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  • insert newline in perl -e statement

    - by lydonchandra
    Hi If I do this in bash perl -e '$x; $y' How can I insert a new line between the character ; and $y? I don't want to re-type the whole line, I just want to move my cursor to the position and then insert a newline ? Is this possible? Many thanks

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  • MySQL select query result set changes based on column order

    - by user197191
    I have a drupal 7 site using the Views module to back-end site content search results. The same query with the same dataset returns different results from MySQL 5.5.28 to MySQL 5.6.14. The results from 5.5.28 are the correct, expected results. The results from 5.6.14 are not. If, however, I simply move a column in the select statement, the query returns the correct results. Here is the code-generated query in question (modified for readability). I apologize for the length; I couldn't find a way to reproduce it without the whole query: SELECT DISTINCT node_node_revision.nid AS node_node_revision_nid, node_revision.title AS node_revision_title, node_field_revision_field_position_institution_ref.nid AS node_field_revision_field_position_institution_ref_nid, node_revision.vid AS vid, node_revision.nid AS node_revision_nid, node_node_revision.title AS node_node_revision_title, SUM(search_index.score * search_total.count) AS score, 'node' AS field_data_field_system_inst_name_node_entity_type, 'node' AS field_revision_field_position_college_division_node_entity_t, 'node' AS field_revision_field_position_department_node_entity_type, 'node' AS field_revision_field_search_lvl_degree_lvls_node_entity_type, 'node' AS field_revision_field_position_app_deadline_node_entity_type, 'node' AS field_revision_field_position_start_date_node_entity_type, 'node' AS field_revision_body_node_entity_type FROM node_revision node_revision LEFT JOIN node node_node_revision ON node_revision.nid = node_node_revision.nid LEFT JOIN field_revision_field_position_institution_ref field_revision_field_position_institution_ref ON node_revision.vid = field_revision_field_position_institution_ref.revision_id AND (field_revision_field_position_institution_ref.entity_type = 'node' AND field_revision_field_position_institution_ref.deleted = '0') LEFT JOIN node node_field_revision_field_position_institution_ref ON field_revision_field_position_institution_ref.field_position_institution_ref_target_id = node_field_revision_field_position_institution_ref.nid LEFT JOIN field_revision_field_position_cip_code field_revision_field_position_cip_code ON node_revision.vid = field_revision_field_position_cip_code.revision_id AND (field_revision_field_position_cip_code.entity_type = 'node' AND field_revision_field_position_cip_code.deleted = '0') LEFT JOIN node node_field_revision_field_position_cip_code ON field_revision_field_position_cip_code.field_position_cip_code_target_id = node_field_revision_field_position_cip_code.nid LEFT JOIN node node_node_revision_1 ON node_revision.nid = node_node_revision_1.nid LEFT JOIN field_revision_field_position_vacancy_status field_revision_field_position_vacancy_status ON node_revision.vid = field_revision_field_position_vacancy_status.revision_id AND (field_revision_field_position_vacancy_status.entity_type = 'node' AND field_revision_field_position_vacancy_status.deleted = '0') LEFT JOIN search_index search_index ON node_revision.nid = search_index.sid LEFT JOIN search_total search_total ON search_index.word = search_total.word WHERE ( ( (node_node_revision.status = '1') AND (node_node_revision.type IN ('position')) AND (field_revision_field_position_vacancy_status.field_position_vacancy_status_target_id IN ('38')) AND( (search_index.type = 'node') AND( (search_index.word = 'accountant') ) ) AND ( (node_revision.vid=node_node_revision.vid AND node_node_revision.status=1) ) ) ) GROUP BY search_index.sid, vid, score, field_data_field_system_inst_name_node_entity_type, field_revision_field_position_college_division_node_entity_t, field_revision_field_position_department_node_entity_type, field_revision_field_search_lvl_degree_lvls_node_entity_type, field_revision_field_position_app_deadline_node_entity_type, field_revision_field_position_start_date_node_entity_type, field_revision_body_node_entity_type HAVING ( ( (COUNT(*) >= '1') ) ) ORDER BY node_node_revision_title ASC LIMIT 20 OFFSET 0; Again, this query returns different sets of results from MySQL 5.5.28 (correct) to 5.6.14 (incorrect). If I move the column named "score" (the SUM() column) to the end of the column list, the query returns the correct set of results in both versions of MySQL. My question is: Is this expected behavior (and why), or is this a bug? I'm on the verge of reverting my entire environment back to 5.5 because of this.

