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  • problem with joomla, php and json

    - by sebastian
    hi, i have a problem with a joomla component. i'm, unsing php and json for some dynamic drop down boxes. here is the code:` jQuery( function () { //jQuery.ajaxSetup({error : function (a,b) {console.dir(a); console.dir(b);}}); jQuery("#util, #loc").change( function() { var locatie = jQuery("#loc").val(); var utilitate = jQuery("#util").val(); if ( (locatie!= '---') && (utilitate!='---') ) jQuery.getJSON( "index.php?option=com_calculator&opt=json_contor&format=raw", { locatie: locatie, utilitate: utilitate }, function (data) { var html = ""; if ( data.success == 'ok' ) for (var i in data.val) html += "<option name=den_contor value ='"+ i+"' >" + data.val[i]+ " </option>"; jQuery("#den_contor").html( html ) } ) }) }); the query works, but only on one PC. we have exactly the same xampp server, exactly the same files. on one pc it works, and on a online server and on my pc it doesn't. EDIT: i have three drop down boxes, the first is populated directly from the database, the second has 4 predefined values. and the third is populated depending on combination of the first two. i have a test site online. http://contor.redxart.com must be logged in to use Calculator in the menu. you can make an new account :) "Adaugare Index" is the part that isn't working any ideas? thanks, sebastian

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  • Rails/mysql SUM distinct records - optimization

    - by pepernik
    Hey. How would you optimize this SQL SELECT SUM(tmp.cost) FROM ( SELECT DISTINCT clients.id as client, countries.credits_cost AS cost FROM countries INNER JOIN clients ON clients.country_id = countries.id INNER JOIN clients_groups ON clients_groups.client_id=clients.id WHERE clients_groups.group_id IN (1,2,3,4,5,6,7,8,9) GROUP BY clients.id ) AS tmp; I'm using this example as part of my Ruby on Rails project. Note that my nested SQL (tmp) can have more then 10 milion records. You can split that in more SQLs if the performance is better. Should I add any indexes to make it quicker (i have it on IDs)?

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  • Add column from another table matching results from first MySQL query

    - by Nemi
    This is my query for available rooms in choosen period: SELECT rooms.room_id FROM rooms WHERE rooms.room_id NOT IN ( SELECT reservations.room_id FROM reservations WHERE ( reservations.arrivaldate >= $arrival_datetime AND reservations.departuredate <= $departure_datetime) OR ( reservations.arrivaldate <= $arrival_datetime AND reservations.departuredate >= $arrival_datetime ) OR ( reservations.arrivaldate <= $departure_datetime AND reservations.departuredate >= $departure_datetime ) ); How to add average room price column for selected period(from $arrival_datetime to $departure_datetime) from another table (room_prices_table), for every room_id returned from above query. So I need to look in columns whos name is same as room_id... room_prices_table: date room0001 room0002 room0003 ... Something like SELECT AVG(room0003) FROM room_prices_table WHERE datum IS BETWEEN $arrival_datetime AND $departure_datetime ??

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  • MySQL select two tables at the same time...

    - by Jerry
    Hi all I have two tables and want to make a query. I tried to get team AA and team BB's image base on table A. I used: SELECT tableA.team1, tableA.team2, tableB.team, tableB.image, FROM tableA LEFT JOIN tableB ON tableA.team1=tableB.team The result only display imageA on the column. Are there any ways to select imageA and image B without using the second query? I appreciate any helps! Thanks a lot! My table structure are: table A team1 team2 ------------ AA BB table B team image ------------- AA imagaA BB imageB

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  • MySQL: fetching a null or an empty string

    - by Oden
    Hey, I know whats the difference between a NULL value and an empty string ("") value, but if I want to get a value by using the OR keyword, I get no result for a NULL value The table i want to query looks like this: titles_and_tags +----+----------+------+ | id | title | tag | +----+----------+------+ | 1 | title1 | NULL | | 2 | title2 | tag1 | | 3 | title3 | tag2 | +----+----------+------+ The query i use looks like this: select * from `titles_and_tags` WHERE `title` LIKE "title" AND `tag` = "tag1" OR `tag` IS NULL So i want to get here a rows (id: 1,2), BUT this results 0 rows. What have i done wrong?

