i have the following code in php:
$host="localhost"; // Host name
$username="***"; // username
$password="***"; // password
$db_name="***"; // Database name
//$rc_profile_table="rc_profile_table"; // Table name
//$rc_profile_relation_table="rc_profile_relation_table"; // Table name
mysql_connect("$host", "$username", "$password");
mysql_select_db("$db_name");
$sql="SELECT created_at FROM rc_profile_table where created_at > 2011-04-19 08:00:00";
$result=mysql_query($sql);
$count=mysql_num_rows($result);
$sql="SELECT created_at FROM rc_profile_relation_table where created_at > 2011-04-19 08:00:00";
$result2=mysql_query($sql);
$count2=mysql_num_rows($result);
mysql_close();
What do you actually want to do?
You have to describe the problem else no one can help you...
You have no proper error handling. The mysql functionality provided with php have a build in function that outputs the error on the screen.
This would be a lot better:
<?php
$host="localhost"; // Host name
$username="***"; // username
$password="***"; // password
$db_name="***"; //db name
$connection = mysql_connect($host, $username, $password) or die("Could not connect to the database: " . mysql_error());
mysql_select_db($db_name, $connection) or die("Could not select database: " . mysql_error());
$sql = "SELECT `created_at` FROM `rc_profile_table` WHERE `created_at` > '2011-04-19 08:00:00'";
$result = mysql_query($sql) or die("Could not execute query: " . $sql . "ERROR: " . mysql_error());
$count = mysql_num_rows($result);
mysql_close($connection) or die(mysql_error());
?>
In addition to the error handling already mentioned, for your second resultset, you might want to ensure that $count2 is the number of rows returned in $result2 rather than in the first resultset ($result)
$sql="SELECT created_at FROM rc_profile_relation_table where created_at > 2011-04-19 08:00:00";
$result2=mysql_query($sql);
$count2=mysql_num_rows($result2);
Related
This is my PHP code written for website. When I executed this, the query doesn't execute and doesn't show any error. I also checked data types of values that are to be inserted.
The database username and password and all credentials are correct. What could be the problem?
<?php
$password ='abcdef';
$host="localhost"; // Host name
$username="futureti_dsatya"; // Mysql username
$password="D2e3e4v1i"; // Mysql password
$db_name="futureti_db"; // Database name
$tbl_name="users"; // Table name
// Connect to server and select databse.
$con = mysqli_connect($host, $username, $password,$db_name);
if(!$con)
{
die('Could not connect: '. mysql_error());
}
else
{
$res = mysqli_query($con,"insert into users values(55555623,'saran1','satya_saran',$password)");
if($res){
print("i am ok");
}
else{
print("bad");
}
}
?>
Wrap $pass in quotes (55555623,'saran1','satya_saran','$pass') as shown below with an explanation about "$password", and change mysql_error()); to mysqli_error()); those two functions don't mix and that is why you did not get the proper error messages to show up.
As already stated, you're using $password twice; change one of the variables to something else.
What you're presently doing is overwriting your $password variable.
I am assuming you want to enter abcdef into your DB. If so, then do this instead:
<?php
$pass ='abcdef';
$host = "localhost"; // Host name
$username = "futureti_dsatya"; // Mysql username
$password = "D2e3e4v1i"; // Mysql password
$db_name = "futureti_db"; // Database name
$tbl_name = "users"; // Table name
// Connect to server and select databse.
$con = mysqli_connect($host, $username, $password,$db_name);
if ( !$con ) {
die('Could not connect: '. mysqli_error());
} else {
$res = mysqli_query($con, "insert into users values (55555623,'saran1','satya_saran','$pass' )");
if( $res ) {
print("i am ok");
} else {
print("bad");
}
}
?>
Also, inserting data into a table without telling it which columns to use is not a recommended method.
Use something to the effect of:
($con, "insert into users (column1, column2, column3, column4) values (55555623,'saran1','satya_saran','$pass' )
Sidenote: If the column for your first value isn't an (int) you will need to wrap that in quotes as well.
Also, if your first column is an AUTO_INCREMENT, you will need to remove the AUTO_INCREMENT from the column's type.
you dont get any error because you are making using mysql. not mysqli.
your code is wroking just wrap password . i guess the connection is not connecting.
replace this:
die('Could not connect: '. mysql_error());
to
die('Could not connect: '. mysqli_error()); //to see the error
I've got two different sites. What I'd like to do is to automatically run a script that sends some of the data inserted into the database in site 1 when a user registers and updates a table in the database for site 2 so that an account is automatically created in site 2 using the same details.
I'm at the stage of trying to create a query that will update the database. I'm the self-made type so don't know that well what I'm doing. Got this query from somewhere but can't make it work. Can anyone tell what's wrong with it? It's not executing the query.
