Search Results

Search found 63904 results on 2557 pages for 'mysql error 1025'.

Page 576/2557 | < Previous Page | 572 573 574 575 576 577 578 579 580 581 582 583  | Next Page >

  • Simple PHP query question: LIKE

    - by pg
    When I replace $ordering = "apples, bananas, cranberries, grapes"; with $ordering = "apples, bananas, grapes"; I no longer want cranberries to be returned by my query, which I've written out like this: $query = "SELECT * from dbname where FruitName LIKE '$ordering'"; Of Course this doesn't work, because I used LIKE wrong. I've read through various manuals that describe how to use LIKE and it doesn't quite make sense to me. If I change the end of the db to "LIKE "apples"" that works for limiting it to just apples. Do I have to explode the ordering on the ", " or is there a way to do this in the query?

    Read the article

  • error in python d not defined.

    - by dtechie
    Hi I am learning python and have this error . I can figure out where\what the error is in the code. File "<string>", line 1, in <module>. Name = "" Desc = "" Gender = "" Race = "" # Prompt user for user-defined information Name = input('What is your Name? ') Desc = input('Describe yourself: ') When i run the program it outputs What is your Name? (i input d ) this gives the error Traceback (most recent call last): File "/python/chargen.py", line 19, in <module> Name = input('What is your Name? ') File "<string>", line 1, in <module> NameError: name 'd' is not defined This is an example code from Python 3 for Absolute Beginners Thank you for your help :)

    Read the article

  • How can I find all records for a model without doing a long list of "OR" conditions?

    - by gomezuk
    I'm having trouble composing a CakePHP find() which returns the records I'm looking for. My associations go like this: User -(has many)- Friends , User -(has many)- Posts I'm trying to display a list of all a user's friends recent posts, in other words, list every post that was created by a friend of the current user logged in. The only way I can think of doing this is by putting all the user's friends' user_ids in a big array, and then looping through each one, so that the find() call would look something like: $posts = $this->Post->find('all',array( 'conditions' => array( 'Post.user_id' => array( 'OR' => array( $user_id_array[0],$user_id_array[1],$user_id_array[2] # .. etc ) ) ) )); I get the impression this isn't the best way of doing things as if that user is popular that's a lot of OR conditions. Can anyone suggest a better alternative?

    Read the article

  • User's Post count from specific category [Wordpress]

    - by morningglory
    Hello, I want to show user's post count from specific category. Currently, I can only be able to query all posts. My code is like this <?php $userpost_count = $wpdb->get_var("SELECT COUNT(*) FROM $wpdb->posts WHERE post_status = 'publish' AND post_type ='post' AND post_author = '".$curauth->ID."'");?> <?php echo "<span>Total post: </b></span>".$userpost_count.""?> I know that, I need to join two table which is post table and term_relationships, but i don't know how to get it. Please kindly help me with that. Thank you.

    Read the article

  • How do you display a PHPmyadmin table onto a web browser/web page?

    - by user1390754
    Just a simple question, i've been searching around and for some reason i cannot find an answer to this. after creating a database/table in Phpmyadmin using xampp, what command do i need to put into my webpage (PHP) to show the table I made? I know the first step involves connecting to the database and i think i've done that properly. This is the code i found from somewhere about connecting (w3 tutorials I believe) $con = mysql_connect("localhost","xxxxxx,""); if (!$con) { die('Could not connect: ' . mysql_error()); }

    Read the article

  • When SET SCAN ON used after END throws error

    - by Karthik
    Hi, Im trying to use SET SCAN ON after as follows.. SET SCAN OFF; DECLARE -- declared a variable BEGIN --update statement END; SET SCAN ON; The use of SET SCAN ON; is causing the error when i try to run the script. The error captured ORA-06550: line 16, column 1: PLS-00103: Encountered the symbol "SET" 06550. 00000 - "line %s, column %s:\n%s" *Cause: Usually a PL/SQL compilation error. *Action:

    Read the article

  • How to skip an empty LIKE operator in a multiple LIKE query?

