Hi I am writing to a remote MSSQL 2005 server from a php application, and have a situation where mssql_num_rows errors out with the "mssql_num_rows() expects parameter 1 to be resource, boolean given" message.... but I can't figure out why
$writeitem = "INSERT INTO RebateSubmissionProducts VALUES ('" . $buyproduct . "'," . $quantity . ",CAST('" . $itemUUID . "' as UNIQUEIDENTIFIER),CAST('" . $eligible . "' as UNIQUEIDENTIFIER),CAST('" . $prodID . "' as UNIQUEIDENTIFIER),CAST('" . $UUID . "' as UNIQUEIDENTIFIER),NULL)";
$itemresult = mssql_query($writeitem);
if (!mssql_num_rows($itemresult)){
echo 'Problem writing to RebateSubmissionProducts';
} else {
echo 'Success writing to RebateSubmissionProducts';
}
mssql_free_result($itemresult);
The upshot is that I get the error message, but the insert works fine.
BTW all the input is run through HTMLPurifier so don't slag me too hard about that. The hosting company can't set up PDO_DBLIB so I can't use PDO/bound params.... I also don't have access to the MS server for creating a stored procedure.
Any ideas why php thinks that $itemresult is a boolean? (both mssql_num_rows and mssql_free_result issue the same error message)
As andrewsi pointed out, mssql_num_rows() responds in a different way depending on the query. In my case, I was running an insert, and so when this ran:
$itemresult = mssql_query($writeitem);
$itemresult was boolean(TRUE) beacuse the insert succeeded, so mssql_num_rows (and mssql_free_result() both issued warnings since there wasn't a result set.
Related
I'm running a LightSail amazon server with Lamp PHP 8.1 instance and I needed a lightweight DB for my server, so I've installed SQLite. I've created a DB and trying to insert into a table from php code
$db = new SQLite3('/home/bitnami/bot.db', SQLITE3_OPEN_READWRITE);
$sql = 'insert into users values("' . $id . '", "' . $platform . '", "' . $name . '", "' . $chatId . '", false)';
logLine($sql);
$db->exec($sql);
$err = $db-> LastErrorCode();
logLine("Error code: " . $err);
For some reason completely unclear to me, I'm getting error 8: attempt to write a readonly database.
I've tried the following with no success:
Give 777 permissions on db file
Change owner of db file to the user which runs apache (daemon)
Change owner to root
Shaman dance
When I run the same code with php cli from console it works! But if it runs under apache it doesn't!
You need to set Pragma journal mode in WAL to use read write functions
$db = new SQLite3('yourfile.db');
$db->query("PRAGMA journal_mode=WAL");
I am trying to restore a database to a different server through PHP. I managed to execute all needed commands but the database keeps hanging in the 'Restoring...' state on the server.
I have searched and followed the answer on SQL Server: Database stuck in “Restoring” state with PHP (which leads to this article) but this did not work for me; I get an error when trying to change the environment to the newly recovered database.
Code:
sqlsrv_configure( "WarningsReturnAsErrors", 0 );
$connOptions = ["Database"=>"master"];
$sqlConnection = sqlsrv_connect("server_name\\SERVER", $connOptions);
sqlsrv_query($sqlConnection, "USE master");
$sql = "IF EXISTS(SELECT name FROM sys.databases
WHERE name = 'db_name')
DROP DATABASE db_name";
sqlsrv_query($sqlConnection, $sql);
$sql = "RESTORE FILELISTONLY FROM DISK='$path'";
$logicalNamesStatement = sqlsrv_query($sqlConnection, $sql);
$moveArray = [];
while($logicalNames = sqlsrv_fetch_array($logicalNamesStatement, SQLSRV_FETCH_ASSOC)){
if($logicalNames['Type'] === "D"){
$moveArray['MDF'] = $logicalNames['LogicalName'];
}
elseif($logicalNames['Type'] === "L"){
$moveArray['LDF'] = $logicalNames['LogicalName'];
}
}
$localDbPath = "c:\\Program Files\\Microsoft SQL Server\\MSSQL11.SERVER\\MSSQL\\DATA\\";
$sql = "RESTORE DATABASE db_name FROM DISK='$path'
WITH
MOVE '" . $moveArray['MDF'] . "' TO '" . $localDbPath . "db_name.mdf',
MOVE '" . $moveArray['LDF'] . "' TO '" . $localDbPath . "db_name_log.ldf',
REPLACE,
STATS=10";
sqlsrv_query($sqlConnection, $sql);
$sql = "RESTORE DATABASE db_name FROM DISK='$path' WITH REPLACE, RECOVERY";
sqlsrv_query($sqlConnection, $sql);
sqlsrv_query($sqlConnection, "USE db_name");
The error that comes back is:
[Microsoft][SQL Server Native Client 11.0][SQL Server]Database 'db_name' cannot be opened. It is in the middle of a restore.
Do you have any ideas on how I can solve this? The only authority on a workaround for the bug described in my previously linked article seems to be that exact article, which does not work as you can see.
Thanks!