Thanks!
Eugenie
<?php
$host = "localhost"; // Host name
$username = "----"; // Mysql username
$password = "----"; // Mysql password
$db_name1 = "------"; // Database name
$db_name2 = "-----"; // Database name
$tbl_name1 = "-----"; // Table name
$tbl_name2 = "---"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name1")or die("cannot select DB");
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name2")or die("cannot select DB");
$query = "USE $db_name2
UPDATE $db_name2.dbo.$tbl_name2
SET email=d2.email FROM $db_name1.dbo.$tbl_name1 d2
WHERE d2.uid = $tbl_name1.uid";
$result = mysql_query($query) or die ("could't execute query.");
?>
<?php
$host = "localhost"; // Host name
$username = "----"; // Mysql username
$password = "----"; // Mysql password
$db_name1 = "------"; // Database name
$db_name2 = "-----"; // Database name
$tbl_name1 = "-----"; // Table name
$tbl_name2 = "---"; // Table name
$conn = mysql_connect($host, $username, $password);
mysql_select_db($db_name1, $conn) or die("cannot select DB");
mysql_select_db($db_name2, $conn) or die("cannot select DB");;
$query1 = "SELECT * FROM `" . $db_name1.$tb1_name1 . "` ";
$query2 = "SELECT * FROM `" . $db_name2.$tb1_name2 . "` ";
You can fetch data of above query from both database as below
$result1 = mysql_query($query1);
while($row = mysql_fetch_assoc($result1)) {
$data1[] = $row;
}
$result2 = mysql_query($query2);
while($row = mysql_fetch_assoc($result2)) {
$data2[] = $row;
}
print_r($data1);
print_r($data2);
?>
Suggestion: Try shifting to mysqli or PDO since mysql is depreciated now.
Recall the documentation for mysql_connect:
Returns a MySQL link identifier on success or FALSE on failure.
... and the documentation for the second parameter for mysql_query:
The MySQL connection. If the link identifier is not specified, the last link opened by mysql_connect() is assumed. If no such link is found, it will try to create one as if mysql_connect() was called with no arguments. If no connection is found or established, an E_WARNING level error is generated.
... should solve your problem. Example:
$link1 = mysql_connect( ... ); // For db 1.
$link2 = mysql_connect( ... ); // For db 2.
$result1 = mysql_query( "some query for db 1", $link1 );
$result2 = mysql_query( "some query for db 2", $link2 );
Well,
first of all, you're not connecting to two different databases, but using two different schemas in the same database. So only a mysql_connect should be used.
Also, if you're using full qualified names to access your tables you don't need to call mysql_select_db, nor the 'use db_name' mysql command.
Your query string is wrong. After USE $db_name2 you should have a semi-colon, and the update sentence is not correct.
Code could be somthing like that:
mysql_connect(...)
$query = "update $db2.$table2, $db1.$table1
I'm fairly new to SQL and PHP.
I'm trying to write a simple login script. I have a form in a HTML document that I have proved posts the correct data into the 2 variables required but my script fails when it executes the SQL...
I've also tested the SQL in mysqlWorkbench and I get the result I want ???
Please help.
Here is my script:
<?PHP
$odbc = mysql_connect('localhost', 'root', '') or die ("could not connect to database");
mysql_select_db('examresults', $odbc) or die("Could not find database");
// username and password sent from form
$username=$_POST['username'];
$password=$_POST['password'];
$sql='SELECT * FROM tuser where username = '.$username.' and password = '.$password.'';
$result = mysql_query($sql, $odbc) or die ("Error in SQL");
// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
//If result matched username and password, table row must only equal 1 row
if($count==1)
{
header("location:exammenu.php");
}
else
{
echo 'username and password do not match';
}
?>
Note: mysql_* functions are deprecated, you should not use them anymore. Your code is also vulnerable to SQL Injections.
Using mysql_error instead of just printing out "Error in SQL" would give us (and you) a more detailed sql error message. But most likely it is failing because you forgot to put " " around your strings in the query.
$sql='SELECT * FROM tuser where username = "'.$username.'" and password = "'.$password.'"';
If you're really going to need to use mysql, at least sanitize your input. Also note the quotes in the $sql variable. This should work (though not tested):
<?PHP
$odbc = mysql_connect('localhost', 'root', '') or die ("could not connect to database");
mysql_select_db('examresults', $odbc) or die("Could not find database");
// username and password sent from form
$username=mysql_real_escape_string($_POST['username'], $odbc);
$password=mysql_real_escape_string($_POST['password'], $odbc);
$sql=sprintf('SELECT * FROM tuser where username = "%s" and password = "%s"', $username, $password);
$result = mysql_query($sql, $odbc) or die ("Error in SQL");
// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
//If result matched username and password, table row must only equal 1 row
if($count==1)
{
header("location:exammenu.php");
}
else
{
echo 'username and password do not match';
}
I suggest using sprintf to format your sql statement to make it easier to spot such errors.