    - by alex
    I notice my query doesn't behave correctly if one of the like variables is empty: SELECT name FROM employee WHERE name LIKE '%a%' AND color LIKE '%A%' AND city LIKE '%b%' AND country LIKE '%B%' AND sport LIKE '%c%' AND hobby LIKE '%C%' Now when a and A are not empty it works but when a, A and c are not empty the c part is not excuted so it seems? How can I fix this?

    Read the article

  • login/logout problem in PHP

    - by user356170
    i have a question. I have a website in which i am giving security like login id and password( as usual). No what i want is that, 1) I don't want to allow a single user to login in different machine at the same time. 2) For this i am using a column in database which is keeping the current status of user(i.e. loging/logout). I am allowing user to login only when has session has not closed and status is login. 3) So my problem is that when i am logging out manually. it is closing the session as well as updating the database with status "logout". 4) but when i am closing the window from Cross buttonat top right corner. it is closing the ssion but table data is still "login". so later on i can't be able to login into the same user. 5) So how could i solve this problem. Please help me!

    Read the article

  • PHP: How to get the days of the week?

    - by fwaokda
    I'm wanting to store items in my database with a DATE value for a certain day. I don't know how to get the current Monday, or Tuesday, etc. from the current week. Here is my current database setup. menuentry id int(10) PK menu_item_id int(10) FK day_of_week date message varchar(255) So I have a class setup that holds all the info then I was going to do something like this... foreach ( $menuEntryArray as $item ) { if ( $item->getDate() == [DONT KNOW WHAT TO PUT HERE] ) { // code to display menu_item information } } So I'm just unsure what to put in "[DONT KNOW WHAT TO PUT HERE]" to compare to see if the date is specified for this week's Monday, or Tuesday, etc. The foreach above runs for each day of the week - so it'll look like this... Monday Item 1 Item 2 Item 3 Tuesday Item 1 Wednesday Item 1 Item 2 ... Thanks!

    Read the article

  • help me with the following sql query

    - by rupeshmalviya
    could somebody correct my following query, i am novice to software development realm, i am to a string builder object in comma separated form to my query but it's not producing desired result qyery is as follows and string cmd = "SELECT * FROM [placed_student] WHERE passout_year=@passout AND company_id=@companyId AND course_id=@courseId AND branch_id IN('" + sb + "')"; StringBuilder sb = new StringBuilder(); foreach (ListItem li in branch.Items) { if (li.Selected == true) { sb.Append(Convert.ToInt32(li.Value) +", "); } } li is integer value of my check box list which are getting generated may be differne at different time ...please also suggest me some good source to learn sql..

    Read the article

  • Worpress WorkFlow Modfications

    - by blgnklc
    Hi All WordPress Lovers, I would like to ask a help about Zensor which is a plugin that you publish a post then a moderator approves the post to be published on the wordpress blog site. When a post is awating for approval, each awaiting post is appearing "waiting moderation". But, I dont want any link appears before moderator approval. Actually I found the joing sentence below; 1- Must be added to the end of JOIN part of any query: LEFT JOIN wp_zensor ON ID = wp_zensor.post_id 2- Must be added to the end of WHERE condition : AND wp_zensor.moderation_status = 'approved' Could you please show me; where should I add these modification on the category link presentation below: <h2>Politics</h2> <?php $recent = new WP_Query("cat=31&showposts=1"); while($recent->have_posts()) : $recent->the_post();?> <b><a href="<?php the_permalink() ?>" rel="bookmark"><?php the_title(); ?></a></b> <?php the_content_limit(140, "devami &raquo;"); ?> <div class="hppostmeta"> <p><?php the_time('j F Y, H:i'); ?> | <?php the_author_posts_link(); ?></p> </div> <?php endwhile; ?> Or any general solutions will be welcomed. Thanks. BK

    Read the article

  • Magento: Add (and retrieve) custom database field for CMS pages

    - by Toby H
    I want to assign custom parameters to CMS pages in Magento (i.e. 'about', 'customer service', etc), so they can be grouped. The end goal is to use the parameters for each page to show (or hide) them in a nav menu. Writing a quick method in the page/html block to retrieve the pages (active only) for the menu was easy, but I can't figure out how to group them so that 'testimonials', 'history', and 'contact' are associated with 'about', and 'return policy', 'shipping', and 'contact' are associated with 'customer service'. Any help to point me in the right direction would be greatly appreciated. Thanks!

