I have been trying to figure this out for hours, I have created a database ( MySql/PHPMyadmin) and i am trying to get user input stored to be able to call back up, however the info is not making it/ saving it to the database, everything shows up okay except this part of code:
$registered = mysqli_affected_rows ($dbc);
echo $registered. "Row is affected";
when run gives me a display of -1 row, I believe this to be a big part of the problem as everything else seems to work okay. I am a complete beginner so could you guys tell me how the best way of debugging this is.
$dbc = $dbc = mysqli_connect ($hostname, $username, $password, $dbname) OR die("Could not Connect");
To input the data to the db i have the following:
if ($_SERVER['REQUEST_METHOD'] == 'POST'){
$comments = $_POST ['Comments'];
if (!empty ($comments)){
include ('mysqldb.php');
mysqli_query ($dbc,"INSERT INTO 'User-Comments' (Comments) VALUES ('$comments')");
$registered = mysqli_affected_rows ($dbc);
echo $registered. "Row is affected";
}else {
echo "Nothing Submitted";
}
}
The line:
mysqli_query ($dbc,"INSERT INTO 'User-Comments' (Comments) VALUES ('$comments')");
should be:
mysqli_query ($dbc,"INSERT INTO `User-Comments` (Comments) VALUES ('$comments')");
Notice the change in the apostrophe character ( ` ) around your table name.
An excerpt from the documentation for function mysqli_stmt_affected_rows(): -1 indicates that the query has returned an error.
You should check the value returned by mysqli_query(). If it returns FALSE then you can get details about the reason (error message) by using function mysqli_error().
Related
I've spent today going through tons of similar questions and trying to figure out what is wrong with my code, lots of issues people had with back ticks, quotes, etc but none seem to help or change my cause. My code is no producing any errors, but when I use echo to print out my query results, it seems that the id is not getting a value.
In my delete.php:
<?
ini_set('display_errors',"1");
$username="xxx";
$password="xxx";
$database="xxx";
$conn = new mysqli(localhost, $username, $password, $database);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$id = (int)$_GET['number'];
mysqli_query($conn,"DELETE FROM tourdates WHERE id=".$id."");
$conn->close();
?>
And the delete button in my main.php (the rest of the php is correctly displaying my table with data):
<td><a href='delete.php?number='".$row['id']."'>Delete</a></td>
Can someone help pick out what is causing my rows not to delete when I hit the delete button that I have created, or maybe something that more clearly can help me debug? (I don't want to use checkboxes for this).
EDIT:
I also tried this code (while defining the function as $sql and I'm getting a "Success" message:
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
EDIT 2:
I changed the structure following the advice that I should use POST, thinking I might have caught something I didn't notice before, but still not working.
echo "<td><form method='post' action='delete.php'>
<input type='hidden' name='row_id' value=".$row['id']." />
<input type='submit' name='delete_row' />
</form>";
-
if(isset($_POST['delete_row'])) {
$stmt = $conn->prepare("DELETE FROM tourdates WHERE ID = ?");
$stmt->bind_param('i', $_REQUEST['row_id']);
$stmt->execute();
}
If I do it the above way, nothing happens. Also tried this way, and get a syntax error:
if(isset($_POST['delete_row'])) {
$id = $_POST['row_id'];
$sql = "DELETE FROM tourdates WHERE id=".$id;
mysqli_query($conn,$sql);
}
A potential problem that I can see, is that you are not quoting localhost so php will look for a constant called localhost:
$conn = new mysqli('localhost', $username, $password, $database);
^ ^ here
You are also not checking for errors so that is why you don't see any. The easiest way to fix that, is to have mysqli throw exceptions. Just add this to the top of your script:
mysqli_report(MYSQLI_REPORT_STRICT);
I also don't know if you can mix procedural and object oriented mysqli like that. You should probably stick to the OOP version.
Apart from that you should not use a link (GET request) for your delete actions. What if a web-crawler or a browser extension tries to fetch the links? Instead you should use a POST request (like a form with a button).
Edit: There is another problem which causes you not to get your ID and as you cast it to int, you will always get 0:
<td><a href='delete.php?number='".$row['id']."'>Delete</a></td>
^ Oooops, closing the href attribute value here...
Your id gets placed after the value / outside of the quote of the href value. You can easily verify this if you look at the source of your page.
