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  • MySQL LEFT JOIN error

    - by Alex
    Hello, I've got some SQL that used to work with an older MySQL version, but after upgrading to a newer MySQL 5 version, I'm getting an error. Here's the SQL: SELECT portfolio.*, projects.*, types.* FROM projects, types LEFT JOIN portfolio ON portfolio.pfProjectID = projects.projectID WHERE projects.projectType = types.typeID AND types.typeID = #URL.a# ORDER BY types.typeSort, projects.projectPriority ASC and the new error I'm receiving: Unknown column 'projects.projectID' in 'on clause' How can I convert this to compatible SQL for the newer MySQL version? Thanks very much!

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  • MySQL forgot about automatically creating an index for a foreign key?

    - by bobo
    After running the following SQL statements, you will see that, MySQL has automatically created the non-unique index question_tag_tag_id_tag_id on the tag_id column for me after the first ALTER TABLE statement has run. But after the second ALTER TABLE statement has run, I think MySQL should also automatically create another non-unique index question_tag_question_id_question_id on the question_id column for me. But as you can see from the SHOW INDEXES statement output, it's not there. Why does MySQL forget about the second ALTER TABLE statement? By the way, since I have already created a unique index question_id_tag_id_idx used by both question_id and tag_id columns. Is creating a separate index for each of them redundant? mysql> DROP DATABASE mydatabase; Query OK, 1 row affected (0.00 sec) mysql> CREATE DATABASE mydatabase; Query OK, 1 row affected (0.00 sec) mysql> USE mydatabase; Database changed mysql> CREATE TABLE question (id BIGINT AUTO_INCREMENT, html TEXT, PRIMARY KEY(id)) ENGINE = INNODB; Query OK, 0 rows affected (0.05 sec) mysql> CREATE TABLE tag (id BIGINT AUTO_INCREMENT, name VARCHAR(10) NOT NULL, UNIQUE INDEX name_idx (name), PRIMARY KEY(id)) ENGINE = INNODB; Query OK, 0 rows affected (0.05 sec) mysql> CREATE TABLE question_tag (question_id BIGINT, tag_id BIGINT, UNIQUE INDEX question_id_tag_id_idx (question_id, tag_id), PRIMARY KEY(question_id, tag_id)) ENGINE = INNODB; Query OK, 0 rows affected (0.00 sec) mysql> ALTER TABLE question_tag ADD CONSTRAINT question_tag_tag_id_tag_id FOREIGN KEY (tag_id) REFERENCES tag(id); Query OK, 0 rows affected (0.10 sec) Records: 0 Duplicates: 0 Warnings: 0 mysql> ALTER TABLE question_tag ADD CONSTRAINT question_tag_question_id_question_id FOREIGN KEY (question_id) REFERENCES question(id); Query OK, 0 rows affected (0.13 sec) Records: 0 Duplicates: 0 Warnings: 0 mysql> SHOW INDEXES FROM question_tag; +--------------+------------+----------------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+ | Table | Non_unique | Key_name | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment | +--------------+------------+----------------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+ | question_tag | 0 | PRIMARY | 1 | question_id | A | 0 | NULL | NULL | | BTREE | | | question_tag | 0 | PRIMARY | 2 | tag_id | A | 0 | NULL | NULL | | BTREE | | | question_tag | 0 | question_id_tag_id_idx | 1 | question_id | A | 0 | NULL | NULL | | BTREE | | | question_tag | 0 | question_id_tag_id_idx | 2 | tag_id | A | 0 | NULL | NULL | | BTREE | | | question_tag | 1 | question_tag_tag_id_tag_id | 1 | tag_id | A | 0 | NULL | NULL | | BTREE | | +--------------+------------+----------------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+ 5 rows in set (0.01 sec) mysql>

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  • how can i have a working dropdownlist with links from a csv in php

