tables not being created in database - php

Ok, I'm following a youtube guide on how to create a very simple blogging system using PHP/MySQL as I'd like to get to learn these 2 languages a bit more. I'm creating this in my localhost, permissions set-up correctly.
The problem is, when I go onto localhost/tables.php, it comes up as white screen which it's supposed to, but it's not creating the relevant tables within the database?
Here's the code I'm using:
mysql.php
<?php
mysql_connect('localhost','username','password'); //where localhost is the host, username is the relevant username and password is the relevant password.
mysql_select_db('database'); //where database is the chosen database in which to drop the tables.
?>
tables.php
<?php
include "mysql.php";
$table = "ENTRIES";
mysql_query ("CREATE TABLE IF NOT EXISTS `$table` (`ID` INT NOT NULL AUTO_INCREMENT , PRIMARY KEY ( `id` ) )");
mysql_query ("ALTER TABLE `$table` ADD `TITLE` TEXT NOT NULL");
mysql_query ("ALTER TABLE `$table` ADD `SUMMARY` TEXT NOT NULL");
mysql_query ("ALTER TABLE `$table` ADD `CONTENT` TEXT NOT NULL");
?>
Nothing is appearing in the error log which is frustrating and not helping me diagnose the problem.
Any help would be much appreciated! Thanks.

As you have no errors run this code to show if you have created created the table ENTRIES.
<?php
include "mysql.php";
$result = mysql_query("SHOW COLUMNS FROM `ENTRIES`");
if (!$result) {
echo 'Could not run query: ' . mysql_error();
exit;
}
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_assoc($result)) {
print_r($row);
}
}
?>
NOTE As you are starting with MySQL you would be advised to use PDO in place of the deprecated mysql_.
Here is a good tutorial
EDIT
Following comments the following code lists databases and tables on server(Note uses deprecated mysql_ function).
Ensure that the proper parameters replace "XXX".
<?php
$host= "localhost";
$username="XXX";
$password="XXX";
$database="XXX";
$link = mysql_connect($host,$username,$password); //where localhost is the host, username is the relevant username and password is the relevant password.
$db_list = mysql_list_dbs($link);
while ($row = mysql_fetch_object($db_list)) {
echo $row->Database . "\n";
echo "<BR>";
}
echo "<BR>";
$result = mysql_list_tables($database);
$num_rows = mysql_num_rows($result);
for ($i = 0; $i < $num_rows; $i++) {
echo "Table: ", mysql_tablename($result, $i), "\n";
echo "<BR>";
}
?>

Nothing is appearing in the error log which is frustrating and not helping me diagnose the problem
So your first poblem is to find out why it's not logging any errors. BTW it would also be a good idea to inject some echo / print statements to find out where it's failing.
Forget about the MySQL stuff and try:
<?php
trigger_error("Testing", E_USER_WARNING);
?>
Even though (once you've got the error reporting sorted out) you should get an error logged it will likely only contain a limited amount of information. Any time you call a mysql function, be via mysql_, mysqli or PDO, you should poll the return value and handle any error - even if it's just to echo the value to the screen.
It's saying that I haven't selected a database
Possibly the user account you are connecting as does not have permission to access the database.

Related

Selecting * from table returns nothing

I wrote this php script that allows me to fetch all the rows in a table in my MySQL database.
I have put the echo "1", etc. to see whether it gets to the code at the very end. The output proves it does. However, it does not output anything when echoing json_encode($resultsArray), which I can't seem to figure out why.
Code:
// Create connection
$connection = mysqli_connect("localhost", "xxx", "xxx");
// Check connection
if (!$connection) { die("Connection failed: " . mysqli_connect_error()); } else { echo "0"; }
// select database
if (!mysqli_select_db($connection, "myDB")) { die('Unable to connect to database. '. mysqli_connect_error()); } else { echo "1"; }
$sql = "select * from myTable";
$result = mysqli_query($connection, $sql) or die(mysqli_error($connection));;
echo "3";
$resultsArray = array();
while($row = mysqli_fetch_assoc($result)) {
// convert to array
$resultsArray[] = $row;
}
echo "4";
// return array w/ contents
echo json_encode($resultsArray);
echo "5";
Output:
01345
I figured, it is not about the json_encode, because I can also try to echo sth. like $result['id'] inside the while loop and it just won't do anything.
For testing, I went into the database using Terminal. I can do select * from myTable without any issues.
Any idea?
After around 20hrs of debugging, I figured out the issue.
As I stated in my question, the code used to work a few hours before posting this question and then suddenly stopped working. #MichaelBerkowski confirmed that the code is functional.
I remembered that at some point, I altered my columns to have a default value of an empty string - I declared them as follows: columnName VARCHAR(50) NOT NULL DEFAULT ''.
I now found that replicating the table and leaving out the NOT NULL DEFAULT '' part makes json_encode() work again, so apparently there's an issue with that.
Thanks to everybody for trying anyway!

