I have a problem. I'm trying to update a mysql table with data grabbed. The problem is there is an invisible \r\n on the end and it's not functioning properly.
DELETE FROM `tracker`.`gains` WHERE `gains`.`runescape_name` = 'Andy FFS\r\n';
DELETE FROM `tracker`.`gains` WHERE `gains`.`runescape_name` = 'Cancerbear\r\n';
DELETE FROM `tracker`.`gains` WHERE `gains`.`runescape_name` = 'Pandora25';
DELETE FROM `tracker`.`gains` WHERE `gains`.`runescape_name` = 'Pether\r\n';
I have a file called memberlist and update.php loops through each line in memberlist.txt and inserts or updates the columns in the table. If you look at the above query (that is just me trouble shooting) I don't actually want to delete it's just to show you the \r\n at the end. It's bringing back 0 in all columns except for the name pandora25 (which happens to be at the bottom of memberlist.txt. What I don't understand is I have a php script to edit memberlist.txt on my website, and that's when the problem occurs. But say I open memberlist.txt via SFTP and upload it to my website, it works correctly. So maybe it's something to do with the editing of the txt file with php and it's adding and invisible \r\n at the end I'm not sure. But I have studied quite a bit on this issue and I've looked that you can insert into a MySQL table with ->prepare in PDO (I don't use mysqli).
I believe the loop variable [$k] plays an important role in this problem. Here is my code that I seem to be having a problem with:
// Connection data (server_address, database, name, poassword)
$hostdb = 'localhost';
$namedb = 'tracker';
$userdb = 'root';
$passdb = 'password';
try {
// Connect and create the PDO object
$conn = new PDO("mysql:host=$hostdb; dbname=$namedb", $userdb, $passdb);
$conn->exec("SET CHARACTER SET utf8"); // Sets encoding UTF-8
// Query
$sqlquery = ("
INSERT IGNORE INTO
gains (runescape_name, hitpoints_starting_exp, magic_starting_exp, range_starting_exp)
VALUES
('" . htmlspecialchars($name[$k]) . "', '" . $hitpointsexp . "', '" . $magicexp . "', '" . $rangedexp . "')
ON DUPLICATE KEY UPDATE
hitpoints_starting_exp='" . $hitpointsexp . "', magic_starting_exp='" . $magicexp . "', range_starting_exp='" . $rangedexp . "'");
if($conn->exec($sqlquery) !== false) echo 'completed'; // If the result is not false, display confirmation
$conn = null; // Disconnect
}
catch(PDOException $e) {
echo $e->getMessage();
}
The purpose of this script is to use start.php to insert the current grabbed data into hitpoints_starting_exp, magic_starting_exp and range_starting_exp. I then run stop.php which is a similar script and insert the data grabbed there into hitpoints_end_exp, magic_end_exp and range_end_exp. I then echo a table on index.php which calculates the difference between the two columns.
Related
Once again I come back to all of you with another question.
I have tried everything in my mind as well as most of the recommendations I have found on the web and here in Stackoverflow but nothing seems to fix this issue for me.
For some reason the sql command in my code is returning false even though it should not.
Here is my php file called (dbRKS-DBTest.php)
<?php
//Gets server connection credentials stored in serConCred.php
//require_once('/../prctrc/servConCred2.php');
require_once('C:\wamp64.2\www\servConCred2.php');
//SQL code for connection w/ error control
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if(!$con){
die('Could not connect: ' . mysqli_connect_error());
}
//Selection of the databse w/ error control
$db_selected = mysqli_select_db($con, DB_NAME);
if(!$db_selected){
die('Can not use ' . DB_NAME . ': ' . mysqli_error($con));
}
//VARIABLES & CONSTANTS
//Principal Investigator Information
$PI_Selected = '6';
//Regulatory Knowledge and Support Core Requests variables
$RKS_REQ_1_Develop = '1';
//This sets a starting point in the rollback process in case of errors along the code
$success = true; //Flag to determine success of transaction
//start transaction
$command = "SET AUTOCOMMIT = 0";
$result = mysqli_query($con, $command);
$command = "BEGIN";
$result = mysqli_query($con, $command);
//Delete this portion of code afyer testing is finished
//Core Requests saved to database
$sql = "INSERT INTO rpgp_form_table_3 (idPI, RKS_REQ_1_Develop)
VALUES ('$PI_Selected', '$RKS_REQ_1_Develop')";
//*************TEsts code for "SCOPE_IDENTITY()" -> insert_id() for mysql
$sqlInsertId = mysqli_insert_id($con); //This value is supposed to be 0 since no queries have been executed.