    - by Mark Dekker
    I have a website that loads a CSV, divides it into parts, and shows these parts. There are 7 parts, and since it is for a music store it is sliced like this: the name of the product the subname the price the stock in one shop the stock of the other shop the group name the brandname What i have now is that it shows 12 products on 1 page, with a next and previous link on top of the page. The pages are made with the group name, every group has it's own page, so you have a drums page, guitar page, speaker page. That all works great. What i programmed in there is a dropdownlist, it drops down a list of brandnames, which should narrow the search for a person who is looking at the products. Problem is right now, that the dropdown menu works, i see the brands, but they are NONE clickable, only the brands that are currently on the page are shown, so NOT all the brands from that group are shown, only the 12 that are currently showing, when you press next, it shows 12 more brands, but i want them to be shown right on the start. And the third problem is, when it is showing the brands, it shows them double or triple, depending on how many products have the same brand. Is there a way what i want, with this code as a basis ? <html> <body bgcolor=#E2E965 link=black vlink=black alink=black text=#D5DF23> <style type="text/css"> #nav, #nav ul { padding: 0; margin: 0; list-style: none; } #nav li { float: left; width: 120px; } #nav ul { position: absolute; width: 120px; left: -1000px; } #nav li:hover ul { left: auto; } <!-- a {text-decoration:none} //--> body { scrollbar-arrow-color: #E2E965; scrollbar-face-color: #D7182A; scrollbar-highlight-color: #000000; scrollbar-3dlight-color: #D6DF23; scrollbar-shadow-color: #00000; scrollbar-darkshadow-color: #00000; scrollbar-track-color: #D6DF23; } input:link {text-decoration: none; color: #E2E965;} input:visited {text-decoration: none; color: #E2E965;} input:active {text-decoration: none; color: #E2E965;} .spacer_black { margin: 0px; padding: 0px; border: 5px; height: 2px; width: 100%; line-height: 0px; font-size: 0px; background-color: #000000;} </style> <table width=800 border=0><td> </html> <?PHP $offset = isset($_GET['offset'])?$_GET['offset']:0; $LinesToDisplay = 12; $row = $offset + $LinesToDisplay; $row2 = $offset - $LinesToDisplay; $file_handle = fopen("web.txt", "rb"); error_reporting( E_ALL ); // DEBUGGING $SelectArray=array(); while ((($parts = fgetcsv($file_handle,4096,"|")) !== FALSE) && ($LinesToDisplay > 0) && (!feof($file_handle))) { //new code //skip first $offset lines $num = count($parts[6]); $SelectArray[]=$parts[6]; if ($parts[5] == 9999) { if ($offset-- > 0) {continue;} $parts[0] = ucwords(strtolower($parts[0])); $parts[1] = ucwords(strtolower($parts[1])); ?> <td> <?php echo "<table BACKGROUND='background.jpg' border=0 width=250><td width='243' height='105'>"; echo "<font size=-1 face='helvetica' color=#812990><b>$parts[0]</b></font>"; echo "<i>"; ?> <html> <div onMouseOver="this.style.color = 'black';" onMouseOut="this.style.color = '#D5DF23';"> </html><?php echo "<font size=2>-$parts[1]</font>"; echo "</div></i>"; ?><html><a href="#" title="Koop nu de <?php echo $parts[0]; ?>" onClick="window.open('form.php?p=<?php echo urlencode($parts[0]); ?>','popuppage','width=400,height=400,top=250,left=250,resizable=0,statusbar=0,titlebar=yes,toolbar=no,scrollbars=no,location=no,directories=no');"> <div><img src='ster.jpg' border=0 width='46' align='right'></a> <a href="#"><img src='envelope.jpg' border=0 width='46' align='right'title="Heeft u een vraag over <?php echo $parts[0]; ?>" onClick="window.open('vraag.php?p=<?php echo urlencode($parts[0]); ?>','popuppage','width=400,height=400,top=250,left=250,resizable=0,statusbar=0,titlebar=yes,toolbar=no,scrollbars=no,location=no,directories=no');"> <div></a> <TABLE BORDER='0' cellpadding='0' CELLSPACING='0'> <TR> <TD WIDTH='70' HEIGHT='20' BACKGROUND='pricebackground.jpg' VALIGN='bottom'> <center> <font size=2 color=white face='helvetica'> <b></html> <?php echo "€ $parts[2]"; ?> </b> </td> </tr> </table> <?php echo "<b><font size=3 color=#D7182A>Op Voorraad In:<br></font>"; echo ("<font color=black> Amsterdam </font>"); if ( $parts[3] >= 1 ) echo ("<IMG SRC =green.gif>").""; if ( $parts[3] <= 0 ) echo ("<IMG SRC =red.gif>").""; echo ("<font color=black> Utrecht </font>"); if ( $parts[4] >= 1 ) echo ("<IMG SRC =green.gif>")."</td></table></b><p style='margin:9px;'>"; if ( $parts[4] <= 0 ) echo ("<IMG SRC =red.gif>")."</td></table></b><p style='margin:9px;'>"; $LinesToDisplay--; if ($LinesToDisplay/3==intval($LinesToDisplay/3)) { echo "<tr><td>"; } }} fclose($file_handle); ?> <tr align=right> <select style="background-color: #FFFFFF; color: #000000; font-family: Arial; font-weight: none; font-size: 12; width: 150px; "> <?php $i=1; foreach ($SelectArray as $val){ echo "<option value=\"$i\">$val</option>\n"; $i++; } ?> </select> <font color=black><body alink=black vlink=black link=black text=black> <center><a href="occasiona.php?offset=<?php echo $row2; ?>" align=center>&laquo; previous</a> || <a href="occasiona.php?offset=<?php echo $row; ?>">next &raquo;</a></table></html>

