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  • PHP error problem.

    - by TaG
    I get the following error on line 8: Undefined index: real_name which is $privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']); I was wondering how can I fix this problem? Here is the PHP. if (isset($_POST['submitted'])) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.* FROM users WHERE user_id=3"); $privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']); if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO users (user_id, privacy_policy) VALUES ('$user_id', '$privacy_policy')"); } if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE users SET privacy_policy = '$privacy_policy' WHERE user_id = '$user_id'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; } if (!$dbc) { print mysqli_error($mysqli); return; } } Here is the HTML. <form method="post" action="index.php"> <fieldset> <ul> <li><input type="checkbox" name="privacy_policy" id="privacy_policy" value="yes" <?php if (isset($_POST['privacy_policy'])) { echo 'checked="checked"'; } else if($privacy_policy == "yes") { echo 'checked="checked"'; } ?> /></li> <li><input type="submit" name="submit" value="Save Changes" class="save-button" /> <input type="hidden" name="submitted" value="true" /> <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li> </ul> </fieldset> </form>

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  • Managing Foreign Keys

    - by jwzk
    So I have a database with a few tables. The first table contains the user ID, first name and last name. The second table contains the user ID, interest ID, and interest rating. There is another table that has all of the interest ID's. For every interest ID (even when new ones are added), I need to make sure that each user has an entry for that interest ID (even if its blank, or has defaults). Will foreign keys help with this scenario? or will I need to use PHP to update each and every record when I add a new key?

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  • How to add SQL elements to an array in PHP

    - by DanLeaningphp
    So this question is probably pretty basic. I am wanting to create an array from selected elements from a SQL table. I am currently using: $rcount = mysql_num_rows($result); for ($j = 0; $j <= $rcount; $j++) { $row = mysql_fetch_row($result); $patients = array($row[0] => $row[2]); } I would like this to return an array like this: $patients = (bob=>1, sam=>2, john=>3, etc...) Unfortunately, in its current form, this code is either copying nothing to the array or only copying the last element.

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  • getting notice like undefined index

    - by user2533308
    $result = mysql_query("SELECT * FROM customers WHERE loginid='$_POST[login]' AND accpassword='$_POST[password]'"); if(mysql_num_rows($result) == 1) { while($recarr = mysql_fetch_array($result)) { $_SESSION[customerid] = $recarr[customerid]; $_SESSION[ifsccode] = $recarr[ifsccode]; $_SESSION[customername] = $recarr[firstname]. " ". $recarr[lastname]; $_SESSION[loginid] = $recarr[loginid]; $_SESSION[accstatus] = $recarr[accstatus]; $_SESSION[accopendate] = $recarr[accopendate]; $_SESSION[lastlogin] = $recarr[lastlogin]; } $_SESSION["loginid"] =$_POST["login"]; header("Location: accountalerts.php"); } else { $logininfo = "Invalid Username or password entered"; } Notice: Undefined index:login and Notice: Undefined index:password try to help me out getting error message in second line

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  • cakephp pagination overide paginate() method

    - by islam
    iam using cakephp pagination and to paginate using custome query i override the paginate() method in my controller the problem is that every other action that use paginate() method not work any more how i can override this method without conflict with other that use this method from the same model

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  • Why does this query only select a single row?

    - by Joe
    SELECT * FROM tbl_houses WHERE (SELECT HousesList FROM tbl_lists WHERE tbl_lists.ID = '123') LIKE CONCAT('% ', tbl_houses.ID, '#') It only selects the row from tbl_houses of the last occuring tbl_houses.ID inside tbl_lists.HousesList I need it to select all the rows where any ID from tbl_houses exists within tbl_lists.HousesList

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  • Converting a certain SQL query into relational algebra

    - by Fumler
    Just doing an assignment for my database course and I just want to double check that I've correctly wrapped my head around relational algebra. The SQL query: SELECT dato, SUM(pris*antall) AS total FROM produkt, ordre WHERE ordre.varenr = produkt.varenr GROUP BY dato HAVING total >= 10000 The relational algebra: stotal >= 10000( ?R(dato, total)( sordre.varenr = produkt.varenr( datoISUM(pris*antall(produkt x ordre)))) Is this the correct way of doing it?