echo "<br>MYSQLi_INSERT_ID() value before query should be 0 and it is:= " . $sqlInsertId;
//Checks for errors in the db connection.
$result = mysqli_query($con, $sql); //Executes query.
if($result == false){ //Checks to see for errors in previews query ($sql)
//die ('<br>Error in query to Main Form: Research Proposal Grant Preparation: ' . mysqli_error($con));
echo "<br>Result for the sql run returned FALSE. Check for error in sql code execution.";
echo "<br>Error given by php is: " . mysqli_error($con);
$success = false; //Chances success to false is it encounted an error in order to rollback transaction to database
}
else{
//*************TEsts code for "SCOPE_IDENTITY()" -> insert_id() for mysql
$sqlInsertId = mysqli_insert_id($con); //Saves the last id entered. This would be for the main table
echo "<br>MYSQLi_INSERT_ID() value after Main form query= " . $sqlInsertId; //Displays id last stored. This is the main forms id
$MAIN_ID = mysqli_insert_id($con); //Sets last entered id in the MAIN Form db to variable
}
//Checks for errors or craches inside the code
// If found, execute rollback
if($success){
$command = "COMMIT";
$result = mysqli_query($con, $command);
echo "<br>Tables have been saved witn 0 errors.";
}
else{
$command = "ROLLBACK";
$result = mysqli_query($con, $command);
echo "<br>Error! Databases could not be saved. <br>
We apologize for any inconvenience this may cause. <br>
Please contact a system administrator at PRCTRC.";
}
$command = "SET AUTOCOMMIT = 1"; //return to autocommit
$result = mysqli_query($con, $command);
//Displays message
//echo '<br>Connection Successfully. ';
//echo '<br>Database have been saved';
//Close the sql connection to dababase
mysqli_close($con);
?>
Here is my php frontend html code named (RPGPHomeQueryTest.php)
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
</head>
<form id="testQuery" name="testQuery" method="post" action="../dbRKS-DBTest.php" enctype = "multipart/form-data">
<input type="submit" value="Submit query"/>
</form>
</html>
And here is how my database looks (rpgp_form_table_3):
So, when I open my html code, All I will see is a button since its all the code there is there. Once you press the button, the form should submit and execute the php code called (dbRKS-DBTest.php). This should take the predetermine values I already declared and saved them to the database called (rpgp_form_table_3). This database is set to InnoDB format.
Now, the output I should be getting is a message saying "Tables have been saved witn 0 errors." but the problem is that the message I am getting is this one bolow:
I honestly don't know why. I am posting this message to find guidance to this issue. I am still learning by myself and its been very did-heartedly to not find a solution this fixing this.
As always, I thank you for your patient and guidance! Let me know what other details I can provide.
Here is the SQL code you run:
$sql = "INSERT INTO rpgp_form_table_3 (idPI, RKS_REQ_1_Develop)
VALUES ('$PI_Selected', '$RKS_REQ_1_Develop')";
You are inserting data into rpgp_form_table_3. From the screenshot, we can see that table has several (7) fields yet you are only inserting 2 fields. The question then is: do you need to specify a value for all fields?
The error you are getting states
Error given by php is: Field 'idCollaRecord_1' doesn't have a default value Error! Databases could not be saved.
It's clear that you have to insert the row by specifying a value for each column, not just the two columns you are interested in.
Try
$sql = "INSERT INTO rpgp_form_table_3 (idPl, RKS_REQ_1_Develop, idCollaRecord_1, idCollaRecord_2, idCollaRecord_3, idCollaRecord_4)
VALUES ('$PI_Selected', '$RKS_REQ_1_Develop',0,0,0,0)";
Try this insert code. If the PI_Selected is NUMERIC use the First one. If it is string use the second one
$sql = "INSERT INTO rpgp_form_table_3 (idPI, RKS_REQ_1_Develop) VALUES (" .
