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I'm working on a simple comment section for a website I'm working on, but there seems to be something awry with my code. Here's the form:
<?php
if($_SESSION['authenticated']) { //// if not logged show registration form
echo "<form name='comment' action='addcomment.php' method='get'>";
echo "<textarea name='CommentText'></textarea>";
echo "<input type='submit' value='Submit' />";
echo "</form><br><br>";}
else {
echo "Please log in to post.";
}
$res=mysqli_query($db, "SELECT * FROM comments");
$row = mysqli_fetch_array($res);
while ($row) {
echo '<p>from</p>';
$row = mysqli_fetch_array($res);
}
?>
And here's the PHP/SQL to add the form data to my database table (the table's named comments)
<?php $dbHost = 'cust-mysql-123-17';
$dbUser = 'user';
$dbPass = 'mypass';
$dbName = 'ollie';
$db = mysqli_connect( $dbHost, $dbUser, $dbPass, $dbName ) or die("Cannot connect");
$CommentText = $_GET['CommentText'];
$email = $_SESSION['authenticated'];
$res=mysqli_query($db, "INSERT INTO 'comments' (email, CommentText) VALUES('$email','$CommentText')");
if ($res){echo"<script lang='javascript' type='text/javascript'>
alert('Post successful');
window.location = 'forum.php';
</script>";}
else{
echo "alert('Post Failed');";
}
?>
Whenever I try to run it it shows the "post failed" alert I set up in the last if statement. Help?
remove quotes from the tablename - if you want to enclose it, use backticks ` (key below escape)
$res = mysqli_query($db, "INSERT INTO comments (email, CommentText) VALUES('$email','$CommentText')");
about security, i recommend you to learn about mysqli::real_espace_string on php.net
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Closed 1 year ago.
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I am facing a problem when i am try to check user and password from database while login it keep
reply an error message :
Notice: Trying to get property 'num_row' of non-object in /Applications/XAMPP/xamppfiles/htdocs/studyact/login.php on line 27
User name or Password is incorrect, please check and try again.
i type user and password correct! enter image description here
php file :
<?php
//html
$user_staff = $_POST["user_staff"];
$pass_staff = $_POST["pass_staff"];
// Create connection
$servername = "localhost";
$username = "root";
$password = "";
$db ="studyact";
$con = new mysqli($servername, $username, $password,$db);
// Check connection
if ($con->connect_error) {
die("Connection failed: " . $con->connect_error);
}else{
$stmt =$con-> prepare("select * from loginstaff where user_staff = ?");
$stmt->bind_param("s",$user_staff);
$stmt->execute();
$stmtresult = $stmt->get_result();
if($stmtresult-> num_row > 0){
$data = $stmtresult-> fetch_assoc();
if($data["pass_staff"] === $pass_staff){
echo "<h2>Login Successfully</h2>";
}
else{
echo "<h2> Sorry User name or Password is incorrect.</h2>";
}
}else{
echo "<h2> User name or Password is incorrect, please check and try again.</h2>";
}
}
?>
You've got a typo on line 27
if($stmtresult->num_rows > 0)
mysqli_stmt::$num_rows — Returns the number of rows fetched from the server
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Closed 7 years ago.
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I have the next code, but it inserts two rows in the mysql database instead of one. Could ypu please take a look to the code?
Regards.
<?php
$name= $_POST['name'];
$password = $_POST['password'];
mysql_connect("localhost","username","mypass");
mysql_select_db("databaseName");
mysql_query($query ="insert into users(name,password) values ('$name','$password')");
if (mysql_query($query) === TRUE) {
echo "Record saved";
} else {
echo "Error";
}
?>
Don't call mysql_query() when you assign the $query variable. And remember to escape your data, since you're not using prepared statements.
mysql_connect("localhost","username","mypass");
$name = mysql_real_escape_string($_POST['name']);
$password = mysql_real_escape_string($_POST['password']);
$query ="insert into users(name,password) values ('$name','$password')";
if (mysql_query($query)) {
echo "Record saved";
} else {
echo "Error: " . mysql_error();
}
Dont use the mysql_query function twice,
you want to check it in if statement, then dont call it before if clause.
See this following code.
$query ="insert into users(name,password) values ('$name','$password')"
if (mysql_query($query) === TRUE) {
echo "Record saved";
} else {
echo "Error";
}
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I am trying to print out one column called Language1 from my Table that is called Mull, in a database called v6e.
At the moment i am getting a blank white screen.
<?php
session_start();
$servername = "localhost";
$user = "xxxx";
$password = "xxxx";
$dbname = "v6e";
// Create connection
$conn = mysqli_connect($servername, $user, $password, $dbname);
// Check connection
if (!$conn)
{
die("Connection failed: " . mysqli_connect_error());
}
else
{
$query = "SELECT Language1 FROM Mull WHERE username = 'Mull'";
$result = mysqli_query($query);
$row = mysqli_fetch_arrary($result);
echo $row['Language1'];
}
mysqli_close($conn);
?>
You have a typo issue. Change the line:
$row = mysqli_fetch_arrary($result);
With:
$row = mysqli_fetch_array($result);
Plus, you're also not connecting to DB with your query
$result = mysqli_query($conn, $query);
Reference:
http://php.net/manual/en/mysqli.query.php
You should also check for errors:
http://php.net/manual/en/mysqli.error.php
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Closed 8 years ago.
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I know I am doing something wrong but I really would like to know what it is. I can echo the
username of the session loggedin user using <?php echo $_SESSION['username']; ?>but I don't know why it doesn't work when I try to query database using the same technique. my codes below
I include this in the page
<?php
session_start();
$username=$_SESSION['username'];
?>
and here is the code that was suppose to display firstname and user_id of the sessions logged in user
<?php
$conn = new mysqli('localhost', 'root', 'browser', 'test');
if (mysqli_connect_errno()) {
exit('Connect failed: '. mysqli_connect_error());
}
$username = '$username';
$sql = "SELECT `user_id`, `firstname` FROM `members` WHERE `username`='$username'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo '<br /> user_id: '. $row['user_id']. ' - firstname: '. $row['firstname'];
}
}
else {
echo '0 results';
}
$conn->close();
?>
$username = '$username';
PHP variables inside single-quotes are not expanded. So now your variable is the literal string '$username', which undoubtedly won't match any user in your database.
You probably need to set $username = $_SESSION['username']; in your second PHP script.
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Closed 8 years ago.
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EDIT/AWNSER: It was a bloody typo, dont mind me facedesking, thanks Sadikhasan
For some reason, the function mysqli_query in my code below, doesnt work, when i open the page, it returns an error.
Fatal error: Call to undefined function msqli_query() in
**/**/**/**/**db.php on line 16
I double checked the script, but couldnt find any typo's or ";" misplacements, the login part works, its purly the query that derps.
<?php
$sqlhost = '*****';
$sqlname = '*****';
$sqlpass = '*****';
$sqldbname = '*****';
$con=mysqli_connect($sqlhost,$sqlname,$sqlpass,$sqldbname);
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}else{
echo "connection successfull!";
}
$result = msqli_query($con,"SELECT * FROM PEOPLE");
while($row = mysqli_fetch_array($result)) {
echo $row['ID'] . "<br>";
echo $row['NAME'] . "<br>";
echo $row['AGE'] . "<br>";
echo $row['SEX'] . "<br>";
echo "<hr>";
}
mysqli_close($con);
?>
The names are in capitals in the database, i checked that too :)
thanks for the help in advance!
Correct spelling to mysqli in this line
$result = msqli_query($con,"SELECT * FROM PEOPLE");