UPDATE two rows - MySQL - php

I'm trying to update two rows in my database using a query (which is going to be run from a PHP script) and there is just one Condition (WHERE). What I've tried is:
$sql = 'UPDATE ' . CANNED_MESSAGES . "
SET canned_message_content = '" . $db->sql_escape($content) . "',
canned_message_title = '" . $db->sql_escape($title) . "'
WHERE id = '" . intval($id) . "'" ;
$db->sql_query($sql);
Can you tell me whats wrong with my query? :)

This may be due to Quotes mismatch. Please use this
$sql = "UPDATE '" . CANNED_MESSAGES ."'
SET canned_message_content = '" . $db->sql_escape($content) . "',
canned_message_title = '" . $db->sql_escape($title) . "'
WHERE id = '" . intval($id) . "' " ;

I highly doubt that two rows can have the same id column. Do they? If not, how could you update 2 rows by specifying a condition on a column with such a constraint?

Related

How to use the result of one query in another query (PHP/MySQL)

I have 2 tables (artist, cd) and I'm trying to use the result of the first query which returns an artID and make it equal to the artID in the 2nd table(cd) where artID is a foreign key but I'm not sure how to do it. Any help would be appreciated.
$strqueryID="SELECT artID FROM artist WHERE artName= '" . $_POST["category"] . "' ";
$resultsID=mysql_query ($strqueryID) or die(mysql_error());
$strqueryCD="INSERT INTO cd SET cdTitle='" . $_POST['title'] . "', artID='" . ??? . "' cdPrice='" . $_POST['price'] . "', cdGenre='" . $_POST['genre'] . "', cdNumTracks='" . $_POST['tracks'] . "'";
$resultsCD=mysql_query ($strqueryCD) or die(mysql_error());
You can use one single query, like this:
$strqueryCD="
INSERT INTO cd (cdTitle, artID, cdPrice, cdGenre, cdNumTracks)
VALUES(
'" . $_POST['title'] . "',
(SELECT artID FROM artist WHERE artName= '" . $_POST["category"] . "'),
'" . $_POST['price'] . "',
'" . $_POST['genre'] . "',
'" . $_POST['tracks'] . "')
";
also, google 'sqlinjection' before you continue
So, first thing's first - you shouldn't be using mysql_* functions now in 2017. I mean, really - they're actually even removed in later versions of PHP (7.0+). Refer to this StackOverflow post for more information.
Now, for your question at hand. Given the fact that you've searched for (and found) a given artID, you'll first have to get the actual "rows" from the $resultsID variable. In this example, we'll do it in a typical while loop:
while ($row = mysql_fetch_assoc($resultsID)) {
$strqueryCD="INSERT INTO cd SET cdTitle='" . $_POST['title'] . "', artID='" . $row['artID'] . "' cdPrice='" . $_POST['price'] . "', cdGenre='" . $_POST['genre'] . "', cdNumTracks='" . $_POST['tracks'] . "'";
$resultsCD=mysql_query ($strqueryCD) or die(mysql_error());
}
That should now loop over the artIDs that you've found in your first query and use them in the subsequent insert(s).
--
Disclaimer: I've disregarded the fact that user input is being passed straight into the query itself, as it's just too much "out of scope" for this post.

Duplicate data is getting updated ob table

I am looking to update one of the table. After I update, all the duplicate data is getting inserted again. Especially, the cloneSQL part of the code. I tried using DISTINCT, NOT EXISTS but no luck.
if(DB_num_rows($checkResult) > 0){
$cloneSQL = "UPDATE DISTINCT pricematrixdiscount SET
discount='" . $vals[3] . "'
WHERE debtorno='" . $_POST['cloneTo'] . "',
product_line='" . $vals[1] . "',
salestype='" . $vals[2] . "' ";
}
else {
$cloneSQL = "INSERT into pricematrixdiscount
(debtorno,
product_line,
salestype,
discount) VALUES
('" . $_POST['cloneTo'] . "',
'" . $vals[1] . "',
'" . $vals[2] . "',
'" . $vals[3] . "')";
How can I insert only distinct values on the pricematricdiscount table without the duplicates being inserted?

How to sql update two conditions

I have a problem with the condition 'where'.
I want one more condition in this code:
$sql="UPDATE
coursegrade
SET
FirstExam='" . mysql_real_escape_string($_POST['FirstExam']) . "',
SecondExam='" . mysql_real_escape_string($_POST['SecondExam']) . "',
ThirdExam='" . mysql_real_escape_string($_POST['ThirdExam']) . "',
Assignments='" . mysql_real_escape_string($_POST['Assignments']) . "',
FinalExam='" . mysql_real_escape_string($_POST['FinalExam']) . "'
WHERE
SID=" . mysql_real_escape_string($_POST['SID']) ;
Tell now I have no problem .. but the problem is that I don't know how to set the second condition.
CourseID=" . mysql_real_escape_string($_POST['CourseID'])
I want the condition to be something like...
WHERE
SID=" . mysql_real_escape_string($_POST['SID'])
AND CourseID=" . mysql_real_escape_string($_POST['CourseID'])
How could I do it?
Unless you use heredoc syntax php will parse strings on a single line.
ie the rendered clause is:
"WHERE SID=19AND CourseID=45"
Basically you missed a space before
"AND CourseID=" . mysql_real_escape_string($_POST['CourseID'])
or you could put quotes around the values
"SID='" . mysql_real_escape_string($_POST['SID']) . "'
AND CourseID='" . mysql_real_escape_string($_POST['CourseID'])."'"
This is may help
WHERE
SID=" . mysql_real_escape_string($_POST['SID'])
OR CourseID=" . mysql_real_escape_string($_POST['CourseID'])
or
WHERE
SID in (
mysql_real_escape_string($_POST['SID']),
mysql_real_escape_string($_POST['CourseID'])
)
its will work for INT value of SID and CourseID
You can try this, and let me know if there is error message
$sql="UPDATE coursegrade
SET
FirstExam = '" . mysql_real_escape_string($_POST[' FirstExam ']) . "',
SecondExam = '" . mysql_real_escape_string($_POST[' SecondExam ']) . "',
ThirdExam = '" . mysql_real_escape_string($_POST[' ThirdExam ']) . "',
Assignments = '" . mysql_real_escape_string($_POST[' Assignments ']) . "',
FinalExam = '" . mysql_real_escape_string($_POST[' FinalExam ']) . "'
WHERE
SID = ". mysql_real_escape_string($_POST['SID'])."
AND
CourseID = " . mysql_real_escape_string($_POST['CourseID']) .";

