Please help me to solve this. As I am just in a learning phase of PHP/Mysql.
I have a php feedback form as a rating system from 1-5. You can find my form here http://innovatrix.co.in/feedback_priyajit/feedback%20form1.html
Every time a user provide feedback it saves form values into a mysql database. Below is my database structure.
Now I want to calculate average data of every row (like waiting) and show it on a php file as a graph and also separate graph for every option but on a same page.
I know I can use query SELECT AVG(waiting) FROM feedback to get an average of "waiting"
But how can I do this for every options from a same file and also show it as a graph. Database will be updated frequently, thus it should reflect the graph also.
Please help me with a concept for achieving this.
Below is my php file which I am using to store form values into database.
<title>process</title>
<?php
$host="localhost";
$user_name="pramir_feedback";
$pwd="feedback";
$database_name="pramir_feedback";
$db=mysql_connect($host, $user_name, $pwd);
if (mysql_error() > "") print mysql_error() . "<br>";
mysql_select_db($database_name, $db);
if (mysql_error() > "") print mysql_error() . "<br>";
$waiting = $_POST['radio1'];
$consultation = $_POST['radio2'];
$preoperative = $_POST['radio3'];
$specialists = $_POST['radio4'];
$assistants = $_POST['radio5'];
$painful = $_POST['radio6'];
$operatingroom = $_POST['radio7'];
$thought = $_POST['radio8'];
$recommend = $_POST['radio9'];
$suggestions = $_POST['suggestions'];
$query = "insert into feedback (waiting, consultation, preoperative, specialists, assistants, painful, operatingroom, thought, recommend, suggestions) values ('" . $waiting . "', '" . $consultation . "', '" . $preoperative . "', '" . $specialists . "', '" . $assistants . "', '" . $painful . "', '" . $operatingroom . "', '" . $thought . "', '" . $recommend . "', '" . $suggestions . "')";
if (mysql_error() > "") print mysql_error() . "<br>";
$qresult = mysql_query($query);
echo "<h1>Thank you for submitting your details!</h1>";
?>
If you want all the averages in one query, you can just delimit them with commas.
SELECT AVG(waiting), AVG(consultation), AVG(preoperative), AVG(specialists), ...... FROM feedback
If you want to know how to put them in a graph, take a look at one of the many jQuery graph or plot makers, like: http://www.jqplot.com/tests/bar-charts.php
Related
found this floating script to dump given feed into sql. however unable to save feed description and images separately in table. below is the part of code which is currently saving some contents but with empty columns.
any help or advise would be appreciated.
$db = mysql_connect($db_hostname,$db_username,$db_password);
if (!$db)
{
die("Could not connect: " . mysql_error());
}
.....snip.............
EDIT#1:
studying simplepie script and its caching. will update should get it resolved tonight.
EDIT#2:
SQL query which was modified to include description and screengrab for the table contents.
CREATE TABLE rss (
............snip........................
enter image description here
EDIT #3:
thx robert for correcting me yes it was a quick copy paste from the original script and havent put correct query init. i had updated sql query and get things fixed however now the moded script needs more juice by retrieving a image link along with the post. i am using xpath to get each post image link however not able to incorporate the respective image with each row dump. both scripts are running perfectly at their individual grounds but i need more help in merging them together.
require_once("./config.php");
.......snip................
EDIT#4:
well it didnt work i mange to updated existing table with another script
i upvoted your answer but havent accepted it as correct answer due to this reason. might i failed to explain properly but my script is working like a charm and just missing some CSS.
thanks though for your help
NOTE: i snip original scripts from the post/
From your screenshot you're missing info inside "item_enclosure" and "item_status" columns, the rest is filled.
Problem is that your query doesn't try to insert those missing entries
$item_insert_sql = "INSERT INTO rssingest(item_id, feed_url, item_title, item_date, item_description, item_url, fetch_date) VALUES ('" . $item_id . "', '" . $feed_url . "', '" . $item_title . "', '" . $item_date . "', '" . $item_description . "', '" . $item_url . "', '" . $fetch_date . "')";
^ Will never insert "item_enclosure" and "item_status" since the 2 columns are not part of the insert query...
You'd need to extract the 2 missing items and modify the insert query to:
$item_insert_sql = "INSERT INTO rssingest(item_id, feed_url, item_title, item_date, item_description, item_url, fetch_date, item_enclosure, item_status) VALUES ('" . $item_id . "', '" . $feed_url . "', '" . $item_title . "', '" . $item_date . "', '" . $item_description . "', '" . $item_url . "', '" . $fetch_date . "', '" . $item_enclosure . "', '" . $item_status . "')";
You'll need to first define:
$item_enclosure
$item_status
Btw the query you pasted in your updated reply is wrong, it should start with "INSERT INTO" not "CREATE TABLE", you're not echoing the right query.
UPDATE:
In your original code there's this line
$has_image = preg_match('/<img.+src=[\'"](?P<src>.+?)[\'"].*>/i', $RSSitem, $image);
So right now if your code finds an image it should be putting it in an array called $image
Try print_r of $image to see in which array item it's stored, I think it will be $image[0][0] or $image[0]
So do the following to define the image URL (I'm guessing it's $image[0][0], but double check just in case):
$img_url = (isset($image)) ? $image[0][0] : "";
Then you'll need to update the insert query (i don't know what column name is used for the image so it's just an example):
$item_insert_sql = "INSERT INTO rssingest(item_id, feed_url, item_title, item_date, item_description, item_url, fetch_date, item_enclosure, item_status, image_column) VALUES ('" . $item_id . "', '" . $feed_url . "', '" . $item_title . "', '" . $item_date . "', '" . $item_description . "', '" . $item_url . "', '" . $fetch_date . "', '" . $item_enclosure . "', '" . $item_status . "', '" . $img_url . "')";
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.
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?
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.
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?