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  • Reset auto increment column value in script mysql

    - by Lucia
    Hi, I have two mysql tables, one needs to start its auto-increment column id with the last value of the last inserted row in the other table (plus 1). According to mysql manual you can restart the value of an auto increment column like this: mysql ALTER TABLE tbl AUTO_INCREMENT = 100; However, this is not possible: mysql ALTER TABLE tb2 AUTO_INCREMENT = (SELECT MAX(id) FROM tbl1); I need to perform something like this because I'm filling the tables using a script. Is there another way to achieve it?

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  • pagination and url encoding help

    - by Sufyan
    <?php $name=$_POST['name']; ?> <form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <input type="text" name="name"> <input type="submit" value="GO" name="submit"> </form> <?php include ('db.php'); if(isset($_POST['submit'])) { mysql_query ("INSERT INTO example (name) VALUES('$name')") or die(mysql_error()); } if (!isset($_GET['startrow']) or !is_numeric($_GET['startrow'])) { $startrow = 0; } else { $startrow = (int)$_GET['startrow']; } $query = "SELECT * FROM example ORDER BY id DESC LIMIT $startrow, 20"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<li>"; echo $row['name'] ." "." <a href= 'like.php?quote=" . urlencode( $row['name'] ) . "'>Click Here</a>"; echo "</li>"; } echo '<a href="'.$_SERVER['PHP_SELF'].'?startrow='.($startrow+10).'">Next</a>'; ?> I want to make my page links hidden , how can i make then hidden so that a user cant edit it. 2nd question, currently i am showing total 10 records on each page and then a next page button , but the next button is keep showing even when there is no more records...! how to remove a next page button when records ended. ?? line number 28 is the link to pages which can be easyily editable by any user, i wnat to make them secure (using ID) and line 35 is n'next' page link , this link should not be appear when number of records ended

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  • Missing parameter error after running MySql query

