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  • Search MySQL Database

    - by McCrum
    Hey, I have put together an iPhone web app and i'm currently reading and displaying data from a MySQL database. I have added a search bar to this page and was wondering what the best way would be to search the content on the page. http://j.mp/c6lV8c

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  • mysql multiple insert - what happens on error?

    - by aviv
    What happens in mysql multiple records insert during an error. I have a table: id | value 2 | 100 UNIQUE(id) Now i try to execute the query: INSERT INTO table(id, value) VALUES (1,10),(2,20),(3,30) I will get a duplicate-key error for the (2,20) BUT... Will the (1,10) get into the database? Will the (3,30) get into the database?

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  • How can I view live MySQL queries?

    - by barfoon
    Hey all, I am just wondering if its possible to trace MySQL queries on my linux server as they happen? For example I'd love to set up some sort of listener, then request a web page and view all of the queries the engine executed, or just view all of the queries being run on a production server. Are there tools to do this? Thank you,

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  • Does MySQL short-circuit the ORDER BY clause?

    - by nickf
    Given this SQL: SELECT * FROM mytable ORDER BY mycolumn, RAND() Assuming that mycolumn happens to only contain unique values (and hence, contains enough information to perform the ORDER BY), does MySQL short-circuit the operation and skip evaluating the rest?

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  • Mysql join two tables and add column values

    - by Mike
    My knowledge of mysql is not very in depth. If I have two tables that for example look like this: Table1 Date v1 v2 v3 05/01/2010 26 abc 45 05/02/2010 31 def 25 05/03/2010 50 ghi 46 Table2 Date v1 v2 v3 05/01/2010 42 jkl 15 05/02/2010 28 mno 14 05/03/2010 12 pqr 64 How can I join them in a query by their date and have the sum of table1.v1 and table2.v1 and also have the sum of table1.v3 and table2.v3. V2 should be ignored.

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  • Returning mySQL error with jQuery & AJAX

    - by kel
    I've got a form inserting data into mySQL. It works but I'm trying to add error handling in case something happens. If I break the Insert statements mySQL dies but I'm still getting a success message on the front end. What am I doing wrong? AJAX function postData(){ var employeeName = jQuery('#employeeName').val(); var hireDate = jQuery('#hireDate').val(); var position = jQuery('#position').val(); var location = jQuery('#location').val(); var interveiwer = jQuery('#interviewersID').val(); var q01 = jQuery('#q01').val(); var q02 = jQuery('#q02').val(); var q03 = jQuery('#q03').val(); var q04 = jQuery('#q04').val(); var q05 = jQuery('#q05').val(); var summary = jQuery('#summary').val(); jQuery.ajax({ type: 'POST', url: 'queryDay.php', data: 'employeeName='+ employeeName +'&hireDate='+ hireDate +'&position='+ position +'&location='+ location +'&interveiwer='+ interveiwer +'&q01='+ q01 +'&q02='+ q02 +'&q03='+ q03 +'&q04='+ q04 +'&q05='+ q05 +'&summary='+ summary, success: function(){ jQuery('#formSubmitted').show(); }, error: function(jqXHR, textStatus, errorThrown){ jQuery('#returnError').html(errorThrown); jQuery('#formError').show(); } }); }; PHP require_once 'config.php'; $employeeName = $_POST['employeeName']; $hireDate = $_POST['hireDate']; $position = $_POST['position']; $location = $_POST['location']; $interviewerID = $_POST['interveiwer']; $q01 = $_POST['q01']; $q02 = $_POST['q02']; $q03 = $_POST['q03']; $q04 = $_POST['q04']; $q05 = $_POST['q05']; $summary = $_POST['summary']; mysql_query("INSERT INTO employee (name, hiredate, position, location) VALUES ('$employeeName', '$hireDate', '$position', '$location')") or die (mysql_error()); $employeeID = mysql_insert_id(); mysql_query("INSERT INTO day (employee, interviewer, datetaken, q01, q02, q03, q04, q05, summary) VALUES ('$employeeID', '$interviewerID', NOW(), '$q01', '$q02', '$q03', '$q04', '$q05', '$summary')") or die (mysql_error());

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  • can we use join inside join in mysql?

    - by I Like PHP
    i have 3 tables structure is below tbl_login login_id | login_name 1 | keshav tbl_role role_id | login_id( refer to tbl_login.login_id) 1 | 1 tbl_stuff stuff_id | role_id( refer to tbl_role.role_id) 1 | 1 i need data in follow format stuff_id | login_name 1 | keshav how to use JOIN to retrive the above data in mysql?