You need:
<td><a href='delete.php?number=".$row['id']."'>Delete</a></td>
Replace these two parts of code in your php file, first write your host in the quotations
$conn = new mysqli('localhost', $username, $password, $database);
in your where condition you wrote id=".$id."" replace it with id=".$id
write it as:
mysqli_query($conn,"DELETE FROM tourdates WHERE id=".$id);
Edited:
If you want to see error in your query then use the below code:
mysqli_query($conn,"DELETE FROM tourdates WHERE id=".$id) or die(mysqli_error($conn));
why not use try and catch to see your error?
anyways try this
$stmt = $conn->prepare("DELETE FROM tourdates WHERE ID = ?");<br>
$stmt->bind_param('i', $_REQUEST['number']);<br>
$stmt->execute();
could this be the problem ?
$id = (int)$_GET['number'];
May be this would be better... ?
$id = intval($_GET['number']);
Anyway if, echo($query) print an empty id, this is probably because your parameter is not an integer.
I am not sure what I am doing wrong, can anybody tell me?
I have one variable - $tally5 - that I want to insert into database jdixon_WC14 table called PREDICTIONS - the field is called TOTAL_POINTS (int 11 with 0 as the default)
Here is the code I am using. I have made sure that the variable $tally5 is being calculated correctly, but the database won't update. I got the following from an online tutorial after trying one that used mysqli, but that left me a scary error I didn't understand at all :)
if(! get_magic_quotes_gpc() )
{
$points = addslashes ($tally5);
}
else
{
$points = $tally5;
}
$sql = "INSERT INTO PREDICTIONS ".
"(TOTAL_POINTS) ".
"VALUES('$points', NOW())";
mysql_select_db('jdixon_WC14');
I amended it to suit my variable name, but I am sure I have really botched this up!
help! :)
I think you just need to learn more about PHP and its relation with MYSQL. I will share a simple example of insertion into a mysql database.
<?php
$con=mysqli_connect("localhost","peter","abc123","my_db");
// Check for errors in connection to database.
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query = "INSERT INTO Persons (FirstName, LastName, Age) VALUES ('Peter', 'Griffin',35)";
mysqli_query($con, $query);
mysqli_close($con); //Close connection
?>
First, you need to connect to the database with the mysqli_connect function. Then you can do the query and close the connection
Briefly,
For every PHP function you use, look it up here first.
(You will learn that it is better to go with mysqli).
http://www.php.net/manual/en/ <---use the search feature
Try working on the SQL statement first. If you have the INSERT process down, proceed.
You need to use mysql_connect() before using mysql_select_db()
Once you have a connection and have selected a database, now you my run a query
with mysql_query()
When you get more advanced, you'll learn how to integrate error checking and response into the connection, database selection, and query routines. Convert to mysqli or other solutions that are not going to be deprecated soon (it is all in the PHP manual). Good luck!
if(! get_magic_quotes_gpc() )
{
$points = addslashes ($tally5);
}
else
{
$points = $tally5;
}
mysql_select_db('jdixon_WC14');
$sql = "INSERT INTO PREDICTIONS (TOTAL_POINTS,DATE) ". //write your date field name instead "DATE"
"VALUES('$points', NOW())";
mysql_query($sql);
I am trying to write to a MySQL Database / Table with the following code - but for some reason it just won't write! I've changed the "INSERT INTO" line quite a few times, trying different things each time - no luck!!!
The DBsettings.php contains variables with the MySQL connection info - which worked for creating the tables and setting the column types and stuff. For your information, it is running the main code (there are no errors with the user info entered), and echoing "Awesome! No errors!", so I'm not too sure what's not working - the MySQL checking line is saying that I'm able to connect properly... Can someone look over my code?
The PasswordHash.php file contains code for hashing and salting passwords - nothing to see here, got it from another site, no errors at all.
I know I'm not 'cleansing' the MySQL code for more security...
if($error == null){
include('DBsettings.php');
$connect = mysqli_connect($dbserver, $dbuser, $dbpass, $dbname);
if (mysqli_connect_errno()) {
echo 'Failed to connect to MySQL Database! Error: '.mysqli_connect_error();
} else {
include('PasswordHash.php');
$passinfo = explode(':', create_hash($password));
$addinfo = "INSERT INTO {$dbprefix}Users (Email, Displayname, Registered, Rank, Status, Password, Salt) VALUES ('{$email}', '{$displayname}', '{date('Y\/m\/d')}', 9999, 1, '{$passinfo[3]}', '{$passinfo[2]}')";
/* format: algorithm:iterations:salt:hash */
mysqli_query($connect, $addinfo);
mysqli_close($connect);
echo 'Salt: '.$passinfo[2];
echo '<br>Hash: '.$passinfo[3];
echo '<br>Awesome! No Errors!';
}
} else {
echo $error;
}
That's the code in question - I've tried adding;
error_reporting(E_ALL);
ini_set('display_errors', '1');
But all that reveals is undefined localhost errors in my DBsettings.php file - and the file worked when I created the MySQL DB tables, so I don't really have that as a priority.