$PI_Selected . ",'" . $RKS_REQ_1_Develop . "')";
$sql = "INSERT INTO rpgp_form_table_3 (idPI, RKS_REQ_1_Develop) VALUES ('" .
$PI_Selected . "','" . $RKS_REQ_1_Develop . "')";
I have a problem with my PHP code.
I am trying to make a level create function for a small game project me and another person are working on.
My code works.. but generates a lot of duplicates.
Here's the code: (Don't laugh at me for how vulnerable this is, this will be fixed eventually, THIS IS JUST TEST CODE AND WILL NEVER BE ON A PUBLIC SERVER IN ANY CIRCUMSTANCE, OK?)
$mysqli = new mysqli("localhost", "Username", "Password", "sys");
$SqlQuery = "INSERT INTO levels (levelname, levelauthor, leveldata)
VALUES(\"" . $_GET["levelName"] . "\", \"" . $_GET["levelAuthor"] . "\", \"" . $_GET["levelData"] . "\");";
$query2 = "SELECT * FROM levels WHERE leveldata = \"" . $_GET["levelData"] . "\";";
//echo "SELECT * FROM levels WHERE leveldata = \"" + $_GET["levelData"] + "\";";
$uresult = $mysqli->query($SqlQuery, MYSQLI_USE_RESULT);
$res2 = $mysqli->query($query2, MYSQLI_USE_RESULT);
if ($uresult) {
while ($row = $res2->fetch_assoc()) {
//This should always work. Lol
echo "(SUC)|ID:" . $row["levelid"];
}
}
After running this code, I expected to just check my database and be able to see the test I wrote, without duplicates.
I started the PHP development server and went to:
http://localhost/Create.php?levelName=PHPTest&levelAuthor=Test3&levelData=[snip]
I expected to see something along the lines of "(SUC)|ID:4" (there were 3 entries in the database at the time), but I saw this:
(SUC)|ID:4(SUC)|ID:5(SUC)|ID:6(SUC)|ID:7(SUC)|ID:8(SUC)|ID:9(SUC)|ID:10(SUC)|ID:11
This was unexpected. I thought it was just an error in my code (keep in mind, the last one had a broken ID grabbing system, but worked), and that it would work, but then, I went to check the database, and saw a ton of duplicates with the same data.
Does anyone know how to fix this code?
Obvious question but autocommit is enabled on database?
Do you have some open transaction?
Use this to check open transactions on MySQL.
This question already has answers here:
When to use single quotes, double quotes, and backticks in MySQL
(13 answers)
Closed 6 years ago.
I've been having some trouble recently with trying to automate new database creations with a php script.
Basically, the script takes the new login username and creates a database (and then insert some tables and data later on, which is also done via a php script).
I used to have to manually create the database, but now need to make it automated.
The issue is that I used to be able to just create a new database using the phpadmin "new database" function from the web GUI and put in names like "test1.siteA", "userb.siteB".
However, now that I've tried to do the same via php script, it keeps giving me the "You have an error in your syntax..." from my last "echo".
Main parameters are:
$name = $user->username;
$servernm = 'localhost';
$usnm = 'user';
$pasd = 'user';
$dbname = $name;
$dbname .= '.site';
I've found that the error would disappear once I remove the .site part from the code (it still exist even if I combine the $dbname into 1 line).
According to some articles that I've found online, it seems that MySQL doesn't allow special characters like "." to be included in the database name.
It just seems very weird to me that the ".site" can be added manually through phpMyadmin while the php/mysqli script doesn't allow this.
