I am trying to run 2 update queries as one `$result, Is this possible, and if so can someone please show me where am I going wrong with the below query?
$query = "INSERT INTO ptb_users (id, user_id, first_name, last_name, email, password )
VALUES('NULL','NULL','" . $firstname . "','" . $lastname . "','" . $email . "',MD5('" . $password . "'))";
mysql_query($query) or dieerr();
$result = mysql_query("UPDATE ptb_users SET ptb_users.user_id=ptb_users.id UPDATE ptb_users SET ptb_users.account_type= \"Client\"");
Mysql* functions are deprecated. Use PDO/mysqli instead.
As a security measure, mysql* functions wont allow you to execute multiple queries. You can with Mysql* however, you should prepare your statement before executing.
Since you have no WHERE clause on your UPDATEs, you can combine them like this:
UPDATE ptb_users SET ptb_users.user_id = ptb_users.id,
ptb_users.account_type = 'Client'
EDIT: If you want to UPDATE the row, you just INSERTed, then just add WHERE ptb_users.id = $id (you can use mysql_insert_id() to get the ID).
$query = "INSERT INTO ptb_users (id, user_id, first_name, last_name, email, password )
VALUES('NULL','NULL','" . $firstname . "','" . $lastname . "','" . $email . "',MD5('" . $password . "'))";
mysql_query($query) or dieerr();
$id = mysql_insert_id();
$result = mysql_query("UPDATE ptb_users SET ptb_users.user_id = ptb_users.id,
ptb_users.account_type = 'Client' WHERE ptb_users.id =".$id);
Related
I have a select statement and an update statement. What I would like to do in the update statement is set the value of 'recipes_saved' to the result of the select statement. I have tried to use this:
$query = "UPDATE `users` SET `recipes_saved` = ('SELECT `recipe_1_name` FROM `carbohydrates`') WHERE `user_id` = '" . $_SESSION['user_id'] . "'";
$data= mysqli_query($dbc,$query) or die('Query failed: ' . mysqli_error());
but the query fails.
Any help would be much appreciated.
I think you have an extra ' in your 'SELECT and also in your FROM carbohydrates' and use LIMIT again like:
Try to copy the query below:
$query = "UPDATE `users` SET `recipes_saved` = (SELECT `recipe_1_name` FROM `carbohydrates` LIMIT 1) WHERE `user_id` = '" . $_SESSION['user_id'] . "'";
You could of course remove the back tick if you want to make it less cluttering, like:
$query = "UPDATE users SET recipes_saved = (SELECT recipe_1_name FROM carbohydrates LIMIT 1) WHERE user_id = '" . $_SESSION['user_id'] . "'";
As far as I know, you do not need so many quotes in your query. Try:
$query = "UPDATE users SET recipes_saved = (SELECT recipe_1_name FROM carbohydrates) WHERE user_id='" . $_SESSION['user_id'] . "'";
It would also be useful to log in to your database directly (either command line or a GUI client) and try running the query:
UPDATE users SET recipes_saved = (SELECT recipe_1_name FROM carbohydrates) WHERE user_id='username'
and see if that works.
Here is my double-minded query:
$Quest = "SELECT * FROM TOAWorkorders";
$FindTechResult = mysql_query($Quest, $cxn)
or die ('The easter bunny is watching you' . mysql_error());
while ($row = mysql_fetch_array($FindTechResult))
{
if (strpos($BBT, 0, 3) != 'Sys')
{
$IdNum = $row['IdNum'];
$BBT = $row['BBT'];
$BBTArray = explode("-", $BBT);
$TechNum = $BBTArray["0"];
$Title = $BBTArray["2"];
$Name = explode(" ", $BBTArray['1']);
$FirstName = $Name["0"];
$LastName = $Name["1"];
}
echo $TechNum . ' !! ' . $FirstName . ' !! ' . $LastName . ' !! ' . $Title . '<br>';
$Quest = "UPDATE TOAWorkorders SET TechNum = '$TechNum', FirstName = '$FirstName', LastName = '$LastName', Title = '$Title' WHERE IdNum = '$IdNum'";
$result = mysql_query($Quest, $cxn) or die(mysql_error());
}
Everything works for about 2/3s of the database. That leaves 33,000 rows that are not updated. I cannot find any difference between the data that works and the data that doesn't.