    Read the article

  • Find node level in a tree

    - by Álvaro G. Vicario
    I have a tree (nested categories) stored as follows: CREATE TABLE `category` ( `category_id` int(10) unsigned NOT NULL AUTO_INCREMENT, `category_name` varchar(100) NOT NULL, `parent_id` int(10) unsigned DEFAULT NULL, PRIMARY KEY (`category_id`), UNIQUE KEY `category_name_UNIQUE` (`category_name`,`parent_id`), KEY `fk_category_category1` (`parent_id`,`category_id`), CONSTRAINT `fk_category_category1` FOREIGN KEY (`parent_id`) REFERENCES `category` (`category_id`) ON DELETE SET NULL ON UPDATE CASCADE ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_spanish_ci I need to feed my client-side language (PHP) with node information (child+parent) so it can build the tree in memory. I can tweak my PHP code but I think the operation would be way simpler if I could just retrieve the rows in such an order that all parents come before their children. I could do that if I knew the level for each node: SELECT category_id, category_name, parent_id FROM category ORDER BY level -- No `level` column so far :( Can you think of a way (view, stored routine or whatever...) to calculate the node level? I guess it's okay if it's not real-time and I need to recalculate it on node modification.

    Read the article

  • Zend Framework Multiple Table Query

    - by Jeff
    I am looking to execute this statement via Zend Framework. As I understand it, I can use Zend_Db_Select. Is it possible to use Zend_Db_Table? Three tables: classes, students, and class_students select classes.name, students.student_id, students.fname, students.lname from students, classes, class_students where class_students.student_id=students.student_id AND class_students.class_id=classes.class_id;

    Read the article

  • sql count function

    - by suryll
    Hi I have three tables and I want to know how much jobs with the wage of 1000 an employee has had The first SQL query gives me the names of all the employees that has recieved 1000 for a job SELECT distinct first_name FROM employee, job, link WHERE job.wage = 1000 AND job.job_id = link.job_id and employee.employee_id = link.employee_id; The second SQL query gives me the total number for all employees of how much jobs they have made for 1000 SELECT count(wage) FROM employee, job, link WHERE job.wage = 1000 AND job.job_id = link.job_id and employee.employee_id = link.employee_id; I was wondering if there was a way of joining both queries and also making the second for each specific employee???

    Read the article

  • Creating dynamic icons based on data entered into database from django forms

    - by John Hoke
    So I'm using Django to create a projects page with multiple forms for each project. Let's call them form 1, 2, 3, and 4. Once you create a project you can fill out any of these forms. I want to create "buttons" or links for each one of the forms that would show up on the main page. Now this is the part I need help with: Step 1. I want it so that if you click on a button for a form (say form 1) and none exists for that project yet a pop up would come up saying "This form does not exist yet, are you sure you want to create one?". And if you'd answer yes you would be directed to the form page. Step 2. But if that form does exist, I don't want any pop up to open and I want the link to take the user directly to that page. Step 3. My next problem is this. These forms are in order, so if you didn't create form 1 but created form 2, I don't want to give the user access to form 1. So in this scenario, if you click on form 1 I want a pop up to open and say "This form can no longer be created", and the link wouldn't function anymore. Basically the button will have 3 function. First it should look at the database and if data for that specific form exists it should do "Step 2", if data for that form and the proceeding forms don't exist it should do "Step 1", and if data for that form doesn't exist but data for proceeding form's does exist is should do "Step 3". Is this possible? Please help as I need to find a solution to this soon. Any help would be highly appreciated. Thank you

