I really want the data in my database to be in the same row.
My problem is that when I submit the start time the end time automatically becomes like this: 00:00:00
Here is my code:
<?php
//Connect and select a database
mysql_connect("localhost", "root" , "" )or die("cannot connect to database server");
mysql_select_db("dbdorm")or die("cannot select the database");
//if form is submitted
if(isset($_POST['submit']))
{
date_default_timezone_set('Asia/Manila');
$query = "INSERT INTO datetime (starttime) VALUES(NOW())";
$submit = mysql_query($query) or die("insertion error");
header("location:loggedin.php?");
}
else{
date_default_timezone_set('Asia/Manila');
$query = "INSERT INTO datetime (endtime) VALUES(NOW())";
$submit2 = mysql_query($query) or die("insertion error");
header("location:loggedin.php?");
}
?>
Related
I am working on a site to share names of songs, and I have made a recommendation form that I include in every page. This recommendation form is in HTML and leads to a PHP action page, where the information received is added to a SQL table. Here is the code:
<?php
ob_start();
$host="localhost"; // Host name
$username="root"; // Mysql username
$password="MYPASSWORD"; // Mysql password
$db_name="DB NAME"; // Database name
$tbl_name="songshare"; // Table name
// Connect to server and select databse.
$link = mysqli_connect("$host", "$username", "$password")or die("cannot connect");
mysqli_select_db($link, "$db_name")or die("cannot select DB");
// Define $myusername and $mypassword
$song=$_POST['song'];
$album=$_POST['album'];
$artist=$_POST['artist'];
$linkitunes=$_POST['linkitunes'];
$artwork=$_POST['albumPic'];
// To protect MySQL injection (more detail about MySQL injection)
$song = stripslashes($song);
$album = stripslashes($album);
$artist = stripslashes($artist);
$song = mysqli_real_escape_string($link, $song);
$album = mysqli_real_escape_string($link, $album);
$artist = mysqli_real_escape_string($link, $artist);
$sql="SELECT * FROM $tbl_name WHERE song='$song'";
$result=mysqli_query($link, $sql);
if ($result->num_rows){
echo "Song already taken" . "<br />";
echo "<a href='/music.php'>music</a>";
exit();
}
$sql="INSERT INTO recommendation (user_id, artist, song, album, artwork, linkitunes)";
$sql = $sql . " VALUES ('$_SESSION['user_id']', '$artist', '$song', '$album'. '$artwork'. '$linkitunes');";
$result=mysqli_query($link, $sql);
if(!$result) {
echo "Recommendation failed" . "<br />";
echo $sql;
} else {
print "$song, $artist, $album";
}
ob_end_flush();
?>
I have checked that every username, password, link is correct and valid. My server does, in fact, run PHP. It doesn't seem to me like the PHP code is even running though.
Thank you so much in advance.
-Cameron
Turn on error reporting by adding this on top of page:
ini_set("display_errors",true);
and change this line:
$link = mysqli_connect("$host", "$username", "$password")
to
$link = mysqli_connect($host, $username, $password,$db_name);
Please have a look how to work with mysqli
Instead of '$album'. '$artwork'. '$linkitunes' Do: '$album', '$artwork', '$linkitunes', while saving data.
Try this :-
$sql = $sql . " VALUES ('".$_SESSION['user_id']."', '$artist', '$song', '$album', '$artwork', '$linkitunes')";
instead of
$sql = $sql . " VALUES ('$_SESSION['user_id']', '$artist', '$song', '$album'. '$artwork'. '$linkitunes');";
You should check the version of local server you are working with. If you are working with a higher of local server and you php was written in a lower version it throws a blank page.
I have a form that posts variables back from a drop down list in php. I then need to take those values and update my database values for each "TeamName". I am using the following code but nothing is updating.
$g1 = mysql_real_escape_string($_POST['g1']);
$conn = mysql_connect("****.com","****","*****") or die("Connection to MYSQL");
mysql_select_db("****_teamlist", $conn) or die("Connection to MYSQL database failed");
$sSQL = "UPDATE Sheet1 Set g1='$g1' WHERE TeamName = 'TeamName' ";
$result = mysql_query($sSQL, $conn) or die(mysql_error());
Where is my error? I am stuck.
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 working on my first PHP/MySQL project, and I've gotten basic logins and INSERT queries working, but not updates. This is my first update, which is just one row with a state and zipcode. Is anything wrong?
$dbc = mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$state=$_POST['state'];
$zip=$_POST['zip'];
$custnum = 0;
$sql="UPDATE $tbl_name SET state = '$state', zip = '$zip', WHERE custnum = '$custnum'";
$result = mysqli_query($dbc, $sql)
or die('Error querying database.');
$sql="UPDATE {$tbl_name} SET state='{$state}', zip='{$zip}' WHERE custnum='{$custnum}'";
Remove the last comma before "WHERE" clause. Also, if you're just starting out it's good to put parenthesis around variables names when using double-quotes for strings. Helps you to distinguish the variables better.
Pekka is also correct in his comments, you are mixing mysql and mysqli functions. Use mysql_query() instead.
I think you need to get rid of the comma just before the WHERE.
$suitno =mysqli_real_escape_string($ecms,$_POST['suitno']);//protecting sql injection
$defendant=mysqli_real_escape_string($ecms,$_POST['defendant']);//protecting sql injection
$casenature=mysqli_real_escape_string($ecms,$_POST['casenature']);//protecting sql injection
$sql="UPDATE causelist SET suitno='{$suitno}',
casenature='{$casenature}' WHERE suitno='{$suitno}'";
$result = mysqli_query($ecms, $sql)
or die('Error querying database.');
$dbc = mysql_connect($host, $username, $password)or die("cannot connect"); //don't need quotes
mysql_select_db($db_name,$dbc)or die("cannot select DB"); //added the $dbc (connection link) as a second parameter
$state=mysql_real_escape_string($_POST['state']); //Should make it safe!
$zip=mysql_real_escape_string($_POST['zip']); //Should make it safe!
$custnum = 0;
$sql="UPDATE $tbl_name SET state = '$state', zip = '$zip' WHERE custnum = '$custnum'";
//removed an extra comma
//Notice that $tbl_name isn't defined!
u
$result = mysql_query($sql)
or die('Error querying database.'); //from mysqli to mysql
Looks like a sql syntax error:Remove the comma before WHERE
if(isset($_POST['update']))
{
$name=$_POST['name'];
//echo $name; die;
$surname=$_POST['surname'];
$upd="update table_name SET name='$name',surname='$surname' where id=$id";
mysql_query($upd);
}
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);