Thanks!
If you echo your query, you will notice this issue. Following is your final query
INSERT INTO Users (Email, Displayname, Registered, Rank,Status, Password, Salt)
VALUES ('', '', '{date('Y\/m\/d')}', 9999, 1, '', '')
Notice that your date was not interpolated like you expected it to, and i'm sure if you have that field in MySQL set as a datetime field, it wont accept that value {date('Y\/m\/d')}, Move the date function call outside the string.
Plus you are not getting any error after the query execution because you are simply not checking for one. One example how to check for that can be
if (!mysqli_query($connect, $addinfo)) {
printf("Error: %s\n", mysqli_error($connect));
}
I saw your INSERT query contains this '{date('Y/m/d')}' ,maybe the single quotes has conflict,You'd better escaping the date('Y/m/d') statement's single quotes.
Almost identical to a previous question however I am now trying to use mySQLi to record some form data.
After I submit, the data does not post to the table. I've been reading through to the mySQLi documentation and some different videos but I can't figure it out.
Thanks in advance!
<?php
include('config.php');
if (
isset($_POST['store_id']) &&
isset($_POST['item_title']) &&
isset($_POST['date']) &&
isset($_POST['price'])
)
{
$store = get_post('store_id');
$item = get_post('item_title');
$date = get_post('date');
$price = get_post('price');
$query = "INSERT INTO ebay_data VALUES('".$store."', '".$item."', '".$date."', '".$price."')";
$input = $db_mysqli->query($query);
}
?>
You want to make sure your config is actually connecting of course. I typically just go to do it on one page if I get stuck, either way try:
$mysqli = new mysqli('localhost', 'my_user', 'my_password', my_databasename);
/* check connection */
if (mysqli_connect_errno()) {
printf('Connect failed: %s\n', mysqli_connect_error());
exit();
}
$query = 'INSERT INTO ebay_data VALUES($store, $item, $date, $price)';
$mysqli->query($query);
printf ('New Record has id %d.\n', $mysqli->insert_id);
/* close connection */
$mysqli->close();
See if you get anything out on the page. Also should double check your post values are not empty. I have also ran into weird problems where setting the post to a var messed up sql queries, maybe try:
if (isset($_POST['store_id']) && isset($_POST['item_title']) && isset($_POST['date']) && isset($_POST['price']))
{
$query = "INSERT INTO ebay_data VALUES($_POST['store_id'], $_POST['item_title'], $_POST['date'], $_POST['price'])";
...
You also 100% need to check those post values, keeping your database and users safe. Prepared statements is a start. You do need a lot more than that though to keep it safe...
http://php.net/manual/en/mysqli.quickstart.prepared-statements.php
I use this code , to log $_SERVER['REMOTE_ADDR']; to my small db
my issue is value never saved to db , cant figure what i missed in the code
Any tips ?
<?php
mysql_connect("localhost", "usr", "passwd");
mysql_select_db("db") or die ( 'Can not select database' );
function initCounter() {
$ip = $_SERVER['REMOTE_ADDR'];
$sql = "INSERT INTO logs(REMOTE_ADDR,) VALUES ('$ip')";
}
echo $_SERVER['REMOTE_ADDR'];
?>
This should work. In addition to the other comments here, you had a comma (,) too much in your query.
<?php
mysql_connect("localhost", "usr", "passwd");
mysql_select_db("db") or die ( 'Can not select database' );
function initCounter() {
$ip = $_SERVER['REMOTE_ADDR'];
$sql = "INSERT INTO logs (REMOTE_ADDR) VALUES ('$ip')";
mysql_query($sql);
}
initCounter();
?>
You aren't actually executing the query. You create the SQL but don't use mysql_query($sql)
You have a comma at this point in the SQL REMOTE_ADDR, <-- remove that
When you execute the query, use mysql_error() to test for an error message (and check the result of mysql_query() for a boolean false.
Finally I would suggest switching to MySQLi or PDO.
If that's you're full code... there is one thing missing you actually need to EXECUTE the query...
mysql_query($sql);
EDIT:
I have just noticed, you're connecting to the DB OUTSIDE of the function trying to run the Query... obviously it will fail as inside the function, it has no awareness of the DB connection.