    - by annelie
    Hello, I'm completely new to MySql and haven't used SqlDataSource with UpdateParameters before, so I'm probably missing something very obvious. When trying to update a record, the update does happen but then throws an error saying "'id' parameter is missing at the statement". So the query works and the database gets updated as it should, but an error is thrown afterwards. These are the update parameters: <UpdateParameters> <asp:Parameter Name="business_name" Type="string" Size="256" /> <asp:Parameter Name="addr_line_1" Type="string" Size="256" /> <asp:Parameter Name="addr_line_2" Type="string" Size="256" /> <asp:Parameter Name="addr_line_3" Type="string" Size="256" /> <asp:Parameter Name="postcode" Type="string" Size="32" /> <asp:Parameter Name="county" Type="string" Size="128" /> <asp:Parameter Name="town_city" Type="string" Size="256" /> <asp:Parameter Name="tl_url" Type="string" Size="256" /> <asp:Parameter Name="customer_id" Type="string" Size="16" /> <asp:Parameter Name="region_id" Type="Int16" /> <asp:Parameter Name="description" Type="string" Size="1024" /> <asp:Parameter Name="approval_status" Type="string" Size="1" /> <asp:Parameter Name="tl_user_name" Type="string" Size="256" /> <asp:Parameter Name="phone" Type="string" Size="50" /> <asp:Parameter Name="uploaders_own" Type="Int16" /> </UpdateParameters> Here's the update statement: UPDATE myTable SET business_name = ?, addr_line_1 = ?, addr_line_2 = ?, addr_line_3 = ?, postcode = ?, county = ?, town_city = ?, tl_url = ?, customer_id = ?, region_id = ?, description = ?, approval_status = ?, tl_user_name = ?, phone = ?, uploaders_own = ? WHERE id = " + id Here's the stack trace: [InvalidOperationException: 'id' parameter is missing at the statement] CoreLab.MySql.r.a() +775 CoreLab.MySql.r.a(Int32& A_0, ArrayList& A_1) +448 CoreLab.MySql.x.e() +398 CoreLab.MySql.x.o() +89 CoreLab.MySql.MySqlCommand.a(CommandBehavior A_0, IDisposable A_1, Int32 A_2, Int32 A_3) +1306 CoreLab.Common.DbCommandBase.ExecuteDbDataReader(CommandBehavior behavior) +310 System.Data.Common.DbCommand.ExecuteReader() +12 CoreLab.Common.DbCommandBase.ExecuteNonQuery() +64 System.Web.UI.WebControls.SqlDataSourceView.ExecuteDbCommand(DbCommand command, DataSourceOperation operation) +386 System.Web.UI.WebControls.SqlDataSourceView.ExecuteUpdate(IDictionary keys, IDictionary values, IDictionary oldValues) +325 System.Web.UI.DataSourceView.Update(IDictionary keys, IDictionary values, IDictionary oldValues, DataSourceViewOperationCallback callback) +92 System.Web.UI.WebControls.DetailsView.HandleUpdate(String commandArg, Boolean causesValidation) +837 System.Web.UI.WebControls.DetailsView.HandleEvent(EventArgs e, Boolean causesValidation, String validationGroup) +509 System.Web.UI.WebControls.DetailsView.OnBubbleEvent(Object source, EventArgs e) +95 System.Web.UI.Control.RaiseBubbleEvent(Object source, EventArgs args) +37 System.Web.UI.WebControls.DetailsViewRow.OnBubbleEvent(Object source, EventArgs e) +113 System.Web.UI.Control.RaiseBubbleEvent(Object source, EventArgs args) +37 System.Web.UI.WebControls.LinkButton.OnCommand(CommandEventArgs e) +118 System.Web.UI.WebControls.LinkButton.RaisePostBackEvent(String eventArgument) +135 System.Web.UI.WebControls.LinkButton.System.Web.UI.IPostBackEventHandler.RaisePostBackEvent(String eventArgument) +10 System.Web.UI.Page.RaisePostBackEvent(IPostBackEventHandler sourceControl, String eventArgument) +13 System.Web.UI.Page.RaisePostBackEvent(NameValueCollection postData) +175 System.Web.UI.Page.ProcessRequestMain(Boolean includeStagesBeforeAsyncPoint, Boolean includeStagesAfterAsyncPoint) +1565 Does anyone know what I'm doing wrong? Thanks, Annelie