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  • how to specify a BIGINT in a rails scaffold?

    - by webdestroya
    I am trying to create a model in ruby that uses a BIGINT datatype (as opposed to the INT done by :integer). I have search all over Google, but all I seem to find is "run an SQL statement to alter the table to a BIGINT" - This seems a bit hack-ish to me, so I wanted to know if there was a way to specify a bigint in the ruby system like :big_int or something Any ideas?

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  • PHP in Wordpress Posts - Is this okay?

    - by Thomas
    I've been working with some long lists of information and I've come up with a good way to post it in various formats on my wordpress blog posts. I installed the exec-PHP plugin, which allows you to run php in posts. I then created a new table (NEWTABLE) in my wordpress database and filled that table with names, scores, and other stuff. I was then able to use some pretty simple code to display the information in a wordpress post. Below is an example, but you could really do whatever you wanted. My question is - is there a problem with doing this? with security? or memory? I could just type out all the information in each post, but this is really much nicer. Any thoughts are appreciated. <?php $theResult = mysql_query("SELECT * FROM NEWTABLE WHERE Score < 100 ORDER BY LastName"); while($row = mysql_fetch_array($theResult)) { echo $row['FirstName']; echo " " . $row['LastName']; echo " " . $row['Score']; echo "<br />"; } ?>

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  • Php mysqli and stored-procedure

    - by Mneva skoko
    I have a stored procedure: Create procedure news(in dt datetime,in title varchar(10),in desc varchar(200)) Begin Insert into news values (dt,title,desc); End Now my php: $db = new mysqli("","","",""); $dt = $_POST['date']; $ttl = $_POST['title']; $desc = $_POST['descrip']; $sql = $db-query("CALL news('$dt','$ttl','$desc')"); if($sql) { echo "data sent"; }else{ echo "data not sent"; } I'm new with php please help thank you

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  • In SQL, in what situation do we want to Index a field in a table, or 2 fields in a table at the same

    - by Jian Lin
    In SQL, it is obvious that whenever we want to do a search on millions of record, say CustomerID in a Transactios table, then we want to add an index for CustomerID. Is another situation we want to add an index to a field when we need to do inner join or outer join using that field as a criteria? Such as Inner join on t1.custumerID = t2.customerID. Then if we don't have an index on customerID on both tables, we are looking at O(n^2) because we need to loop through the 2 tables sequentially. If we have index on customerID on both tables, then it becomes O( (log n) ^ 2 ) and it is much faster. Any other situation where we want to add an index to a field in a table? What about adding index for 2 fields combined in a table. That is, one index, for 2 fields together?

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  • Can you automatically create a mysqldump file that doesn't enforce foreign key constraints?

    - by Tai Squared
    When I run a mysqldump command on my database and then try to import it, it fails as it attempts to create the tables alphabetically, even though they may have a foreign key that references a table later in the file. There doesn't appear to be anything in the documentation and I've found answers like this that say to update the file after it's created to include: set FOREIGN_KEY_CHECKS = 0; ...original mysqldump file contents... set FOREIGN_KEY_CHECKS = 1; Is there no way to automatically set those lines or export the tables in the necessary order (without having to manually specify all table names as that can be tedious and error prone)? I could wrap those lines in a script, but was wondering if there is an easy way to ensure I can dump a file and then import it without manually updating it.

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  • Use Zip to Pre-Populate City/State Form with jQuery AJAX

    - by Paul
    I'm running into a problem that I can solve fine by just submitting a form and calling a db to retrieve/echo the information, but AJAX seems to be a bit different for doing this (and is what I need). Earlier in a form process I ask for the zip code like so: <input type="text" maxlength="5" size="5" id="zip" /> Then I have a button to continue, but this button just runs a javascript function that shows the rest of the form. When the rest of the form shows, I want to pre-populate the City input with their city, and pre-populate the State dropdown with their state. I figured I would have to find a way to set city/state to variables, and echo the variables into the form. But I can't figure out how to get/set those variables with AJAX as opposed to a form submit. Here's how I did it without ajax: $zip = mysql_real_escape_string($_POST['zip']); $q = " SELECT city FROM citystatezip WHERE zip = $zip"; $r = mysql_query($q); $row = mysql_fetch_assoc($r); $city = $row['city']; Can anybody help me out with using AJAX to set these variables? Thanks!

