I've encountered a certain problem. I want to create a table in my database through PHP but it does not work properly.. Even though I copy the text which serves to create a database into PHPMYADMIN, it works perfectly there. Can anyone find any mistake? I have already checked my connection parameters.
if(empty($errors)) {
$namess = mysqli_real_escape_string($conn, $_POST['topicname']);
mysqli_select_db($conn,"d197243_games");
$sql = "CREATE TABLE {$namess} (id INT UNSIGNED AUTO_INCREMENT, topicname VARCHAR(255) NOT NULL, topicdescription VARCHAR(255), topictext TEXT ,primary key (id));";
$result = mysqli_query($conn, $sql);
if(!$result) {
$errors[] = 'Unknown error while creating a post.';
}
}
In your query have you used
{$namess}
if yes then remove those flower brackets
and simply right $namess
as below
$sql = "CREATE TABLE $namess (id INT UNSIGNED AUTO_INCREMENT, topicname VARCHAR(255) NOT NULL, topicdescription VARCHAR(255), topictext TEXT ,primary key (id));";
and hope you have defined your table name in $namess varible
Alright, so I've solved it myself, and thanks for that ''show error'' command. It showed my problem. I have not had enough permissions to access the database.. I had to use different login credentials.. Anyways, thank you for your help & answers. Appriciate it.
Related
so, i'm trying to show a certain value in a table row, if the value recieved by post is null.
So far i've got this as my small database (it's in spanish):
DROP DATABASE IF EXISTS fantasmas;
CREATE DATABASE fantasmas;
USE fantasmas;
CREATE TABLE tipos(
ID tinyint(2) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
TIPO VARCHAR(45) UNIQUE
);
CREATE TABLE datos(
ID tinyint(2) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
NOMBRE VARCHAR(50) NOT NULL,
ARCHIVO VARCHAR(30),
AVISTAMIENTO VARCHAR(50) NOT NULL,
LOCALIDAD VARCHAR(50) NOT NULL,
INFO VARCHAR(200),
FKTIPOS TINYINT(2) UNSIGNED,
FOREIGN KEY (FKTIPOS) REFERENCES tipos(ID)
);
INSERT INTO tipos (TIPO) VALUES('Vapor'), ('Forma Animal'), ('Forma Humanoide'), ('Dios/Semidios'),('No Catalogado');
Those inserted values into "tipos" are then showed in options like this:
<select name="tipos">
<option value="sintipo">---------</option>
<?php
while( $row = mysqli_fetch_assoc($rta)):
$tipo = $row['TIPO'];
echo '<option value="'.$row['ID'].'">'.$tipo.'</option>';
endwhile;
?>
</select>
After, when the form is sent, the inset to the table, looks like this, it assigns null value if someone select the "------" option:
$query = "INSERT INTO datos SET NOMBRE='$nombre', AVISTAMIENTO='$lugar', LOCALIDAD='$localidad', INFO='$info', ARCHIVO='$ruta', FKTIPOS = NULLIF('$tipo','sintipo')";
All the data filled in the form, is then showed in a different row of a table.
What I need now, is that, if someone selects the "------" option, the value shown in screen is "No catalogado"
So far I have not been able to do it.
Can anyone help me?
Before the insert happens you can do a null check on the PHP side. Or... you can just use SQL's built in ISNULL()
You can also use CASE statements in SQL to accomplish this behavior across multiple values.
I know this isn't the most lengthy answer, but it should work.
Is this what your asking for?
$query = "INSERT INTO datos SET NOMBRE='$nombre', AVISTAMIENTO='$lugar', LOCALIDAD='$localidad', INFO='$info', ARCHIVO='$ruta', FKTIPOS = " . ($tipo == 'sintipo' ? 'NULL' : "'$tipo'");
Be aware that you have very dirty code, you don't even filter and escape income values, as I guess. Try to google php mysql escape values.
I'm having trouble, as title says, when I INSERT a record in a table that has got a 1-1 relationship with another.
First things first, the SQL code that generates the tables:
DROP TABLE IF EXISTS Facebook_Info;
DROP TABLE IF EXISTS Conversations;
CREATE TABLE IF NOT EXISTS Conversations(
c_id INT AUTO_INCREMENT NOT NULL,
c_start TIMESTAMP DEFAULT CURRENT_TIMESTAMP,
channel ENUM('desktop', 'facebook'),
u_name VARCHAR(20) DEFAULT NULL,
u_email VARCHAR(50) DEFAULT NULL,
PRIMARY KEY(c_id)
);
CREATE TABLE IF NOT EXISTS Facebook_Info (
c_id INT AUTO_INCREMENT NOT NULL,
f_id INT(12) NOT NULL,
PRIMARY KEY(c_id),
FOREIGN KEY(c_id) REFERENCES Conversations(c_id)
);
I assure you this code works: I tested it. I hope this is the best way to provide a 1-1 relationship between Conversations and Facebook_Info.
In any case, now I can introduce you my nightmare: I'm trying to insert a new record in Conversations via PHP (procedural style).
public function create_new_id_conv($channel = 1) {
$w_ch = '';
if ($channel == 2) {
$w_ch = 'facebook';
} else {
$w_ch = 'desktop';
}
$query = "INSERT INTO Conversations (c_id, c_start, channel) VALUES (NULL, CURRENT_TIMESTAMP,'$w_ch')";
$conn = mysqli_connect("localhost", Wrapper::DB_AGENT, Wrapper::DB_PSW, Wrapper::DB_NAME);
$res = mysqli_query($conn, $query);
$id_conv= mysqli_insert_id($conn);
mysqli_free_result($res);
return $id_conv;
}
The Wrapper:: * variables are all set well, in fact, an INSERT operation is done, but not only one! I'm having this situation after I call this function:
This is the content of Conversations table:
And here's the content of Facebook_Info:
What's happening?
