I have created this page, I have put dummy content into MySQL database, I have ran the same query in phpMyAdmin and it brings back data, yet on the web page it is bringing up "No Data".
Have I missed something? (obviously blanked out database details)
<?php
// Connection Info
$host ='localhost';
$user ='-----';
$pass ='-----';
$dbname ='-----';
$conn = new mysqli($host, $user, $pass, $dbname);
if($conn->connection_errno){
echo "Connection Failed" . $conn->connect_error;
exit();
}
?>
<?php
$query = "SELECT * FROM posts";
$result = $conn->query($query);
if($results->num_rows > 0){
while($row = $result->fetch_assoc()){
echo $row['title'];
}
echo"We Have Data";
} else {
echo"No Data";
}
?>
Change the following line
if($results->num_rows > 0){
to
if($result->num_rows > 0){
He typo mistake friend
Make following changes in to your code
From
if($results->num_rows > 0){
To
if($result->num_rows > 0){
Related
I have been trying to echo out database data through a while loop now for awhile but it doesn't work, I can't find where the issue is. I have tried echoing out data manually and that works just fine.
<?php $results = mysqli_query($con,"SELECT * FROM guestbook ORDER BY id DESC"); ?>
<?php while ($row = mysqli_fetch_assoc($results)) : ?>
<li>
<?php echo $row['message']; ?>
</li>
<?php endwhile ?>
First of all make sure you are connected to the database. I don't know if you omitted that on purpose or not. Also, check if the connection is established.
<?php
//Insert your server info here
$servername = "servername";
$username = "root";
$password = "root";
$dbname = "test_database";
// Create and check your connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM guestbook ORDER BY id DESC";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
echo '<li>' . row["message"] . '</li>';
}
} else {
// Your query produced 0 results
echo "Empty result!";
}
// Remember to close the connection
mysqli_close($conn);
?>
I have been working with this script for a while and I do not understand why it is giving me no results when the username and password exist. I have checked the capitalization and everything. I have attempted to edit my script to work with this link but alas no results. Here is a screenshot of the database user:. Thanks for your help!
$sql = "SELECT Status FROM Users WHERE Username = '".$_GET["username"]."' AND Password = '".$_GET["password"]."'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row or remove while loop if you wish
while($row = $result->fetch_assoc()) {
echo $row['Status'] ;
}
} else {
echo "0 results";
}
$conn->close();
Edit: I also tried this code, which gave me the error "Getting property of non-object on line 16 (which here is line 9):
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Status FROM Users WHERE Username = '".$_GET["username"]." AND Password = '".$_GET["password"]."'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row or remove while loop if you wish
while($row = $result->fetch_assoc()) {
echo $row['Status'] ;
}
} else {
echo "0 results";
}
$conn->close();
?>
Edit: I did fix my SQL injections problem, thanks for the advice!
I tried your code as followed, Its working on my machine. Can you make sure of the $conn object creation. Also please enable the errors if not to see whats going wrong.
$conn = new mysqli('localhost','user','password','db');
$sql = "SELECT Status FROM Users WHERE Username = '".$_GET["username"]."'
AND Password = '".$_GET["password"]."'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row or remove while loop if you wish
while($row = $result->fetch_assoc()) {
echo $row['Status'] ;
}
} else {
echo "0 results";
}
$conn->close();
Im unable to find out the solution , please help. below is the code. Thanks in advance
<?php
require_once('connect.php');
$sql = "select * from projet";
$result = $conn->query($sql);
$xml = new SimpleXMLElement('<xml/>');
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$mydata = $xml->addChild('mydata');
$mydata->addChild('Id',$row['idProjet']);
}
} else {
echo "0 results";
}
$conn->close();
header ("Content-Type:text/xml");
echo($xml->asXML());
?>
and the file connect.php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "mtocrowdrise";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "connected succesfully";
Meanwhile i keep getting this error:
This page contains the following errors:
error on line 1 at column 1: Document is empty
Below is a rendering of the page up to the first error.
You shouldn't be writing/outputting any HTML to the page when said page is being cast as an XML Document (header ("Content-Type:text/xml");).
Remove echo "connected succesfully"; from connect.php.
You'll also (eventually) get the same error if:
...
} else {
echo "0 results";
}
...
header ("Content-Type:text/xml");
satisfies. So you should only cast the document to XML if there are no errors and there is actually some XML to display.
Something like the following would only set the Document to XML if there are results to display (per your original code):
require_once('connect.php');
$sql = "select * from projet";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
$xml = new SimpleXMLElement('<xml/>');
// output data of each row
while($row = $result->fetch_assoc()) {
$mydata = $xml->addChild('mydata');
$mydata->addChild('Id',$row['idProjet']);
}
header ("Content-Type:text/xml");
echo($xml->asXML());
} else {
echo "0 results";
}
$conn->close();
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();
I am trying to query data from a table using the following script:
//connect file
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//connect file included above - ECHO tested and it is connected as $conn
$sql = "SELECT * FROM userInfo";
$results = $conn->query($sql);
if (!$results) {
printf("Errormessage: %s\n", $conn->error);
exit;
} else {
echo $row['username'];
}
UPDATE --
It now no longer tries to throw an error and seems to go to the else section; however, no echo - and the spelling is correct this time and the column is filled.
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo $row["username"];
}
} else {
echo "0 results";
}
This now returns results. Thank you #Fred -ii- especially for your help on this.
Also thanks #jjczopek for the error checking advice!