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  • PHP IF stops IE content from loading

    - by Rsmithy
    scenario I have a PHP if statement to test if a user is logging in for the first time ever. This then displays, a demo in a greybox popup box. Sadly when the box loads in IE, the content of the website doesn't. This means when I user closes the box, they are left the with background. Code - I'm now using PHP include. greybox.php <?php if ($fli == 0) {echo " <script type='text/javascript'> window.onload = function() { GB_showCenter('Your first login', '../video.php'); }; </script> ";} else echo "";?> <!-- GB scrip --> <script type="text/javascript"> var GB_ROOT_DIR = "greybox/"; </script> <script type="text/javascript" src="greybox/AJS.js"></script> <script type="text/javascript" src="greybox/AJS_fx.js"></script> <script type="text/javascript" src="greybox/gb_scripts.js"></script> <link href="greybox/gb_styles.css" rel="stylesheet" type="text/css" /> <!-- GB --> RELEVANT Script on main site <?php $fli = $_SESSION["USER"]["fli"]; ?> <?php include "greybox.php" ?> I would deeply appreicate any help at all please! :)

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  • Host...not allowed error in ODBC connection, while connecting to an online MySQL database

    - by sqlchild
    I have downloaded MySQL ODBC Connector 5.1. Now am trying to setup the DSN. But am getting the error: Connection Failed : [HY000] [MySQL] [ODBC 5.1 Driver]Host '117.x.x.x' is not allowed to connect to this MySQL server My server url is server.myweb.com - this name am entering in the TCP/IP Server and Port =3306. I have also entered the userid and password , which is the one which i enter when i open www.myweb.com/cpanel Is this a version problem? Should the version of MySQL on my server also be 5.1, i.e. the one of the ODBC? Please help.

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  • Why i get everytime the error-message that i've already sent the headers

    - by mikep
    Hey, i've another question about web-programming. I programmed a login script, but everytime when i try to login it says that i've send the header informations already. Here are the 2 files: <?php if($_GET['logout'] == 1) { setcookie('authorized', 1, time()-3600); } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Login - photoAdminSite</title> </head> <style type="text/css"> body { text-align: center; font-family: helvetica; } #loginForm { padding: 1em; background: #e3e3e3; width: 260px; margin: 3em auto 0; text-align: left; } </style> <body> <div id="loginForm"> <form method="post" action="confirm_login_credentials.php"> <h2>LOGIN</h2> <p>Username: <input type="text" name="username" /></p> <p>Password: <input type="password" name="password" /></p> <p><input type="submit" value="Login" name="submit" /></p> </form> </div> </body> </html> <?php $username = $_POST['username']; $password = $_POST['password']; require 'database.php'; $q = "SELECT id FROM users_photoadminsite WHERE user_name = '$username' AND password = '$password'"; $result = $mysqli->query($q) or die(mysqli_error()); if (mysqli_num_rows($result) == 1) { setcookie('authorized', 1, 0); header("Location: index.php"); } else { header("Location: login.php"); } ?> i would be really happy about some helpful answers.

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  • Configuring PHP in IIS with Tomcat

    - by Silent Walker
    I have my Java site running under IIS 7. I need to install wordpress blog in it. I've installed and configured PHP in IIS. I have tested the PHP handler by creating a separate site, everything works fine, phpinfo() gives the desired output. However, I'm having problem running the PHP files inside my Java web application. I've put my test PHP file inside a folder called blog. When I access this folder in the browser as /mysite/blog I get a 404 page from my Java application. When I try to invoke the php page directly, http://mysite/blog/index.php, I get an unprocessed php page. I'm using isapi_handler for the reidrects. How do I tell my isapi_handler to ignore /blog folder? In my IIS handler mapping, *.php is mapped with Fast CGI. I'm not sure how to approach this problem and any help on this would be much appreciated. Thanks in advance.