The full script is as follows (I'm sure it can be heavily improved, so any suggestions regarding that are also welcome):
<?php
define("_VALID_PHP", true);
require_once(APPPATH. "/libraries/init.php");
include (BASEPATH . "/database/DB_temp.php");
$row = $user->getUserData();
$name = $user->username;
$servernm = 'localhost';
$usnm = 'user';
$pasd = 'user';
$dbname = $name;
$dbname .= '.site';
// Create connection
$conn = mysqli_connect($servernm, $usnm, $pasd);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
// Check if DB exist
$sql = "SELECT count(SCHEMA_NAME) FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = '$dbname'";
$check = mysqli_query($conn,$sql)
or die("Connection failed: " . mysqli_connect_error());
while($row = mysqli_fetch_array($check,MYSQLI_NUM))
{
$dbval = $row[0];
}
if ($dbval == "0")
{
$createsql = "CREATE DATABASE '$dbname' ";
}
if ($dbval == "1")
{
$createsql = "SELECT count(SCHEMA_NAME) FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = '$dbname'";
}
if (mysqli_query($conn, $createsql)) {
Echo "Completed. DBVAL= " .$dbval ;
}
else
{
echo "Error creating database: " . mysqli_error($conn);
}
?>
PHP version: 5.6.18
phpmyadmin: 4.5.4.1
Ubuntu 14.04
Apologies if I've made some posting errors on here. Do let me know about them and I'll try to correct it as much as I can. Any help is greatly appreciated!
. is a meta character in SQL, use to separate db/table/field names:
SELECT foo.bar.baz FROM sometable
^---------- database 'foo'
^------- table 'bar'
^--- field 'baz'
You should NOT be using metacharacters in any identifiers. It just leads to pain later on, and having to do stuff like:
SELECT `foo.bar`.baz.qux FROM ...
^^^^^^^^^--------- database 'foo.bar'
^------ table 'baz'
^-- field 'qux'
So you can use backticks if you absolutely have to, but you shouldn't be doing this in the first place.
try wrapping the database name with back ticks.
$dbname .= '`.site`';
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 seem to be missing something quite fundamental here and yet my code doesn't seem to be any different to any of the numerous online tutorials that I have looked at.
What I would like is for someone to look at this and say....Oh you have forgotten to...etc;
This is what I have on a separate update page which is intended to perform the update then cycle back to the main admin page:
require_once('../Connections/MyConn.php');
$sql_statement = "UPDATE skyscrapers SET ";
$sql_image = "Ad_image = '" . $_REQUEST['image'] . "', ";
$sql_expire = "Ad_Expires = '" . $_REQUEST['expire'] . "'";
$result = mysql_query($sql_statement . $sql_image . $sql_expire . " WHERE Ad_ID=" . $_REQUEST['ADID']);
if (!$result) {
echo("<p>Error performing query: " . mysql_error() . "</p>");
exit();
}
mysql_close ($MyConn);
header("location:Admin_skyscrapers.php");
However when I run this I get the following error:-
"Error performing query: No database selected"
Well, haven't I selected the database in the connection script which already works everywhere else?
I realise the code isn't very pretty and I am being naughty using the url to pass variables at the moment - I do promise to change this when I get it to work :)
So, any pointers would be helpful, thanks in advance.
Edit to add...
This is the connection script with the sensitive stuff redacted:-
# FileName="Connection_php_mysql.htm"
# Type="MYSQL"
# HTTP="true"
$hostname_MyConn = "*************.co.uk";
$database_MyConn = "db**********";
$username_MyConn = "dbo*********";
$password_MyConn = "*****";
$MyConn = mysql_pconnect($hostname_MyConn, $username_MyConn, $password_MyConn) or trigger_error(mysql_error(),E_USER_ERROR);
This may or may not be declared in your MyConn.php but all you need is a line:
mysql_select_db($db_name);
Where $db_name is the name of your database.
This should come before you attempt to execute the query.
For DB select you have to add mysql_select_db(DatabaseName); or
$dbconn=mysql_select_db($dbname,$MyConn);in MyConn.php
For update in Database you have to use connection variable which is in MyConn.php i.e.$MyConn as follows
$result = mysql_query($sql_statement . $sql_image . $sql_expire . " WHERE Ad_ID=" . $_REQUEST['ADID'],$MyConn);
or
$result = mysql_query($sql_statement . $sql_image . $sql_expire . " WHERE Ad_ID=" . $_REQUEST['ADID'],$dbconn);
respectively
Hope It Helps!!!!!!!