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  • Spring-mvc project can't select from a particular mysql table

    - by Dan Ray
    I'm building a Spring-mvc project (using JPA and Hibernate for DB access) that is running just great locally, on my dev box, with a local MySQL database. Now I'm trying to put a snapshot up on a staging server for my client to play with, and I'm having trouble. Tomcat (after some wrestling) deploys my war file without complaint, and I can get some response from the application over the browser. When I hit my main page, which is behind Spring Security authentication, it redirects me to the login page, which works perfectly. I have Security configured to query the database for user details, and that works fine. In fact, a change to a password in the database is reflected in the behavior of the login form, so I'm confident it IS reaching the database and querying the user table. Once authenticated, we go to the first "real" page of the app, and I get a "data access failure" error. The server's console log gets this line (redacted): ERROR org.hibernate.util.JDBCExceptionReporter - SELECT command denied to user 'myDbUser'@'localhost' for table 'asset' However, if I go to MySQL from the shell using exactly the same creds, I have no problem at all selecting from the asset table: [development@tomcat01stg]$ mysql -u myDbUser -pmyDbPwd dbName ... mysql> \s -------------- mysql Ver 14.12 Distrib 5.0.77, for redhat-linux-gnu (i686) using readline 5.1 Connection id: 199 Current database: dbName Current user: myDbUser@localhost ... UNIX socket: /var/lib/mysql/mysql.sock -------------- mysql> select count(*) from asset; +----------+ | count(*) | +----------+ | 19 | +----------+ 1 row in set (0.00 sec) I've broken down my MySQL access settings, cleaned out the user and re-run the grant commands, set up a version of the user from 'localhost' and another from '%', making sure to flush permissions.... Nothing is changing the behavior of this thing. What gives?

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  • zsh conditional statement help

    - by Roy Rico
    Feeling kinda dumb right now: Why is my contional always true? I've tried # this should let me know what's not a directory or # symbolic link. whoa=`find ${MUSICDIR} ! -type l ! -type d | wc -l` # I would expect if it's 0 (meaning nothing was found) that # one of these statements would evaluate to false, but so far # it's always evaluating to true if [[ "${whoa}" != "0" ]] if [[ ${whoa} -gt 0 ]] What am I missing?

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  • Microsoft Excel IF/THEN statement for Words

    - by user1667462
    Right now I have an excel spreadsheet. In Cells A1 and down I have cities listed. In Cells B1 and down I have some generic information with the word CITY in it. What I want to know if there is a formula to replace the word CITY that is in the contents of Cell B1 with the contents in Cell A1. For instance... Cell A1 has "Daytona Beach, FL" Cell B1 has "Compare mortgage & refinance rates from different mortgage lenders and brokers in CITY. Find the home refinance rate you were looking for in CITY." I need a formula that replaces "CITY" in B1 with "Dayton Beach, FL" in A1. Is this possible? Thank you for your help!

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  • Nginx 'if' statement in http context?

    - by andy
    I want to set a variable in the http context of nginx so it applies to all my servers. However, 'if' is only supported in server & location. How can I set a variable in the http context so it will affect all servers? Might the lua module be able to help with this (although I'd rather have a pure nginx solution). If so, please provide an example. I just want to set the following variable so it applies to all servers: # only allow gzip encoding. For all other cases, use identity (uncompressed) if ($http_accept_encoding ~* "gzip") { set $sanitised_accept_encoding "gzip"; } if ($http_accept_encoding != "gzip") { set $sanitised_accept_encoding ""; }

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  • Nested IF statement on Google Spreadsheet, second part same as the first [migrated]

    - by lazfish
    I have a spreadsheet for my budget. Payments are either drawn from my bank or my Amex card and then my Amex card is drawn from my bank as well. So I add up all my monthly total like this: =sumif(I3:I20,"<>AMEX",D3:D20) Where I3:I20 = account bill is paid from and D3:D20 is monthly amount due. So I am not including bills that come from my Amex card in the total since the Amex bill itself covers those. Next I have a column that has the day of the month 1-10 (when everything gets paid) and it does this: =sumif(H3:H20,E24:E33,D3:D20) Where H3:H20 = date bill is paid and E25:E35 = range from 1-10. What I want to do is make this second part do the same check as the first. Something like this: =sumif(H3:H19,E24:E33,IF(I3:I19"<>SPG",D3:D19,0)) But I get error: "Parse error" What am I doing wrong?