I don't have 50 reputation yet, so feel free to move this to the comments section. I don't see your database names including brackets "[ ]". If you have any database names with spaces or special characters, you'll need the brackets to identify the object properly. And as Mitch already stated, based on what you are doing, there is no reason for two restore operations; just add the RECOVERY option into the first and be done with it.
i am very new at MYSQL and after i created this script to update a row in the table of a MYSQL Database and run it i get this error
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '265'', Employer_VAT_number = ''45698'', Employer_Name = ''Namtax_Ltd'', Employer' at line 3
here is the code
// username and password sent from form
$Numb=$_POST["Numb"];
$VAT=$_POST["VAT"];
$Name=$_POST["Name"];
$Addr=$_POST["Addr"];
$PO=$_POST["PO"];
// To protect MySQL injection (more detail about MySQL injection )
$Numb = stripslashes($Numb);
$VAT = stripslashes($VAT) ;
$Name = stripslashes($Name) ;
$Addr = stripslashes($Addr) ;
$PO = stripslashes($PO) ;
$Numb = "'" . mysql_real_escape_string($Numb) . "'";
$VAT = "'" . mysql_real_escape_string($VAT) . "'";
$Name = "'" . mysql_real_escape_string($Name) . "'";
$Addr = "'" . mysql_real_escape_string($Addr) . "'";
$PO = "'" . mysql_real_escape_string($PO) . "'";
$sql=("UPDATE $tb1_name SET Employer_Registration_Number ='".$Numb."', Employer_VAT_number = '".$VAT."', Employer_Name = '".$Name."', Employer_Address = '".$Addr."', Employer_Postal_Address = '".$PO."' WHERE Employer_Name = '".$Name."' ");
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "Successfully Updated";
mysqli_close($con);
?>
</body>
This here:
$Numb = "'" . mysql_real_escape_string($Numb) . "'";
Firstly, that isn't proper syntax and you're using mysqli_ to connect with, least I sure hope you are.
Those different MySQL APIs do not intermix with each other.
That should read as:
$Numb = mysqli_real_escape_string($con,$Numb);
while doing the same for the rest of your variables, following the same method outlined here.
Footnotes:
Seeing you didn't post what $tb1_name is, doubt that would be causing an issue. But just for the sake of argument, wrap that variable in ticks, just so if your table name changes to something containing a hyphen or a space, or anything that MySQL will complain about.
UPDATE `$tb1_name` SET...
Plus, since you didn't mention which MySQL API you're using to connect with, make sure it is in fact mysqli_ and not mysql_ or PDO.
It doesn't look like it, but I have to be 100% sure.
Your connection should resemble something like this:
$con = mysqli_connect("yourhost","user","pass","your_DB")
or die("Error " . mysqli_error($con));
Again, those different MySQL APIs do not intermix with each other.
Consult (PHP: Choosing an API - Manual): https://php.net/mysqlinfo.api.choosing
"I am very new at MYSQL..."
Seeing you're new to this:
Use mysqli with prepared statements, or PDO with prepared statements.
Additional notes. (as an edit)
I noticed another question you posted earlier:
https://stackoverflow.com/q/30191388/
where you said "Thank you it worked " in the answer given https://stackoverflow.com/a/30191647/
I don't get that.
How could that possibly work where you're using if (!mysqli_query($con,$sql))?
You'll need to show us the way you're connecting with here.
If you truly want to see if your query was successful, use mysqli_affected_rows().
if(mysqli_affected_rows($con)){
echo "Successfully updated.";
}
else{
echo "Not updated.";
}
Add error reporting to the top of your file(s) which will help find errors.
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
// rest of your code
Sidenote: Error reporting should only be done in staging, and never production.
I'm currently working on creating a login system, one part of which is of course registration. It's been going smoothly up until this point, where I'm getting an error.
I've researched this as thoroughly as I can, but I can't find the solution as it is giving me an incorrect line number.
The error I'm getting is:
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1
My SQL query is
$token = (round(microtime(true) * 1000));
$query = mysql_query("INSERT INTO "
. "`users` "
. "(name, password, email, token) "
. "VALUES "
. "('$_POST[user]'"
. ",'".hash('sha512',$_POST['pass'])."'"
. ",'$_POST[email]'"
. ",'$token')") or die(mysql_error());
if (mysql_query($query) === TRUE) {
//echo "Sucsessfuly registered! Check your email for a confirmation link.";
} else {
echo "Error: " . mysql_error();
}
(this is not the first line of the file, it's the 22d)
When the code runs, even though it throws the error it still is inserting the values into the table correctly.
Also when I run the same query in phpmyadmin, it runs just fine with no errors.
I've been trying to solve this error for the last 3 hours so any help would be appreciated ;)
You're calling mysql_query twice: first with the SQL, and then you're using the result of the query as if it were a query. The error you're getting is because $query is true, which gets turned into 1 when treated as a string.
Either you should just set $query to the SQL string:
$query = "INSERT INTO ...";
if (mysql_query($query)) {
...
} else {
...
}
or you should just check the value of $query:
$query = mysql_query(...);
if ($query) {
...
} else {
...
}
if (!$mysqli->query("UPDATE custom_fav SET credits = credits + 1 WHERE user_id = $current")) {
echo "Credit addition failed: (" . $mysqli->errno . ") " . $mysqli->error;
} else {
echo 'hi';
}
Does anybody know why the mysqli query keeps running twice?
I am getting 1 of the echo statements but I am getting + 2 added to the database instead of 1.
For example i changed it to 5 and it added 10 into the database.
Most likely you are calling this script twice. It this script called via home-brewed SEO-friendly URL? Do not make it act as a 404 handler then.