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  • How to run a set of SQL queries from a file, in PHP?

    - by Harish Kurup
    I have some set of SQL queries which is in a file(i.e query.sql), and i want to run those queries in files using PHP, the code that i have wrote is not working, //database config's... $file_name="query.sql"; $query==file($file_name); $array_length=count($query); for($i=0;$i<$array_length;$i++) { $data .= $query[$i]; } echo $data; mysql_query($data); it echos the SQL Query from the file but throws an error at mysql_query() function...

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  • getting sql records

    - by droidus
    when i run this code, it returns the topic fine... $query = mysql_query("SELECT topic FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; } but when I change it to this, it doesn't run at all. why is this happening? $query = mysql_query("SELECT topic, lock FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; $lockedThread = $row['lock']; echo "here: " . $lockedThread; }

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  • Find the % character in a LIKE query

    - by Jensen
    Hi, I've an SQL database and I would like to do a query who show all the datas containing the sign "%". Normally, to find a character (for example: "z") in a database I use a query like this : mysql_query("SELECT * FROM mytable WHERE tag LIKE '%z%'"); But here, I want to found the % character, but in SQL it's a joker so when I write: mysql_query("SELECT * FROM mytable WHERE tag LIKE '%%%'"); It show me all my datas. So how to found the % character in my SQL datas ? Thanks

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  • How does py2exe actually -and simply explained- work? :)

    - by sandra
    Hi folks, I have a c++ app that calls another python one (bundled into an exe with py2exe) So I have 2 apps. So I was wondering: What if my c++ did what py2exe does? i.e. embed the python app in the c++ one. This way I won't depend on py2exe and its configurations nighmares (yes, it has some) Hence my questions: how does py2exe work (so I can do its job with my c++ app) What about just embedding the whole python app with the c++? I read the python doc about embedding, did an example (a very simple one that does PyRun_SimpleString) but what about a whole python app with tons of modules? (zipimport maybe?) I'd love to hear how you'd do that. Thanks a lot! :)

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  • User has many computers, computers have many attributes in different tables, best way to JOIN?

    - by krismeld
    I have a table for users: USERS: ID | NAME | ---------------- 1 | JOHN | 2 | STEVE | a table for computers: COMPUTERS: ID | USER_ID | ------------------ 13 | 1 | 14 | 1 | a table for processors: PROCESSORS: ID | NAME | --------------------------- 27 | PROCESSOR TYPE 1 | 28 | PROCESSOR TYPE 2 | and a table for harddrives: HARDDRIVES: ID | NAME | ---------------------------| 35 | HARDDRIVE TYPE 25 | 36 | HARDDRIVE TYPE 90 | Each computer can have many attributes from the different attributes tables (processors, harddrives etc), so I have intersection tables like this, to link the attributes to the computers: COMPUTER_PROCESSORS: C_ID | P_ID | --------------| 13 | 27 | 13 | 28 | 14 | 27 | COMPUTER_HARDDRIVES: C_ID | H_ID | --------------| 13 | 35 | So user JOHN, with id 1 owns computer 13 and 14. Computer 13 has processor 27 and 28, and computer 13 has harddrive 35. Computer 14 has processor 27 and no harddrive. Given a user's id, I would like to retrieve a list of that user's computers with each computers attributes. I have figured out a query that gives me a somewhat of a result: SELECT computers.id, processors.id AS p_id, processors.name AS p_name, harddrives.id AS h_id, harddrives.name AS h_name, FROM computers JOIN computer_processors ON (computer_processors.c_id = computers.id) JOIN processors ON (processors.id = computer_processors.p_id) JOIN computer_harddrives ON (computer_harddrives.c_id = computers.id) JOIN harddrives ON (harddrives.id = computer_harddrives.h_id) WHERE computers.user_id = 1 Result: ID | P_ID | P_NAME | H_ID | H_NAME | ----------------------------------------------------------- 13 | 27 | PROCESSOR TYPE 1 | 35 | HARDDRIVE TYPE 25 | 13 | 28 | PROCESSOR TYPE 2 | 35 | HARDDRIVE TYPE 25 | But this has several problems... Computer 14 doesnt show up, because it has no harddrive. Can I somehow make an OUTER JOIN to make sure that all computers show up, even if there a some attributes they don't have? Computer 13 shows up twice, with the same harddrive listet for both. When more attributes are added to a computer (like 3 blocks of ram), the number of rows returned for that computer gets pretty big, and it makes it had to sort the result out in application code. Can I somehow make a query, that groups the two returned rows together? Or a query that returns NULL in the h_name column in the second row, so that all values returned are unique? EDIT: What I would like to return is something like this: ID | P_ID | P_NAME | H_ID | H_NAME | ----------------------------------------------------------- 13 | 27 | PROCESSOR TYPE 1 | 35 | HARDDRIVE TYPE 25 | 13 | 28 | PROCESSOR TYPE 2 | 35 | NULL | 14 | 27 | PROCESSOR TYPE 1 | NULL | NULL | Or whatever result that make it easy to turn it into an array like this [13] => [P_NAME] => [0] => PROCESSOR TYPE 1 [1] => PROCESSOR TYPE 2 [H_NAME] => [0] => HARDDRIVE TYPE 25 [14] => [P_NAME] => [0] => PROCESSOR TYPE 1