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  • Loading Ajax within a Google Maps InfoWindow

    - by McCrum
    I have put together a 5 star rating system using PHP and Ajax which will write the rating into a database. (See link below) http://andrewmccrum.com/maps/rate/5star.php I want this rating system to work within a Google Maps InfoWindow but at the minute I can only get the InfoWindow to read and display the rating. It will not let the user vote like the top link. And I have no idea why www.andrewmccrum.com/maps/database/index.php

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  • Hibernate Communications Link Failure in Restlet-Hibernate Based Java application powered by MySQL

    - by Vatsala
    Let me describe my question - I have a Java application - Hibernate as the DB interfacing layer over MySQL. I get the communications link failure error in my application. The occurence of this error is a very specific case. I get this error , When I leave mysql server unattended for more than approximately 6 hours (i.e. when there are no queries issued to MySQL for more than approximately 6 hours). I am pasting a top 'exception' level description below, and adding a pastebin link for a detailed stacktrace description. javax.persistence.PersistenceException: org.hibernate.exception.JDBCConnectionException: Cannot open connection - Caused by: org.hibernate.exception.JDBCConnectionException: Cannot open connection - Caused by: com.mysql.jdbc.exceptions.jdbc4.CommunicationsException: Communications link failure - The last packet successfully received from the server was 1,274,868,181,212 milliseconds ago. The last packet sent successfully to the server was 0 milliseconds ago. - Caused by: com.mysql.jdbc.exceptions.jdbc4.CommunicationsException: Communications link failure - The last packet successfully received from the server was 1,274,868,181,212 milliseconds ago. The last packet sent successfully to the server was 0 milliseconds ago. - Caused by: java.net.ConnectException: Connection refused: connect the link to the pastebin for further investigation - http://pastebin.com/4KujAmgD What I understand from these exception statements is that MySQL is refusing to take in any connections after a period of idle/nil activity. I have been reading up a bit about this via google search, and came to know that one of the possible ways to overcome this is to set values for c3p0 properties as c3p0 comes bundled with Hibernate. Specifically, I read from here http://www.mchange.com/projects/c3p0/index.html that setting two properties idleConnectionTestPeriod and preferredTestQuery will solve this for me. But these values dont seem to have had an effect. Is this the correct approach to fixing this? If not, what is the right way to get over this? The following are related Communications Link Failure questions at stackoverflow.com, but I've not found a satisfactory answer in their answers. http://stackoverflow.com/questions/2121829/java-db-communications-link-failure http://stackoverflow.com/questions/298988/how-to-handle-communication-link-failure Note 1 - i dont get this error when I am using my application continuosly. Note 2 - I use JPA with Hibernate and hence my hibernate.dialect,etc hibernate properties reside within the persistence.xml in the META-INF folder (does that prevent the c3p0 properties from working?)

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  • Adding a Line Break in mySQL INSERT INTO text

    - by james
    Hi Could someone tell me how to add a new line in a text that i enter in a mySql table. I tried using the '\n in the line i entered with INSERT INTO statement but '\n is shown as it is. Actually i have created a table in Ms Access with some data. Ms Access adds new line with '\n. I am converting Ms Access table data into mySql . But when i convert, the '\n is ignored and all the text is shown in one single line when i display it from mySql table on a php form. Can anyone tell me how mySQL can add a new line in a text ? Awaiting response Thanks !!

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  • A typical mysql query( how to use subquery column into main query)