how to insert value from radio button into mysql using php

i have tried this code to insert value into database, but i don't Know why, the value was not send into the databases. The table i have created in the mysql :
<?php
require_once "connection.php";
$conn = connect();
$db = connectdb();
mysql_select_db($db,$conn) or die (mysql_error() . "\n");
$query_usr = "select * from soalselidik";
$usr = mysql_query($query_usr,$conn) or die(mysql_error()."\n".$query_usr);
$row_usr=mysql_fetch_assoc($usr);
//to insert in database
$a1=$_POST['a1'];
$a2=$_POST['a2'];
$a3=$_POST['a3'];
$a4=$_POST['a4'];
$b1=$_POST['b1'];
$b2=$_POST['b2'];
$b3=$_POST['b3'];
$b4=$_POST['b4'];
$c1=$_POST['c1'];
$c2=$_POST['c2'];
$c3=$_POST['c3'];
$c4=$_POST['c4'];
$d1=$_POST['d1'];
$d2=$_POST['d2'];
$d3=$_POST['d3'];
$d4=$_POST['d4'];
$e1=$_POST['e1'];
$f1=$_POST['f1'];
echo $query ="insert into soalselidik (a1,a2,a3,a4,b1,b2,b3,b4,c1,c2,c3,c4,d1,d2,d3,d4,e1,f1) values('$a1','$a2','$a3','$a4','$b1','$b2','$b3','$b4','$c1','$c2','$c3','$c4''$d1','$d2','$d3','$d4','$e1','$f1')";
$result = mysql_query($query);
echo "<script languange = 'Javascript'>
alert('thankyou ! Penilaian anda diterima ');
location.href = 'home.php';</script>";
?>
'$c4''$d1'
Find that in your query and fix it :) And please do some error checking, and please stop using MySQL_* for your own good. Why should people not run any error checking mechanism that's already provided in the language and expect others to debug typos?
In case you didn't get it, there's a comma missing
How can I prevent SQL injection in PHP?

No Database Selected Php

This is a somewhat simple task, I am trying to compare a username to that in the database. I am testing it out in a single php file before I do it properly. My code is below. Basically I have a user, which is in the database and I am checking if it is in there.
<?php
$db_connect = mysql_connect("127.0.0.1", "root", "anwesha01", "mis");
//Check if connection worked.
if(!$db_connect)
{
die("Unable to connect to database: " . mysql_error());
}
//For testing purposes only.
else
{
echo "Database connect success!";
}
$user = "alj001";
$query = "SELECT Username FROM `user` WHERE `Username` = '".$user."'";
//What is being passed through the database.
echo "<p><b>This is what is being queried:</b> " . $query;
//Result
if (mysql_query($query, $db_connect))
{
echo "<br> Query worked!";
}
else
{
echo "<p><b>MySQL error: </b><br>". mysql_error();
}
?>
The result I get is:
Database connect success!
This is what is being queried: SELECT Username FROM user WHERE Username = 'alj001'
MySQL error: No database selected
First I had my mysql_query without the $db_connect as it is above, but I put it in and I still get "no database selected".
Ive looked at the w3c schools for the mysql_query function, I believe I have done everything correctly. http://www.w3schools.com/php/func_mysql_query.asp
Because you haven't called mysql_select_db. Note that the 4th parameter to mysql_connect is not what you think it is.
That said, you really should be using PDO or mysqli, not the plain mysql_ functions, since they're deprecated.

Blank Field in Mysql After Uploading Data

I was uploading data from my android application to a PHP file then inserting it to Mysql database, the problem i am having that i buy a new hosting plan and when i configured everything in the new hosting, i try to upload data from the application it shows only blank fields in the table, i am sure that there's no problem in the PHP or the android code, cause it was working fine and great with the old hosting.. i tried to change the encoding but same issue.
Here's the PHP file:
<?php
$con = mysql_connect("HOST","USER","PASS");
if (!$con)
{
die('Could not Connect:'. mysql_error());
}
mysql_select_db("TABLE",$con);
mysql_query ("INSERT INTO table (rep_desc,dateT) VALUES ('".$_REQUEST['report_Desc']."','".$_REQUEST['Date_Time']."')");
mysql_close($con);
?>
Thanks in advance
If there is any auto_increment column in table. Better check if its auto_increment flag not got uncheck.
Please also check length of database fields and data that you actually trying to insert.
Name of all request variables, columns, tables should be in proper case if it is a UNIX system.
get all the tables by something like this query ...
$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);
if (!$result) {
echo "DB Error, could not list tables\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}
while ($row = mysql_fetch_row($result)) {
echo "Table: {$row[0]}\n";
}

PHP SQL Truncate

I'm having a problem trying to truncate the 'requestID' field from my requests table.
This is my code.
<?php
include 'mysql_connect.php';
USE fypmysqldb;
TRUNCATE TABLE requestID;
echo "Request ID table has been truncated";
?>
I'm using server side scripting so no idea what error is coming back.
Anyone got an idea?
You aren't executing queries, you're just putting SQL code inside PHP which is invalid. This assumes you are using the mysql_*() api (which I kind of suspect after viewing one of your earlier questions), but can be adjusted if you are using MySQLi or PDO.
// Assuming a successful connection was made in this inclusion:
include 'mysql_connect.php';
// Select the database
mysql_select_db('fypmysqldb');
// Execute the query.
$result = mysql_query('TRUNCATE TABLE requestID');
if ($result) {
echo "Request ID table has been truncated";
}
else echo "Something went wrong: " . mysql_error();
Take a look at the function mysql_query which performs the query execution. The code to execute a query should look something like this.
$link = mysql_connect('host', 'username', 'password') or die(mysql_error());
mysql_select_db("fypmysqldb", $link) or die(mysql_error());
mysql_query("TRUNCATE TABLE requestID", $link) or die(mysql_error());
mysql_close($link);

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