The query should be as below:
$sql='SELECT * FROM tuser where username = "'.$username.'" and password = "'.$password.'"';
you can try this code. i think it will work correctly.
<?PHP
$odbc = mysql_connect('localhost', 'root', '') or die ("could not connect to database");
mysql_select_db('examresults', $odbc) or die("Could not find database");
// username and password sent from form
$username=$_POST['username'];
$password=$_POST['password'];
$sql="SELECT * FROM tuser where username = '".$username."' and password = '".$password."'";
$result = mysql_query($sql, $odbc) or die ("Error in SQL");
// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
//If result matched username and password, table row must only equal 1 row
if($count==1)
{
header("location:exammenu.php");
}
else
{
echo 'username and password do not match';
}
?>
Here is my php page which is called on button click.
<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="parth"; // Database name
$con = mysql_connect("localhost","root","") or die('Could not connect to MySQL server: ' . mysql_error());
//echo"Connected to MySQL server";
mysql_select_db("parth") or die('Could not connect to database' . mysql_error());
//echo "Connected to Database";
$sql="CREATE EVENT gold ON SCHEDULE AT CURRENT_TIMESTAMP + INTERVAL 3 SECOND
DO
DELETE FROM table LIMIT 1;";
mysql_query($sql);
echo "done";
?>
the echo message "done" is displayed
But the event query does not get executed. I have given the privileges too.
Just used 3 SECOND for development purpose.
Try this:
<?php
error_reporting(E_All);
$sHost = "localhost";
$sUsername = "root";
$sPassword = "";
$sDb = "parth";
$con = mysql_connect($sHost, $sUsername, $sPassword) or die('Could not connect to MySQL server: ' . mysql_error());
echo"Connected to MySQL server";
mysql_select_db($sDb) or die('Could not connect to database' . mysql_error());
echo "Connected to Database";
$sResult = mysql_query("CREATE EVENT gold ON SCHEDULE EVERY 3 MINUTE STARTS CURRENT_TIMESTAMP + INTERVAL 3 SECOND DO DELETE FROM table LIMIT 1;") or die(mysql_error());
echo "Done";
I've spent quite a long time trying to figure out why this block of code is not working. After logging in (on an html page that uses post to send the data to this php document) it says: "cannot select the database!!" (without quotes). Please help! Thank you!
<?php
$host="localhost";
$username="root";
$password="";
$db_name="firstTestLogins";
$tbl_name="members";
mysql_connect("$host", "$username, $password") or die("cannot connect to the database!!");
mysql_select_db("firstTestLogins") or die("cannot select the database!!");
$username=$_POST['username'];
$password=$_POST['password'];
$username=stripslashes($username);
$password=stripslashes($password);
$username=mysql_real_escape_string($username);
$password=mysql_real_escape_string($password);
$password = md5($password);
echo "This is a debug statement. User = $username and password = $password <br><br><br>";
if(!($sql="SELECT * FROM $tbl_name WHERE username = '$username' and password = '$password")) {die(mysql_error());}
$result = mysql_query($sql);
$test3 = mysql_num_rows(/*$result*/mysql_query($sql));
echo "$test3";
if(mysql_num_rows(/*$result*/mysql_query($sql))) { //should be true because only one row should match the user and pass
$_SESSION['username'] =1;
$_SESSION['password'] =1;
header("location:login_success.php");
}
else {
echo "The incorrect username or password was inserted";
}
?>
This is wrong:
mysql_connect("$host", "$username, $password");
it needs to be
mysql_connect($host, $username, $password);
you will see the exact problem by outputting mysql_error().
mysql_select_db("firstTestLogins")
or die("cannot select the database! ".mysql_error());
please change the code from:
mysql_select_db("firstTestLogins") or die("cannot select the database!!");
to:
mysql_select_db("firstTestLogins") or die(mysql_error());
This will help you to figure out why MYSQL cannot select the DB
In addition to the comments about selecting the database:
if(!($sql="SELECT * FROM $tbl_name WHERE username = '$username' and password = '$password")) {die(mysql_error());}
Silly doing a test with a mysql_error if the $sql string isn't being set... this won't generate a mysql error
You're also missing a closing ' in your $sql
$sql="SELECT * FROM $tbl_name WHERE username = '$username' and password = '$password'";
$result = mysql_query($sql);
if(!$result)
die(mysql_error());
Only thing I can think of is that the firstTestLogins database either doesn't exist on the database or is not accessible by the username/password combination you're using.