mysql replace into with SET-syntax overrides all fields - php

I have the following query:
REPLACE INTO `oxarticles`
SET
OXID = '10-1010',
oxartnum = '10-1010',
oxtitle = 'Dummy',
oxprice = '10.000000',
oxstock = '100',
importstatus = 1"
This works so far as expected, but the fields I do not specifiy, are just overwritten with ' ' / empty string. From what I read, should this syntax work identically like the UPDATE-command.
Am I missing something? How can I prevent that fields are replaced with '' ?
Edit 1
Just to clarify, I can't just use UPDATE. I am setting a flag (importstatus) to 0 before every run and during the import to 1. After the import finishes, I delete all articles, which are still on status 0.
// Just for the compeletion, here is the PHP-snippet:
while (!feof($this->handle))
{
$row = fgetcsv($this->handle, 0, ";");
$sSql = "REPLACE INTO oxarticles SET "
. " OXID = '" . $row[0] . "', "
. " oxartnum = '" . $row[0] . "', "
. " oxtitle = '" . $row[1] . "', "
. " oxprice = '" . str_replace(",", ".", $row[4]) . "', "
. " oxstock = '" . str_replace(",", ".", $row[5]) . "', "
. " importstatus = 1";
// $sSql = "UPDATE oxarticles SET oxtitle ='" . $row[1] . "', oxprice='" . $row[4] . "', oxstock='" . $row[5] . "' WHERE oxartnum ='".$row[0]."'";
$this->db->execute($sSql);
}

From the mysql documentation:
REPLACE works exactly like INSERT, except that if an old row in the
table has the same value as a new row for a PRIMARY KEY or a UNIQUE
index, the old row is deleted before the new row is inserted. See
Section 13.2.5, “INSERT Syntax”.
In other words, the row is being deleted and then inserted, hence your old values aren't staying intact. Perhaps you could select the original row first, and feed those values back in where appropriate.

You query will replace old data into new data if you do not provide data for a field it will set to null . If you do not want to loose your data just want to update field use on duplicate key update.
If did't found any match it will insert new row
If found it will replace data if provide
INSERT INTO table (id,a,b,c,d,e,f,g) VALUES (1,2,3,4,5,6,7,8) ON
DUPLICATE KEY
UPDATE a=a, b=b, c=c, d=d, e=e, f=f, g=g;

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.

check column if empty then do update

guys i have a large database so when i want to fully update my some column i have timeout error (i attempt some method to increase timeout and fail) But my question is i want to bypass this problem i want to update my empty column in some table. so i want to use this query code but i have blank page with no error can some one tell me what problem with that or if possible pleas tell me a good method.
$sql = 'SELECT topic_first_poster_avatar FROM ' . TOPICS_TABLE . ' WHERE topic_poster = ' . (int) $row['user_id'] . 'IF topic_first_poster_avatar = ""
SET topic_first_poster_avatar = \'' . $db->sql_escape($avatar_info);
$db->sql_query($sql);
As plalx said in the comments, you don't need a SELECT or IF, you can specify WHERE for the update statement and have multiple contraints with AND/OR.
$sql = "UPDATE ". TOPICS_TABLE ."
SET topic_first_poster_avatar = '" . $db->sql_escape($avatar_info) ."'
WHERE topic_poster = " . (int) $row['user_id'] . " AND topic_first_poster_avatar = ''";

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.

Using mysql_fetch_array in INSERT statement

I am trying to use the result of one mysql query in another mysql query, but I'm obviously doing something wrong. This is what I have:
<?php
$result = mysql_query('SELECT panel_product_no
FROM panelProduct
WHERE length_mm = "' . ($_POST["p_length_mm"]) . '"
AND width_mm = "' . ($_POST["p_width_mm"]) . '"
AND veneer_type = "' . ($_POST["p_veneer"]) . '"
AND lipping = "' . ($_POST["p_lipping"]) . '"');
$panel = mysql_fetch_array($result);
?>
And then I want to use that in this bit:
<?php
if(!empty($_POST[p_length_mm]) && !empty($_POST[p_width_mm]) && !empty($_POST[p_aperture]))
{
$sql3="INSERT INTO estimateDescribesPanelProduct (estimate_no, panel_product_no, quantity)
VALUES ('$_GET[estimate_no]','$panel','$_POST[p_quantity]')";
if (!mysql_query($sql3,$con))
{
die('Error: ' . mysql_error());
}
}
?>
The query is basically working in that it is inserting the posted estimate_no and quantity into the DB, but not the correct panel_product_no (it just inserts '0'). How can I get it to insert the $result value?
P.S. I know that I should not be using mysql functions and I will not be in future, however I am so nearly finished with this project that at this point I am not in a position change.
Your are basicly copying content from one table to another.
Wy not use the MySQL INSERT .. SELECT syntax?
as #Dmitry Makovetskiyd wrote, mysql_fetch_array() returns a resource, not manipulatable results.
For example:
$result = mysql_query('SELECT panel_product_no
FROM panelProduct
WHERE length_mm = "' . ($_POST["p_length_mm"]) . '"
AND width_mm = "' . ($_POST["p_width_mm"]) . '"
AND veneer_type = "' . ($_POST["p_veneer"]) . '"
AND lipping = "' . ($_POST["p_lipping"]) . '"');
$resource = mysql_fetch_object($result);
You need to add in:
$panel = $resource->'panel_product_no';
You can then continue with your second query.
Note the change from mysql_fetch_array() to mysql_fetch_object() - as your query suggests you are only retrieving a singular value from the table (assuming there is only a singular panel with the specified length, width, veneer type and lipping), the object method will work fine.

UPDATE two rows - MySQL

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?

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