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  • Unknown Column?

    - by Kenny
    ok im trying to get mutual friends between these Two users, user1 and user92 This is the sql that is successful in displaying them SELECT IF(user_a = 1 OR user_a = 92, user_b, user_a) friend FROM friendship WHERE (user_a = 1 OR user_a = 92) OR (user_b = 1 OR user_b = 92) GROUP BY 1 HAVING COUNT(*) > 1 THis is how it looks friend 61 72 73 74 75 76 77 78 79 80 81 So now i want to select all users after the number 72, and i try to do it with this sql but its not working? It gives me the error, "unknown coulum name friend in where clause" SELECT IF(user_a = 1 OR user_a = 92, user_b, user_a) friend FROM friendship WHERE friend > 72 and (user_a = 1 OR user_a = 92) OR (user_b = 1 OR user_b = 92) GROUP BY 1 HAVING COUNT(*) > 1 what am i doing wrong? or what is the correct way?? thx

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  • What database field should this be?

    - by alex
    I have a table called "Event". A column has "duration"--which is...how long that event lasts. What database type should this be? Integer? (in seconds)? In the future, I will do statements such as: If now() > event.duration THEN don't display the event.

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  • Send Email Using Classic ASP with an Embeded Image

    - by Khilen
    hi i am making NewsLetter using the wysiwyg Editor.. it allows me to upload the Image Path and Image Path is stored in the Upload Directory.. Not When i retrieve that Image using it works in website.. the editor's value is stored in database example <br> hi <img src="upload/acb.gif"> <br> Hello i am sending Email and the detail of this email is received from database and this detail is sent to visitor he is gettion all text value but not able to see Image so suggest me what to do..?

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  • SQL Querying for Threaded Messages

    - by Harper
    My site has a messaging feature where one user may message another. The messages support threading - a parent message may have any number of children but only one level deep. The messages table looks like this: Messages - Id (PK, Auto-increment int) - UserId (FK, Users.Id) - FromUserId (FK, Users.Id) - ParentMessageId (FK to Messages.Id) - MessageText (varchar 200) I'd like to show messages on a page with each 'parent' message followed by a collapsed view of the children messages. Can I use the GROUP BY clause or similar construct to retrieve parent messages and children messages all in one query? Right now I am retrieving parent messages only, then looping through them and performing another query for each to get all related children messages. I'd like to get messages like this: Parent1 Child1 Child2 Child3 Parent2 Child1 Parent3 Child1 Child2

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  • How to handle customers with multiple addresses in CakePHP

    - by Ryan
    I'm putting together a system to track customer orders. Each order will have three addresses; a Main contact address, a billing address and a shipping address. I do not want to have columns in my orders table for the three addresses, I'd like to reference them from a separate table and have some way to enumerate the entry so I can determine if the addressing is main, shipping or billing. Does it make sense to create a column in the address table for AddressType and enumerate that or create another table - AddressTypes - that defines the address enumeration and link to that table? I have found other questions that touch on this topic and that is where I've taken my model. The problem I'm having is taking that into the cakePHP convention. I've been struggling to internalize the direction OneToMany relationships are formed - the way the documentation states feels backwards to me. Any help would be appreciated, Thanks!

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  • How to retrieve column total when rows are paginated?