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  • Strange Photoshop Problem: Can not select, zoom, paint, option button 'locked'

    - by nikcub
    I have a very strange problem with Photoshop. I can not use any of the tools, since the cursor appears 'locked'. If I select v on my keyboard, it goes to the zoom tool, but the cursor does not change. If I select the paintbrush tool, I can only paint if I hold down the option key. This is what the cursor look like (I had to paint it since I couldn't capture it). It is a rectangle with two lines through it. I am running Photoshop CS4 on a Macbook Pro with Mac OS X 10.6.6. Using both the trackpad and an external Logitech MX5000 mouse I see the same issue. This started when I fired up Photoshop today for the first time in a while. I can't remember changing any options or doing anything that could cause this. Is it possible that the option key is somehow locked in place, or there is some equivalent of num lock on? Very strange problem, I would appreciate any help anybody can offer. Edit: To add, the icon remains the same within all the menu options - it never goes back to being just a normal mouse cursor. Also, right click works fine, and if I hold down option, the cursor goes back to normal and I can paint with it. I can't use Marquee, Lasso, Crop, Type etc. even with option held down. When I go into Bridge, it is the same icon.

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  • How to configure in crontab with condition statement for checks

    - by chz
    We like to monitor the NAS storage mounted on a linux box. We only like to be notified via mail when the usage exceeds a certain number say 80. We have only seen in linux books where most of them are calling shell scripts at certain times. How do we write inside crontab to only mail us if it exceeds 80 ? Usual eg 2 2 * * * /home/someUser/script.sh 2&1 | mail [email protected] Looking for solution like below 2 2 * * * if [ someNumber "80" ] ; then /home/someUser/script.sh | mail [email protected] Sincerely

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  • IF Statement using dates for a budget template

    - by Leah Allen
    I am working on a budget and want to automatically account for increases in rent in the correct month, I would also like to account for dates tenants move in or out. I may also sometimes have a tenant in a space all year with no changes to rent. Below is an example of my budget with all three scenarios. SQFT BaseRentperSQft BaseRentIncrease DateofIncrease CommencementDate TermDate Jan-Decbymonth 10,000 $15.00 $15.25 05/01/2013 11/30/2013 10,000 $15.00 04/01/2013 10,000 $15.00 I would like to build a formula to accomplish all criteria. Thanks in advance, I can only write simple IF statements, this one is out of my league.

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  • Dangers of the pyton eval() statement

    - by LukeP
    I am creating a game. Specifically it is a pokemon battle simulator. I have an sqlite database of moves in which a row looks something like: name | type | Power | Accuracy | PP | Description However, there are some special moves. For said special moves, their damage (and other attributes not shown above, like status effects) may be dependant on certian factors. Rather than create a huge if/else in one of my classes covering the formulas for every one of these moves. I'd rather include another column in the DB that contains a formula in string form, like 'self.health/2'(simplified example). I could then just plug that into eval. I always see people saying to stay away from eval, but from what I can tell, this would be considered an acceptable use, as the dangers of eval only come into play when accepting user input. Am I correct in this assumption, or is there somthing i'm not seeing.

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  • vbs - 800A0401 - Expected End of Statement

    - by user93200
    The bat has >>%vbs% echo oShellLink.IconLocation = "%1, 0"to produce oShellLink.IconLocation = ""C:\WINDOWS\NOTEPAD.exe", 0"` where %1 is a quoted path However, unless I remove the quotes from path like here: oShellLink.IconLocation = "C:\WINDOWS\NOTEPAD.exe, 0" I get the titular error. What do you suggest? (Note that %1 is always supplied with quotes) Also, I'm not very familiar with vbscript, why does it require no quotes next to each other? ... Still not clear why vbscript can't interpret this assignment, but found a fix: %~1 - Expands %1 and removes any surrounding quotation marks ("").

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  • SQL Statement Help... Ignore already existing rows

    - by Funchy
    I have a table with a foreign key constraint and the command below gives me an error because it's trying to set a value that already in the provider table. How do I update this command to ignore those rows that already exist in the provider table? UPDATE b SET b.iProvider_PVN = a.POIN FROM dbo.ASPVNTOPOIN_stg a INNER JOIN dbo.Provider b ON a.ASPVN = b.iProvider_PVN AND b.vcProv_Type = 'IPA' LEFT JOIN dbo.Provider c ON a.POIN = c.iProvider_PVN WHERE c.iProvider_PVN IS NULL

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  • IsNumeric() Broken? Only up to a point.