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  • how can I select data from MySQL based on date (unix time record)

    - by bn
    I have a record of data with unix time date in it i want to select the row based on the date/month/year only (not with time) currently Im using something like this select * from tablename where date > '$today' and date < '$tomorow' LIMIT 1; how ever this is not that accurate if the $today and $tomorrow have different time (but same date) is there any better way to do this?

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  • Timeout on Large mySQL Query

    - by Bob Stewart
    I have this query: $theQuery = mysql_query("SELECT phrase, date from wordList WHERE group='nouns'"); while($getWords=mysql_fetch_array($theQuery)) { echo "$getWords[phrase] created on $getWords[date]<br>"; } The data table "wordList" contains 75,000 records in the group "nouns" and every time I load the code I am returned an error. Help!

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  • Multiple OR Clauses in MySQL

    - by Grant
    I'm trying to grab content where id = 3 OR id = 9 OR id = 100... Keep in mind, I can have a few hundred of these ids. What is the most efficient way to write my query? $sql = "SELECT name FROM artists WHERE (id=3 OR id=9 OR .... id-100)"

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  • Add results from MySQL?

    - by NardCake
    Not quite sure how descriptive that title was but this is what I want to do. I scripted a URL shortener today and it's working fine, I just want to add some stats on the bottom of it saying how many links there are and how many clicks there are. Now everytime a user clicks one of the links it +1 the column in the database then it redirects the user. I want to query that and add all of those numbers together from each row. I attempted a while loop which im not surprised didn't work: while($rows = mysql_fetch_assoc($check_count)){ $clicks = $rows['clicks']; $clicks = $clicks+$clicks; } If you don't understand please ask and if you do understand it means alot for any response!

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  • connecting to phpMyAdmin database with PHP/MySQL

    - by user303955
    I've made a database using phpMyAdmin , now I want to make a register form for my site where peaple can register .I know how to work with input tags in HTML and I know how to insert data into a database but my problem is that I don't know how I can connect to the database that is already made in phpMyAdmin.

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  • mysql query that has array

    - by Xainee Khan
    //get all id's of ur friend that has installed your application $friend_pics=$facebook->api( array( 'method' => 'fql.query', 'query' => "SELECT uid FROM user WHERE uid IN(SELECT uid2 from friend WHERE uid1='$user') AND is_app_user = 1" ) ); // this query work fine //your top10 friends in app $result="SELECT * FROM fb_user WHERE user_id IN($friend_pics) ORDER BY oldscore DESC LIMIT 0,10"; db_execute($result); i want to retrive ten top scorer from my database stored in oldscore but in my second query the array name $friend_pics is not working i guess,plz help me thanks

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  • How to return array of C++ objects from a PHP extension

    - by John Factorial
    I need to have my PHP extension return an array of objects, but I can't seem to figure out how to do this. I have a Graph object written in C++. Graph.getNodes() returns a std::map<int, Node*>. Here's the code I have currently: struct node_object { zend_object std; Node *node; }; zend_class_entry *node_ce; then PHP_METHOD(Graph, getNodes) { Graph *graph; GET_GRAPH(graph, obj) // a macro I wrote to populate graph node_object* n; zval* node_zval; if (obj == NULL) { RETURN_NULL(); } if (object_init_ex(node_zval, node_ce) != SUCCESS) { RETURN_NULL(); } std::map nodes = graph-getNodes(); array_init(return_value); for (std::map::iterator i = nodes.begin(); i != nodes.end(); ++i) { php_printf("X"); n = (node_object*) zend_object_store_get_object(node_zval TSRMLS_CC); n-node = i-second; add_index_zval(return_value, i-first, node_zval); } php_printf("]"); } When i run php -r '$g = new Graph(); $g->getNodes();' I get the output XX]Segmentation fault meaning the getNodes() function loops successfully through my 2-node list, returns, then segfaults. What am I doing wrong?