    - by Rick
    Hey guys I have a column "price" in a table and I used a pagination script to generate the data displayed. Now the pagination is working perfectly however I am trying to have a final row in my HTML table to show the total of all the price. So I wrote a script to do just that with a foreach loop and it sort of works where it does give me the total of all the price summed up together however it is the sum of all the rows, even the ones that are on following pages. How can I retrieve just the sum of the rows displayed within the pagination? Thank you! Here is the query.. SELECT purchase_log.id, purchase_log.date_purchased, purchase_log.total_cost, purchase_log.payment_status, cart_contents.product_name, members.first_name, members.last_name, members.email FROM purchase_log LEFT JOIN cart_contents ON purchase_log.id = cart_contents.purchase_id LEFT JOIN members ON purchase_log.member_id = members.id GROUP BY id ORDER BY id DESC LIMIT 0,30";

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  • Very simple shopping cart, remove button

    - by Kynian
    Im writing sales software that will be walking through a set of pages and on certain pages there are items listed to sell and when you click buy it basically just passes a hidden variable to the next page to be set as a session variable, and then when you get to the end it call gets reported to a database. However my employer wanted me to include a shopping cart, and this shopping cart should display the item name, sku, and price of whatever you're buying, as well as a remove button so the person doing the script doesnt need to go back through the entire thing to remove one item. At the moment I have the cart set to display everything, which was fairly simple. but I cant figure out how to get the remove button to work. Here is the code for the shopping cart: $total = 0; //TEST CODE: $_SESSION['itemname-addon'] = "Test addon"; $_SESSION ['price-addon'] = 10.00; $_SESSION ['sku-addon'] = "1234h"; $_SESSION['itemname-addon1'] = "Test addon1"; $_SESSION ['price-addon1'] = 99.90; $_SESSION ['sku-addon1'] = "1111"; $_SESSION['itemname-addon2'] = "Test addon2"; $_SESSION ['price-addon2'] = 19.10; $_SESSION ['sku-addon2'] = "123"; //end test code $items = Array ( "0"=> Array ( "name" => $_SESSION['itemname-mo'], "price" => $_SESSION ['price-mo'], "sku" => $_SESSION ['sku-mo'] ), "1" => Array ( "name" => $_SESSION['itemname-addon'], "price" => $_SESSION ['price-addon'], "sku" => $_SESSION ['sku-addon'] ), "2" => Array ( "name" => $_SESSION['itemname-addon1'], "price" => $_SESSION ['price-addon1'], "sku" => $_SESSION ['sku-addon1'] ), "3" => Array ( "name" => $_SESSION['itemname-addon2'], "price" => $_SESSION ['price-addon2'], "sku" => $_SESSION ['sku-addon2'] ) ); $a_length = count($items); for($x = 0; $x<$a_length; $x++){ $total +=$items[$x]['price']; } $formattedtotal = number_format($total,2,'.',''); for($i = 0; $i < $a_length; $i++){ $name = $items[$i]['name']; $price = $items[$i]['price']; $sku = $items[$i]['sku']; displaycart($name,$price,$sku); } echo "<br /> <b>Sub Total:</b> $$formattedtotal"; function displaycart($name,$price,$sku){ if($name != null || $price != null || $sku != null){ if ($name == "no sale" || $price == "no sale" || $sku == "no sale"){ echo ""; } else{ $formattedprice = number_format($price,2,'.',''); echo "$name: $$formattedprice ($sku)"; echo "<form action=\"\" method=\"post\">"; echo "<button type=\"submit\" />Remove</button><br />"; echo "</form>"; } } } So at this point Im not sure where to go from here for the remove button. Any suggestions would be appreciated.

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  • Need help with many-to-many relationships....

    - by yuudachi
    I have a student and faculty table. The primary key for student is studendID (SID) and faculty's primary key is facultyID, naturally. Student has an advisor column and a requested advisor column, which are foreign key to faculty. That's simple enough, right? However, now I have to throw in dates. I want to be able to view who their advisor was for a certain quarter (such as 2009 Winter) and who they had requested. The result will be a table like this: Year | Term | SID | Current | Requested ------------------------------------------------ 2009 | Winter | 860123456 | 1 | NULL 2009 | Winter | 860445566 | 3 | NULL 2009 | Winter | 860369147 | 5 | 1 And then if I feel like it, I could also go ahead and view a different year and a different term. I am not sure how these new table(s) will look like. Will there be a year table with three columns that are Fall, Spring and Winter? And what will the Fall, Spring, Winter table have? I am new to the art of tables, so this is baffling me... Also, I feel I should clarify how the site works so far now. Admin can approve student requests, and what happens is that the student's current advisor gets overwritten with their request. However, I think I should not do that anymore, right?