    - by I Like PHP
    I HAVE TWO TABLES shown below table_joining id join_id(PK) transfer_id(FK) unit_id transfer_date joining_date 1 j_1 t_1 u_1 2010-06-05 2010-03-05 2 j_2 t_2 u_3 2010-05-10 2010-03-10 3 j_3 t_3 u_6 2010-04-10 2010-01-01 4 j_5 NULL u_3 NULL 2010-06-05 5 j_6 NULL u_4 NULL 2010-05-05 table_transfer id transfer_id(PK) pastUnitId futureUnitId effective_transfer_date 1 t_1 u_3 u_1 2010-06-05 2 t_2 u_6 u_1 2010-05-10 3 t_3 u_5 u_3 2010-04-10 now i want to know total employee detalis( using join_id) which are currently working on unit u_3 . means i want only join_id j_1 (has transfered but effective_transfer_date is future date, right now in u_3) j_2 ( tansfered and right now in `u_3` bcoz effective_transfer_date has been passed) j_6 ( right now in `u_3` and never transfered) what i need to take care of below steps( as far as i know ) <1> first need to check from table_joining whether transfer_id is NULL or not <2> if transfer_id= is NULL then see unit_id=u_3 where joining_date <=CURDATE() ( means that person already joined u_3) <3> if transfer_id is NOT NULL then go to table_transfer using transfer_id (foreign key reference) <4> now see the effective_transfer_date regrading that transfer_id whether effective_transfer_date<=CURDATE() <5> if transfer date has been passed(means transfer has been done) then return futureUnitID otherwise return pastUnitID i used two separate query but don't know how to join those query?? for step <1 ans <2 SELECT unit_id FROM table_joining WHERE joining_date<=CURDATE() AND transfer_id IS NULL AND unit_id='u_3' for step<5 SELECT IF(effective_transfer_date <= CURDATE(),futureUnitId,pastUnitId) AS currentUnitID FROM table_transfer // here how do we select only those rows which have currentUnitID='u_3' ?? please guide me the process?? i m just confused with JOINS. i think using LEFT JOIN can return the data i need, or if we use subquery value to main query? but i m not getting how to implement ...please help me. Thanks for helping me alwayz

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  • MySQL and SQLite differences in SQL

    - by Darth
    I'm writing java application that is using both SQLite and MySQL using JDBC. Are there any differences in SQL for those databases? Can I use same queries for both SQLite and MySQL, or is there any db specific stuff, that doesn't work on the other one? As far I've worked only with MySQL, so I don't really know much about SQLite.

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  • Rails 3.2.3 mysql error "max_prepared_stmt_count"

    - by Rob Momary
    I am running a Rails 3.2.3 app deployed with apache2/passenger on a virtual host with a mysql database server. I got this error after a lot of traffic was hitting the site: ActiveRecord::StatementInvalid (Mysql::Error: Can't create more than max_prepared_stmt_count statements (current value: 16382) I'm thinking it has something to do with the amount of traffic, but if so I have to find a way around this. Anyone had this error before? I can't figure out how to stop it. Here's what i see in mysql: mysql show global status like 'com_stmt%'; | Com_stmt_close | 1720319 | Com_stmt_execute | 2094137 | | Com_stmt_fetch | 0 | | Com_stmt_prepare | 1768924 | | Com_stmt_reprepare | 0 | | Com_stmt_reset | 0 | | Com_stmt_send_long_data | 0 | +-------------------------+---------+ I am running resque gem.

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  • Copying a mysql database from localhost to remote server using mysqldump.exe

    - by Ankur
    I want to copy a mysql database from my local computer to a remote server. I am trying to use the mysql dump command. All the examples on the internet suggest doing something like The initial mysql> is just the prompt I get after logging in. mysql> mysqldump -u user -p pass myDBName | NewDBName.out; But when I do this I get You have an error in your SQL syntax; check the manual that corresponds ... to use near 'mysqldump -u user -p pass myDBName | NewDBName.out' Since I have already logged in do I need to use -u and -p? Not doing so gives me the same error. Can you see what is wrong?

