassword) {
echo "You are in";
}
else {
echo "Sorry $username_login. Incorrect password!";
}
}
This is the code I am using to check if a user matches the password (same row) in a table.
I am getting the error:
Warning: mysql_query(): Access denied for user 'apache'#'localhost'
(using password: NO) in
/home/csc4370FA14_18/public_html/program/assignments/group
project3/login.php on line 14 Warning: mysql_query(): A link to the
server could not be established in
/home/csc4370FA14_18/public_html/program/assignments/group
project3/login.php on line 14 Warning: mysql_num_rows() expects
parameter 1 to be resource, boolean given in
/home/csc4370FA14_18/public_html/program/assignments/group
project3/login.php on line 15
I have not a clue why this might be incorrect as the login credentials, etc work fine. I think it has something to do with mysqli, but I don't have a very good grasp of this versus the mysql_* functions. I know for a fact this is the correct connect info.
Try this code.
$servername = "localhost";
$username = "csc4370FA14_18";
$password = "1db23";
$dbname = "csc4370FA14_18";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$username_login = $_POST["username"];
$password_login = $_POST["pw"];
$query2 = mysqli_query($conn,"SELECT * FROM users WHERE name='$username_login'");
$numrow = mysqli_num_rows($query2);
if ($numrow != 0) {
while ($row = mysqli_fetch_assoc($query2)) {
$dbusername = $row['name'];
$dbpassword = $row['password'];
}
// Check to see if username and password match
if ($username_login==$dbusername && $password_login==$dbpassword) {
echo "You are in";
}
else {
echo "Sorry $username_login. Incorrect password!";
}
}

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