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  • How can I get columns name from select query in php?

    - by Farshad Mehrvarzan
    I want to execute a SELECT query but I don't how many columns to select. Like: select name, family from persons; How can I know which columns to select? "I am currently designing a site for the execute query by users. So when the user executes this query, I won't know which columns selected. But when I want to show the results and draw a table for the user I should know which columns selected."

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  • question with its query

    - by user329820
    Hi this is my homework and the question is this: List the average balance of customers by city and short zip code (the first five digits of thezip code). Only include customers residing in Washington State (‘WA’). also the Customer table has 5 columns(Name,Family,CustZip,CustCity,CustAVGBal) I wrote the query like below is this correct? SELECT CustCity,LEFT(CustZip,5) AS NewCustZip,CustAVGBal FROM Customer WHERE CustCity = 'WA' THANKS!!

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  • Understanding Nested If.. Else statements

    - by user1174762
    For some reason my PHP login script keeps returning "invalid email/password combination", yet i know I am entering the correct email and password. Does anyone see what I might be doing wrong? <?php $email= $_POST['email']; $password= $_POST['password']; if (!empty($email) && !empty($password)) { $connect= mysqli_connect("localhost", "root", "", "si") or die('error connecting with the database'); $query= "SELECT user_id, email, password FROM users WHERE email='$email' AND password='$password'"; $result= mysqli_query($connect, $query) or die('error with query'); if (mysqli_num_rows($result) == 1) { $row= mysqli_fetch_array($result); setcookie('user_id', $row['user_id']); echo "you are now logged in"; } else { echo "invalid username/password combination"; } } else { echo" you must fill out both username and password"; } ?>

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  • 500 internal server error at form connection

    - by klox
    hi..all..i've a problem i can't connect to database what's wrong with my code?this is my code: $("#mod").change(function() { var barcode; barCode=$("#mod").val(); var data=barCode.split(" "); $("#mod").val(data[0]); $("#seri").val(data[1]); var str=data[0]; var matches=str.match(/(EE|[EJU]).*(D)/i); $.ajax({ type:"post", url:"process1.php", data:"value="+matches+"action=tunermatches", cache:false, async:false, success: function(res){ $('#rslt').replaceWith( "<div id='value'><h6>Tuner range is" + res + " .</h6></div>" ); } }); }); and this is my process file: switch(postVar('action')) { case 'tunermatches' : tunermatches(postVar('tuner')); break; function tunermatches($tuner)){ $Tuner=mysql_real_escape_string($tuner); $sql= "SELECT remark FROM settingdata WHERE itemname="Tuner_range" AND itemdata="$Tunermatches"; $res=mysql_query($sql); $dat=mysql_fetch_array($res,MYSQL_NUM); if($dat[0]>0) { echo $dat[0]; } mysql_close($dbc); }

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  • need help on my query.

    - by Dharmendra
    i have one table : nobel(yr, subject, winner) and i have this query : In which years was the Physics prize awarded but no Chemistry prize. this is what i tried : select distinct yr from nobel where subject='physics' and subject!='chemistry' but is not working where i am going wrong. see, i am not here to make my homework from someone. i am here to learn something. so, please give me suggetion.

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  • SQL: Find the max record per group

    - by user319088
    I have one table, which has three fields and data. Name , Top , Total cat , 1 , 10 dog , 2 , 7 cat , 3 , 20 horse , 4 , 4 cat , 5 , 10 dog , 6 , 9 I want to select the record which has highest value of Total for each Name, so my result should be like this: Name , Top , Total cat , 3 , 20 horse , 4 , 4 Dog , 6 , 9 I tried group by name order by total, but it give top most record of group by result. Can anyone guide me, please?

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  • mysql_fetch_array() not displaying all results

    - by user1666995
    I have a database with a calendar table (each row represents one day) with 4 years of rows (2012, 2013, 2014, 2015). I use the column name calyear for the year. I use the following code to find values for distinct years then display it: $year = mysql_query("SELECT DISTINCT calyear FROM calendar"); while($yeararray = mysql_fetch_array($year)) { echo($yeararray['calyear']."<br />"); } The problem is it only displays the years 2013, 2014, 2015 even though when I use echo(mysql_num_rows($year); it displays the value 4 which I take to mean all 4 years are there. I'm not quite sure where I'm going wrong with this.

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