Since you're doing an UPDATE, and you say the rest of the code works (meaning, I hope, that you get 109,112 echo'ed results), it must be that the ID isn't being found (WHERE IdNum = '$IdNum').
Try preceding that command with "SELECT COUNT(*) from TOAWorkorders WHERE IdNum = '$IdNum'" and see if you get 33,000 zeros when the program runs. If you do, then you have missing IdNum values in your table.
If you don't, please provide details and I'll let you know.
I have a column in my MySQL database for deleted values are usually 0 or 1. Usually when doing searches I omit the deleted by doing something like "and deleted = "0"" but I cant figure out how to get this query below to omit my deleted column. any ideas would be appreciated thanks!
$query = "SELECT Status, COUNT(Status) FROM Assets GROUP BY Status";
$result = mysql_query($query) or die(mysql_error());
// Print out result
while($row = mysql_fetch_array($result))
{
echo "". $row['COUNT(Status)'] ." systems ". $row['Status'];
echo ", ";
}
This should work:
$query = "SELECT Status, COUNT(Status) FROM Assets WHERE Assets.deleted = 0 GROUP BY Status";
Also, you should use mysqli or PDO rather than mysql (the php mysql library is obsolete).
A simple question!:
Can I use my field value in an insert query?
For example, I've a field named id, it is an auto_increment field. i want to add this field value to another field. while I'm inserting it.
A simple php code of my need:
$query_1 = mysql_query("INSERT INTO table (name) VALUES ('abcd')"); // id automatically increment
$query_2 = mysql_query("SELECT id FROM table WHERE name = 'abcd'"); // selects previous id
// fetches result //
$query_3 = mysql_query("UPDATE table SET code = '" . $id + 1000 . "' WHERE id = '" . $id . "');
Can you convert it to just 1 query?
Thanks ...
I don't think you can do it in one query but you can do it in 2 for sure:
mysql_query("INSERT INTO table (name) VALUES ('abcd')");
$id = mysql_insert_id();
mysql_query("UPDATE table SET code = '" . $id + 1000 . "' WHERE id = '" . $id . "');
Here's how you do it with one query:
$query_1 = "INSERT INTO table (note) VALUES ('abcd'); UPDATE table SET code = LAST_INSERT_ID() + 1000 where id = LAST_INSERT_ID()";
I can't remember if mysql allows multiple queries in one command - I think maybe not, so try this:
$query_1 = "INSERT INTO table (note) VALUES ('abcd')";
$query_2 = "UPDATE table SET code = id + 1000 where id = LAST_INSERT_ID()";
I'm trying to a single value in my DB...When I run it through the console, it works correctly (as I'm replacing the variables with numbers and text).. However, My query is not returning a value for book ID when I insert the PHP variable for it.. It's because the book_id is unpopulated...
$query = "UPDATE books "
. "SET readstatus='".$readstatus."' "
. "WHERE book_id=".$book_id;
echo $query
The echoed query states:
UPDATE books SET readstatus='half' WHERE book_id=0
The book ID is stored in the URI as bookstatusupdate.php?book_id=
Just cannot figure this one out!
It would help to know the error. Firstly, echo out the query:
$query = "UPDATE books "
. "SET readstatus='".$readstatus."' "
. "WHERE book_id=".$book_id;
echo $query;
I would guess that $book_id is unpopulated, so the query fails. What you should really be doing to make it secure is casting integers with (int) and wrapping strings in mysqli_real_escape_string().
$query = "UPDATE books "
."SET readstatus='". mysqli_real_escape_string( $readstatus )."' "
."WHERE book_id=". (int) $book_id;
If you're trying to get data from the URL, do it like so:
$book_id = (int) $_GET['book_id'];
$query = "UPDATE books "
."SET readstatus='". mysqli_real_escape_string( $readstatus )."' "
."WHERE book_id=". (int) $book_id;
echo $query;