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  • whats wrong with this php mysql_real_escape_string

    - by skyhigh
    Hi Atomic Number Latin English Abbreviation * check the variables for content */ /*** a list of filters ***/ $filters = array( 'searchtext' => array( 'filter' => FILTER_CALLBACK, 'options' => 'mysql_real_escape_string'), 'fieldname' => array( 'filter' => FILTER_CALLBACK, 'options' => 'mysql_real_escape_string') ); /*** escape all POST variables ***/ $input = filter_input_array(INPUT_POST, $filters); /*** check the values are not empty ***/ if(empty($input['fieldname']) || empty($input['searchtext'])) { echo 'Invalid search'; } else { /*** mysql hostname ***/ $hostname = 'localhost'; /*** mysql username ***/ $username = 'username'; /*** mysql password ***/ $password = 'password'; /*** mysql database name ***/ $dbname = 'periodic_table'; /*** connect to the database ***/ $link = @mysql_connect($hostname, $username, $password); /*** check if the link is a valid resource ***/ if(is_resource($link)) { /*** select the database we wish to use ***/ if(mysql_select_db($dbname, $link) === TRUE) { /*** sql to SELECT information***/ $sql = sprintf("SELECT * FROM elements WHERE %s = '%s'", $input['fieldname'], $input['searchtext']); /*** echo the sql query ***/ echo '<h3>'.$sql.'</h3>'; /*** run the query ***/ $result = mysql_query($sql); /*** check if the result is a valid resource ***/ if(is_resource($result)) { /*** check if we have more than zero rows ***/ if(mysql_num_rows($result) !== 0) { echo '<table>'; while($row=mysql_fetch_array($result)) { echo '<tr> <td>'.$row['atomicnumber'].'</td> <td>'.$row['latin'].'</td> <td>'.$row['english'].'</td> <td>'.$row['abbr'].'</td> </tr>'; } echo '</table>'; } else { /*** if zero results are found.. ***/ echo 'Zero results found'; } } else { /*** if the resource is not valid ***/ 'No valid resource found'; } } /*** if we are unable to select the database show an error ****/ else { echo 'Unable to select database '.$dbname; } /*** close the connection ***/ mysql_close($link); } else { /*** if we fail to connect ***/ echo 'Unable to connect'; } } } else { echo 'Please Choose An Element'; } ? I got this code from phppro.org tutorials site and i tried to run it. It gives Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established. .... Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'ODBC'@'localhost' (using password: NO).... I went to php.net and look it up "Note: A MySQL connection is required before using mysql_real_escape_string() otherwise an error of level E_WARNING is generated, and FALSE is returned. If link_identifier isn't defined, the last MySQL connection is used." My questions are: 1-why they put single quotation around mysql_real_escape_string ? 2-They should establish a connection first, then use the $filter array statement with mysql_real_escape_string ?

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  • Accessing MySQL server via VPN in python

    - by user210481
    Hi I have a MySQL server that I need access through a VPN. I use MySQLdb package to access MySQL server in Python. When I can access the server without VPN, it works fine, but when I'm at certain locations, I need to connect through VPN. My computer is connected to the VPN and I can access the database through PHPMyAdmin, but MySQLdb gives me an error message: OperationalError: (2003, "Can't connect to MySQL server on 'MY_IP' (10061)") Any ideas on why it's not working? Thanks

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  • UDF function with library dependencies on MySQL 5.1

    - by Fredrik
    I'm having a problem with a MySQL UDF function (mychem.sourceforge.net) that's dependent on a large library (openbabel.org) which is in turn plugin based. The problem is that the format plugins to openbabel doesn't seem to load in MySQL 5.1 and I suspect it might be due to the plugin_dir setting. I have set plugin_dir to /usr/lib/ which is the location for both libmychem.so and libopenbabel.so as well as the directory openbabel that contains the format plugins. Is there a way to turn off the plugin_dir restriction in MySQL (preferably without compiling MySQL from sources) so that I can test this hypothesis or do you have a different idea on what might cause the problem? All this is done on Ubuntu 10.04 (but I had the same kind of problems on 8.04, where I managed to get it working after a lot of steps that I unfortunately have forgotten...) I have turned off apparmor during testing and it doesn't help either.

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  • Run multiple sql files in mysql batch

    - by hmak
    to run a single file you can run in mysql .\ filename or you outside of mysql you can run mysql < filename I have a directory of sql files so I'm trying to run them all at once by using a wildcard *.sql but it doesn't work. Any ideas?

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