    Read the article

  • How to get time from db depending upon conditions

    - by Somebody is in trouble
    I have a table in which the value are Table_hello date col2 2012-01-31 23:01:01 a 2012-06-2 12:01:01 b 2012-06-3 20:01:01 c Now i want to select date in days if it is 3 days before or less in hours if it is 24 hours before or less in minutes if it is 60 minutes before or less in seconds if it is 60 seconds before or less in simple format if it is before 3days or more OUTPUT for row1 2012-01-31 23:01:01 for row2 1 day ago for row3 1 hour ago UPDATE My sql query select case when TIMESTAMPDIFF(SECOND, `date`,current_timestamp) <= 60 then concat(TIMESTAMPDIFF(SECOND, `date`,current_timestamp), ' seconds') when TIMESTAMPDIFF(DAY, `date`,current_timestamp) <= 3 then concat(TIMESTAMPDIFF(DAY, `date`,current_timestamp), ' days')end when TIMESTAMPDIFF(HOUR, `date`,current_timestamp) <= 60 then concat(TIMESTAMPDIFF(HOUR, `date`,current_timestamp), ' hours') when TIMESTAMPDIFF(MINUTE, `date`,current_timestamp) <= 60 then concat(TIMESTAMPDIFF(MINUTE, `date`,current_timestamp), ' minutes') from table_hello Only problem is i am unable to use break and default in sql like switch case in c++

    Read the article

  • Binary string search on one field.

    - by CrazyJoe
    I have 300 boolean fields in one table, and im trying to do somithing like that: One string field: 10000010000100100100100100010001 Ha a simple way to do a simple search os this field like: select * from table where field xor "10000010000100100100000000010001" Im tring this but is to long: select * from teste where mid(info,2,1) and mid(info,3,1) :) Help!!

    Read the article

  • Select past date from database x days from now

    - by Pr0no
    Consider the following table daterange _date trading_day ------------------------ 2011-08-01 1 2011-07-31 0 2011-07-30 0 2011-07-29 1 2011-07-28 1 2011-07-27 1 2011-07-26 1 2011-07-25 1 2011-07-24 0 2011-07-23 0 2011-07-22 1 2011-07-21 1 2011-07-20 1 2011-07-19 1 2011-07-18 1 2011-07-17 0 I'm in need of a query that returns a _date, x days before a given _date. When counting back, _days with trading_day = 0 should be ignored. A few examples: input | output -------------------------+------------ 1 day before 2011-07-19 | 2011-07-18 2 days before 2011-08-01 | 2011-07-28 (trading_day = 0 don't count) 3 days before 2011-07-29 | 2001-07-26 The first one is easy: SELECT _date FROM daterange WHERE trading_day = 0 AND _date < '2011-07-19' LIMIT 1 But I don't know how to query for the other examples. Do you?

    Read the article

  • Php random row help...

    - by Skillman
    I've created some code that will return a random row, (well, all the rows in a random order) But i'm assuming its VERY uneffiecent and is gonna be a problem in a big database... Anyone know of a better way? Here is my current code: $count3 = 1; $count4 = 1; //Civilian stuff... $query = ("SELECT * FROM `*Table Name*` ORDER BY `Id` ASC"); $result = mysql_query($query); while($row = mysql_fetch_array($result)) { $count = $count + 1; $civilianid = $row['Id']; $arrayofids[$count] = $civilianid; //echo $arrayofids[$count]; } while($alldone != true) { $randomnum = (rand()%$count) + 1; //echo $randomnum . "<br>"; //echo $arrayofids[$randomnum] . "<br>"; $currentuserid = $arrayofids[$randomnum]; $count3 += 1; while($count4 < $count3) { $count4 += 1; $currentarrayid = $listdone[$count4]; //echo "<b>" . $currentarrayid . ":" . $currentuserid . "</b> "; if ($currentarrayid == $currentuserid){ $found = true; //echo " '" .$found. "' "; } } if ($found == true) { //Reset array/variables... $count4 = 1; $found = false; } else { $listdone[$count3] = $currentuserid; //echo "<u>" . $count3 .";". $listdone[$count3] . "</u> "; $query = ("SELECT * FROM `*Tablesname*` WHERE Id = '$currentuserid'"); $result = mysql_query($query); $row = mysql_fetch_array($result); $username = $row['Username']; echo $username . "<br>"; $count4 = 1; $amountdone += 1; if ($amountdone == $count) { //$count $alldone = true; } } } Basically it will loop until its gets an id (randomly) that hasnt been chosen yet. -So the last username could take hours :P Is this 'bad' code? :P :(

    Read the article

< Previous Page | 572 573 574 575 576 577 578 579 580 581 582 583  | Next Page >