I searched and searched...
Then I started to think about what I'm getting here: 2147483647. What represents this number? What's that? Seems like a big number!
And what if my script and my queries were correct but the mistake is the skeleton of my tables?
I must register a 14 digit integer, that is too large for the INT type.
So using BIGINT to store the f_id field made all correct and working!
Hope my mistake helps someone!
I am working on a project, and I have to use sql. The variable $file_name needs to be the table name, but when i try this:
$sqlTableCreate = "CREATE TABLE ". $file_name . "(
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
The table does not create. I checked by using this:
if ($sqlConnection->query($sqlTableCreate) === TRUE) {
echo 'Created Sucessfully';
} else {
echo 'Table does not create.';
}
I get 'Table does not create' when trying to use this. Help would be greatly appreciated. Thanks in advance!
Your filename contains a extension, but I suspect you just want to use the name without the extension as the name of the table. You can use the basename function to remove the extension.
$sqlTableCreate = "CREATE TABLE ". basename($file_name, ".csv") . "(
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
If there can be different extensions, and you want to remove them more generally, see
How to remove extension from string (only real extension!)
I don't see any issue with your posted query but couple things may be wrong
Make sure that there is no table exists with that same name. You can use IF NOT EXISTS marker to be sure like
CREATE TABLE IF NOT EXISTS". $file_name . "(
make sure that the variable $file_name is not empty. Else, you are passing a null identifier in CREATE TABLE statement; which will not succeed.
Per your comment: you have $file_name = 'currentScan.csv';
That's the problem here. You are trying to create a table named currentScan.csv which your DB engine thinking that currentscan is the DB name and .csv is the table name which obviously doesn't exits and so the error.
first check your database connection and change your query with given below :
$sqlTableCreate = "CREATE TABLE ". $file_name . " (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
I try to update an existing table in mysql, but I get strange results, I explain my problem:
My table looks like this:
TABLE `myTable` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`photoName` varchar(255) COLLATE latin1_general_ci NOT NULL,
`vote` int(11) NOT NULL,
PRIMARY KEY (`id`),
UNIQUE KEY `photoName_2` (`photoName`),
)
and im trying to use saveVote.php that look like this:
$namePhoto = $_POST['name'];
$likePhoto = $_POST['like'];
mysql_connect("host","dbUser","psw");
mysql_select_db("db_is");
mysql_query("INSERT INTO `myTable` (`photoName`,`vote`) VALUES('$namePhoto','$likePhoto') ON DUPLICATE KEY UPDATE vote = vote + 1");
the 'vote' value is updated but every time when i call the "saveVote.php", for the first time he create an empty entry in my table with only the vote value and after, each time the "saveVote.php" is called
the vote value is updated for the right photoName but the vote value for the empty entry is also updated.
Why my request created this empty entry ?
Thanks for help.
It seems like your $namePhoto = $_POST['name']; is also returning a empty value. Try this:
if(!empty($_POST['name'])){
mysql_query("INSERT INTO `myTable` (`photoName`,`vote`) VALUES('$namePhoto','$likePhoto') ON DUPLICATE KEY UPDATE vote = vote + 1");
}
Keep in mind that this is just to test. This is not a fix. You need to figure out why you are sending a empty value.
I have a weird problem. Every time i execute this query in php i get the output "Challenge" even if the query is empty (should get "emptyq" if empty) when i test it in phpmyadmin everything is great and query is empty when it should be. I also tried to echo $detectChallengeRes[0][1] and got nothing. I cant find the problem, any help is very appreciated.
The script is suppose to look in the database and check if there is any challenges associated with the current userID, its basically a script that checks if a user has been challenged by another user, the gameID on the current page is the same as the one in the database and that the user hasnt completed the challenge already ($yourscore==0).
$detectChallengeRes = query("SELECT * FROM `AMCMS_challenges` WHERE `gameid`=$gameid AND `winner`=0 AND (`userkey1`=$user OR `userkey2`=$user);");
if($detectChallengeRes[0][1]!=$user && $detectChallengeRes[0][2]==$user) {
$yourscore = $detectChallengeRes[0][6]; //Check your score to see if you've already played
} elseif ($detectChallengeRes[0][2]!=$user && $detectChallengeRes[0][1]==$user) {
$yourscore = $detectChallengeRes[0][5]; //Check your score to see if you've already played
}
if ($detectChallengeRes!=NULL && $yourscore==0) {
echo 'Challenge';
} else {
echo 'emptyq';
}
Table structure:
CREATE TABLE IF NOT EXISTS `AMCMS_challenges` (
`primkey` int(11) NOT NULL auto_increment,
`userkey1` int(11) NOT NULL,
`userkey2` int(11) NOT NULL,
`gameid` int(11) NOT NULL,
`winner` int(11) NOT NULL,
`score1` int(11) NOT NULL,
`score2` int(11) NOT NULL,
PRIMARY KEY (`primkey`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ;
$detectChallengeRes will be boolean false or mysql result resource. It will not be ever null.
This might not solve your question but It looks like it is showing an previous data. Put this before your script
unset($detectChallengeRes);
Test for the number of rows returned by your query before trying to process it