Updating Mysql table data with PHP

$updateSeats = mysql_query("UPDATE FORM_dateAndSeating SET NumberOfSeats = " . $removeSeatingNumber . " WHERE DATE = " . $revertToStandardDate);
In the code above I am trying to update the value within the MYSQL table.
When I echo the variables they show the data I am expecting, however the database is not being updated.
There is no error being returned either.
What are other possibilities for the sql not to update properly??
This will work:
$updateSeats = mysql_query("UPDATE FORM_dateAndSeating
SET NumberOfSeats = '" . $removeSeatingNumber . "'
WHERE DATE = '" . $revertToStandardDate . "'");
Long form:
$updateSeats = mysql_query("UPDATE FORM_dateAndSeating SET NumberOfSeats = '" . $removeSeatingNumber . "' WHERE DATE = '" . $revertToStandardDate . "'");
The variables need to be inside double quotes including single quotes
I.e.: '" . $removeSeatingNumber . "' WHERE DATE = '" . $revertToStandardDate . "'
-------^ --------------------------------------------^ -----------------------^ ----------------------------------------------^
Add apostrophes around your column values.

hacking an INSERT query to become a mysql UPDATE query

I am a MySQL noob and basically hacking an insert query to become an update query instead. So I am sure it's something simple with the grammar. But what's wrong with this?
// Save data
$mySQLQuery = 'update `'. $fl['mysql_table']. '` SET '. $fl['mysql_query']. "' WHERE speres = '" . mysql_real_escape_string($_POST['speres']);
$rs = #mysql_query($mySQLQuery);
the original INSERT query (working) was
// Save data
$mySQLQuery = 'INSERT INTO `'. $fl['mysql_table']. '` SET '. $fl['mysql_query'];
$rs = #mysql_query($mySQLQuery);
The data is generated here:
$fl['mysql_query'] = "menrecin = '" . mysql_real_escape_string(YDFLValue($_SESSION['form']['item_17'])) . "', menrecvej = '" . mysql_real_escape_string(YDFLValue($_SESSION['form']['item_18'])) . "', menrecser = '" . mysql_real_escape_string(YDFLValue($_SESSION['form']['item_19'])) . "', menrecud = '" . mysql_real_escape_string(YDFLValue($_SESSION['form']['item_20'])) . "', menresmor = '" . mysql_real_escape_string(YDFLValue($_SESSION['form']['item_22'])) . "', menresfro = '" . mysql_real_escape_string(YDFLValue($_SESSION['form']['item_23'])) . "', menresmid = '" . mysql_real_escape_string(YDFLValue($_SESSION['form']['item_24'])) . "', menresres = '" . mysql_real_escape_string(YDFLValue($_SESSION['form']['item_25'])) . "', menrumind = '" . mysql_real_escape_string(YDFLValue($_SESSION['form']['item_28'])) . "', menrumren = '" . mysql_real_escape_string(YDFLValue($_SESSION['form']['item_29'])) . "', menrumved = '" . mysql_real_escape_string(YDFLValue($_SESSION['form']['item_30'])) . "', tekip = '" . $_SERVER['REMOTE_ADDR'] . "', tekbro = '" . $_SERVER['HTTP_USER_AGENT'] . "', tektid = NOW()";
I have an entry with speres = 100525 in the database, so please try:
http://www.konferencer.nu/form/index.php?speres=100525
Good practices of troubleshooting dynamic SQL:
Look at the SQL, not the code that builds the SQL. In other words, echo out $mySQLQuery to see the final SQL, and most of the time you can see the error right away.
Don't suppress errors. Error-checking is helpful and necessary in any code.
It looks to me like your query ends up being:
update `tablename` SET ..., tektid = NOW()' WHERE speres = '...;
So you have a spurious quote after the NOW() and a missing quote at the end.
If you had checked for errors, you'd get something like this:
ERROR 1064 (42000): You have an error in your SQL syntax;
check the manual that corresponds to your MySQL server version for
the right syntax to use near '' WHERE speres = '...' at line 1
The quoting around the start of the WHERE clause looks odd:
UPDATE `...some table...` SET ...some query... 'WHERE speres = ' ... some criterion ...
Note the single quote placement. Maybe you want to remove the single quotes from inside the double quotes?
you query should look like
$mySQLQuery = 'update'. $fl['mysql_table'].'SET'. $fl['mysql_query'].'= <some value>' ' WHERE speres = '.mysql_real_escape_string($_POST['speres']);
$rs = #mysql_query($mySQLQuery);

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