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  • MySQL query : all records of one table plus count of another table

    - by Ricardo
    Hello Guys! I have 2 tables: User and Picture. The Picture table has the key of the user. So basically each user can have multiple pictures, and each picture belongs to one user. Now, I am trying to make the following query: I want to select all the user info plus the total number of pictures that he has (even if it's 0). How can I do that? Probably it sounds quite simple, but I am trying and trying and can't seem to find the right query. The only thing I could select is this info, but only for users that have at least 1 picture, meaning that the Pictures table has at least one record for that key... But I also wanna consider the users that don't have any. Any idea? Thanks!

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  • MySQL: selecting all but one field?

    - by gsquare567
    instead of SELECT * FROM mytable, i would like to select all fields EXCEPT one (namely, the 'serialized' field, which stores a serialized object). this is because i think that losing that field will speed up my query by a lot. however, i have so many fields and am quite the lazy guy. is there a way to say... `SELECT ALL_ROWS_EXCEPT(serialized) FROM mytable` ? thanks!

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  • mysql IF EXISTS

    - by cosy
    What is wrong with this ? mysql_query("IF EXISTS(SELECT * FROM y where 1=1 ) THEN do something ELSE do something END IF"); Thanks!

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  • MySql - set of time stamped data (timestamp,event) calculating events per day

    - by Kevin Ohashi
    I have a table: id, datetime, event i also have table dates: date (Y-m-d format) the problem is some days don't have any events, I would like them to show 0 (or null) SELECT DATE_FORMAT(table.timestamp, '%Y-%m-%d') ydm, count(table.fkUID) FROM `table` where table.fkUID=$var group by ydm; is there some way to join or use conditional statements to make the result show: date|count ---------- 2010-05-23| 5 2010-05-24| 0 <--- this line just doesn't exist in my query. 2010-05-26| 3

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  • Php/Mysql - need help to insert and update multiple rows with a single query

    - by Guanche
    Hello, is there any way how in this situation insert and update DB with single queries? $message = 'Hello to all group members'; $userdata = mysql_query("SELECT memberid, membernick FROM members WHERE groupid='$cid'") or die('Error'); while(list($memberid, $membernick) = mysql_fetch_row($userdata)) { $result1 = mysql_query("INSERT INTO messages VALUES (NULL,'$membernick', '$memberid', '$message')") or die('Error'); $result2 = mysql_query("UPDATE users SET new_messages=new_messages+1, total_messages=total_messages+1 WHERE id='$memberid'") or die('Error'); }

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  • php cookies block access to directories

    - by user342391
    I have a mysql database of users that can login to my site and view content. I would like to block a couple of directories from certain users. What is the best way to do this. Currently when a user logs in a cookie is created with their customer id and the customer is is used to display their content. How would I block entire directories from my users???

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  • Difference between two PHP times as years, months and days for PHP version 5.2

    - by Dominor Novus
    Forward: I've scanned through the existing questions/answers on this matter. This is not a duplicitous question; I cannot find a working solution from the accepted answers. The main questions/answers I've reviewed can be found here: How to calculate the difference between two dates using PHP? What I need: A calucalation of the difference between two dates expressed as years, months and days that works with PHP version: 5.2. <?php $current_date = date('d-M-Y'); $future_date = '2012-11-01'; ?> What I've tried: Most answers I find online don't seem to exact in that they don't factor in leap years. This highly rated answer won't work because DateTime-diff() is php 5.3+. This accepted answer results in (i.e. the second block of code aimed at PHP 5.2) results in the following being parsed: Array ( [y] = 25 [m] = 11 [d] = 7 [h] = 3 [i] = 15 [s] = 19 [invert] = 0 [days] = 9473 ) Array ( [y] = 25 [m] = 11 [d] = 7 [h] = 3 [i] = 15 [s] = 19 [invert] = 1 [days] = 9473 ) I can't tell if I've incorrectly applied the code or it's simply a case of me not knowing how to manipulate the array.

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