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  • How to use where condition for the for a selected column using subquery?

    - by Holicreature
    I have two columns as company and product. I use the following query to get the products matching particular string... select id,(select name from company where product.cid=company.id) as company,name,selling_price,mrp from product where name like '$qry_string%' But when i need to list products of specific company how can i do? i tried the following but in vein select id,(select name from company where product.cid=company.id) as company,name,selling_price,mrp from product where company like '$qry_string%' Help me

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  • Getting dynamic childs for a parent in SQL

    - by Islam
    I have a table called Categories which contains category_Id and parent_category_Id, so each category can has a child and the child can has a child and so on (it is dynamic). So if i have category A and category A has child B and child B has child C and child C has child D. I want to get all the child tree of A using SQL so when I give this query the id of A its result will be the ids of A's child which is B,C & D.....any ideas. Thanks in regards,

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  • checking if records exists in DB, in single step or 2 steps?

    - by Sinan
    Suppose you want to get a record from database which returns a large data and requires multiple joins. So my question would be is it better to use a single query to check if data exists and get the result if it exists. Or do a more simple query to check if data exists then id record exists, query once again to get the result knowing that it exists. Example: 3 tables a, b and ab(junction table) select * from from a, b, ab where condition and condition and condition and condition etc... or select id from a, b ab where condition then if exists do the query above. So I don't know if there is any reason to do the second. Any ideas how this affects DB performance or does it matter at all?

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  • Unknown column even thoug it exits

    - by george
    I have SELECT servisler.geo_location, servisler.ADRES_MERKEZ, servisler.ADRES_ILCE, servisler.ADRES_IL, servisler.FIRMA_UNVANI, servisler.ADRES_ISTEL, servisler.YETKILI_ADISOYADI, urun_gruplari.GRUP_ADI FROM servisler INNER JOIN urun_gruplari ON kullanici_cihaz.URUN_GRUP_NO= urun_gruplari.RECNO INNER JOIN kullanici ON kullanici.SERVIS_RECNO = servisler.RECNO INNER JOIN kullanici_cihaz ON kullanici.RECNO = kullanici_cihaz.KUL_RECNO AND kullanici_cihaz.URUN_GRUP_NO = urun_gruplari.RECNO where kullanici.kullanici = 'MAR.EDI.003' but it says [Err] 1054 - Unknown column 'kullanici_cihaz.URUN_GRUP_NO' in 'on clause' enen though the column exits. What is its problem? schema Server version: 5.1.33-community-log

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  • How to select DISTINCT rows without having the ORDER BY field selected

    - by JannieT
    So I have two tables students (PK sID) and mentors (PK pID). This query SELECT s.pID FROM students s JOIN mentors m ON s.pID = m.pID WHERE m.tags LIKE '%a%' ORDER BY s.sID DESC; delivers this result pID ------------- 9 9 3 9 3 9 9 9 10 9 3 10 etc... I am trying to get a list of distinct mentor ID's with this ordering so I am looking for the SQL to produce pID ------------- 9 3 10 If I simply insert a DISTINCT in the SELECT clause I get an unexpected result of 10, 9, 3 (wrong order). Any help much appreciated.

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  • Adding to a multidimensional array in PHP

    - by b. e. hollenbeck
    I have an array being returned from the database that looks like so: $data = array(201 => array('description' => blah, 'hours' => 0), 222 => array('description' => feh, 'hours' => 0); In the next bit of code, I'm using a foreach and checking the for the key in another table. If the next query returns data, I want to update the 'hours' value in that key's array with a new hours value: foreach ($data as $row => $value){ $query = $db->query($sql); if ($result){ $value['hours'] = $result['hours']; } I've tried just about every combination of declarations for the foreach loop, but I keep getting the error that it's a non-object. Surely this is easier than my brain is perceiving it.

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