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  • ADO.NET Data Services for MySQL

    - by Shalan
    Hey! I've searched high and low for this, and no luck. Is there a way that CRUD methods for a MySQL install (Linux box) be exposed via ADO.NET WCF Data Services? I would really love to leverage this in my WPF app :) Thank u!

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  • PHP code to convert a MySQL query to CSV

    - by Reilly
    What is the most efficient way to convert a MySQL query to CSV in PHP please? It would be best to avoid temp files as this reduces portability (dir paths and setting file-system permissions required). The CSV should also include one top line of field names. Cheers.

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  • How can I allow NULL values in MySQL

    - by peakUC
    I was wondering how can I allow NULL values in the following code below along with keeping WHERE username = '$username'. here is the mysql code. "SELECT * FROM users WHERE username = '$username' AND user_id <> '$user_id'" I'm trying to check for usernames with the same username but I want all users to have a NULL value if they want.

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  • Create Chart using PHP-MySQL

    - by Ajith
    I have a mysql table - request_events with three fields; request_eventsid,datetime,type.this table will track all the activities of my website day wise and also type wise.thus,type may be 1 or 2.I need to display an open-chart for understanding the progress.So I need to retrieve the ratio of type2/type1 as input day wise.How can I get all these input for last 30 days from this table.Please give me some idea....It already kill my week end.Please help me

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  • SQLite vs MySQL

    - by Teifion
    SQLite is a flat-file database and MySQL is a normal database. That's great but I'm not sure which is faster where or better for what? What are the pros and cons of each option?

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  • Creating a MySQL view with an auto-incrementing id column

    - by hmemcpy
    I have a MySQL database from which a view is created. Is is possible to add an auto-incrementing id for each row in the view? I tried CREATE ALGORITHM=UNDEFINED DEFINER=`database_name`@`%` SQL SECURITY DEFINER VIEW `MyView` AS set @i = 0; select @i:=@i+1 as `id` ... but that doesn't work in a View. Sorry, my SQL is weak, any help is appreciated.

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  • MySQL: Which is faster — INSTR or LIKE?

    - by Grekker
    If your goal is to test if a string exists in a MySQL column (of type 'varchar', 'text', 'blob', etc) which of the following is faster / more efficient / better to use, and why? Or, is there some other method that tops either of these? INSTR( columnname, 'mystring' ) > 0 vs columnname LIKE '%mystring%'

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  • PHP & MySQL Undefined variable problem

    - by comma
    I keep getting the following error Undefined variable: id on line 91 can some one help me correct this problem? The error is on this line. $query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())"; MySQL database tables. CREATE TABLE tags ( id INT UNSIGNED NOT NULL AUTO_INCREMENT, skill VARCHAR(255) NOT NULL, experience VARCHAR(255) NOT NULL, years VARCHAR(255) NOT NULL, PRIMARY KEY (id) ); CREATE TABLE users_skills ( id INT UNSIGNED NOT NULL AUTO_INCREMENT, skill_id INT UNSIGNED NOT NULL, user_id INT UNSIGNED NOT NULL, date_created DATETIME UNSIGNED NOT NULL, PRIMARY KEY (id) ); Here is the PHP & MySQL code. if (isset($_POST['info_submitted'])) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.* FROM learned_skills INNER JOIN users_skills ON learned_skills.id = users_skills.skill_id WHERE user_id='$user_id'"); if (!$dbc) { print mysqli_error($mysqli); return; } $user_id = '5'; $skill = $_POST['skill']; $experience = $_POST['experience']; $years = $_POST['years']; $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.* FROM learned_skills INNER JOIN users_skills ON users_skills.skill_id = learned_skills.id WHERE users_skills.user_id='$user_id'"); if (mysqli_num_rows($dbc) == 0) { if (isset($_POST['skill']) && trim($_POST['skill'])!=='') { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $query1 = mysqli_query($mysqli,"INSERT INTO learned_skills (skill, experience, years) VALUES ('" . $skill . "', '" . $experience . "', '" . $years . "')"); if (mysqli_query($mysqli, $query1)) { print mysqli_error($mysqli); return; } $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT id FROM learned_skills WHERE id='" . $skill . "' AND experience='" . $experience . "' AND years='" . $years . "'"); if (!$dbc) { print mysqli_error($mysqli); } else { while($row = mysqli_fetch_array($dbc)){ $id = $row["id"]; } } $query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())"; } }

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  • PHP mySQL - replace some string inside string

    - by apis17
    i want to replace ALL comma , into ,<space> in all address table in my mysql table. For example, +----------------+----------------+ | Name | Address | +----------------+----------------+ | Someone name | A1,Street Name | +----------------+----------------+ Into +----------------+----------------+ | Name | Address | +----------------+----------------+ | Someone name | A1, Street Name| +----------------+----------------+ Thanks in advance.

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