    - by Phil Factor
    In SQL Server, probably the best-known 'broken' function is poor ISNUMERIC() . The documentation says 'ISNUMERIC returns 1 when the input expression evaluates to a valid numeric data type; otherwise it returns 0. ISNUMERIC returns 1 for some characters that are not numbers, such as plus (+), minus (-), and valid currency symbols such as the dollar sign ($).'Although it will take numeric data types (No, I don't understand why either), its main use is supposed to be to test strings to make sure that you can convert them to whatever numeric datatype you are using (int, numeric, bigint, money, smallint, smallmoney, tinyint, float, decimal, or real). It wouldn't actually be of much use anyway, since each datatype has different rules. You actually need a RegEx to do a reasonably safe check. The other snag is that the IsNumeric() function  is a bit broken. SELECT ISNUMERIC(',')This cheerfully returns 1, since it believes that a comma is a currency symbol (not a thousands-separator) and you meant to say 0, in this strange currency.  However, SELECT ISNUMERIC(N'£')isn't recognized as currency.  '+' and  '-' is seen to be numeric, which is stretching it a bit. You'll see that what it allows isn't really broken except that it doesn't recognize Unicode currency symbols: It just tells you that one numeric type is likely to accept the string if you do an explicit conversion to it using the string. Both these work fine, so poor IsNumeric has to follow suit. SELECT  CAST('0E0' AS FLOAT)SELECT  CAST (',' AS MONEY) but it is harder to predict which data type will accept a '+' sign. SELECT  CAST ('+' AS money) --0.00SELECT  CAST ('+' AS INT)   --0SELECT  CAST ('+' AS numeric)/* Msg 8115, Level 16, State 6, Line 4 Arithmetic overflow error converting varchar to data type numeric.*/SELECT  CAST ('+' AS FLOAT)/*Msg 8114, Level 16, State 5, Line 5Error converting data type varchar to float.*/> So we can begin to say that the maybe IsNumeric isn't really broken, but is answering a silly question 'Is there some numeric datatype to which i can convert this string? Almost, but not quite. The bug is that it doesn't understand Unicode currency characters such as the euro or franc which are actually valid when used in the CAST function. (perhaps they're delaying fixing the euro bug just in case it isn't necessary).SELECT ISNUMERIC (N'?23.67') --0SELECT  CAST (N'?23.67' AS money) --23.67SELECT ISNUMERIC (N'£100.20') --1SELECT  CAST (N'£100.20' AS money) --100.20 Also the CAST function itself is quirky in that it cannot convert perfectly reasonable string-representations of integers into integersSELECT ISNUMERIC('200,000')       --1SELECT  CAST ('200,000' AS INT)   --0/*Msg 245, Level 16, State 1, Line 2Conversion failed when converting the varchar value '200,000' to data type int.*/  A more sensible question is 'Is this an integer or decimal number'. This cuts out a lot of the apparent quirkiness. We do this by the '+E0' trick. If we want to include floats in the check, we'll need to make it a bit more complicated. Here is a small test-rig. SELECT  PossibleNumber,         ISNUMERIC(CAST(PossibleNumber AS NVARCHAR(20)) + 'E+00') AS Hack,        ISNUMERIC (PossibleNumber + CASE WHEN PossibleNumber LIKE '%E%'                                          THEN '' ELSE 'E+00' END) AS Hackier,        ISNUMERIC(PossibleNumber) AS RawIsNumericFROM    (SELECT CAST(',' AS NVARCHAR(10)) AS PossibleNumber          UNION SELECT '£' UNION SELECT '.'         UNION SELECT '56' UNION SELECT '456.67890'         UNION SELECT '0E0' UNION SELECT '-'         UNION SELECT '-' UNION SELECT '.'         UNION  SELECT N'?' UNION SELECT N'¢'        UNION  SELECT N'?' UNION SELECT N'?34.56'         UNION SELECT '-345' UNION SELECT '3.332228E+09') AS examples Which gives the result ... PossibleNumber Hack Hackier RawIsNumeric-------------- ----------- ----------- ------------? 0 0 0- 0 0 1, 0 0 1. 0 0 1¢ 0 0 1£ 0 0 1? 0 0 0?34.56 0 0 00E0 0 1 13.332228E+09 0 1 1-345 1 1 1456.67890 1 1 156 1 1 1 I suspect that this is as far as you'll get before you abandon IsNumeric in favour of a regex. You can only get part of the way with the LIKE wildcards, because you cannot specify quantifiers. You'll need full-blown Regex strings like these ..[-+]?\b[0-9]+(\.[0-9]+)?\b #INT or REAL[-+]?\b[0-9]{1,3}\b #TINYINT[-+]?\b[0-9]{1,5}\b #SMALLINT.. but you'll get even these to fail to catch numbers out of range.So is IsNumeric() an out and out rogue function? Not really, I'd say, but then it would need a damned good lawyer.

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  • PHP PDO MySQL IN (?,?,?...

    - by nute
    I want to write a MySQL statement like: SELECT * FROM someTable WHERE someId IN (value1, value2, value3, ...) The trick here is that I do not know ahead of time how many values there will be in the IN(). Obviously I know I can generate the query on the go with string manipulations, however since this will run in a loop, I was wondering if I could do it with a PDO PreparedStatement. Something like: $query = $PDO->prepare('SELECT * FROM someTable WHERE someId IN (:idList)'); $query->bindValue(':idList', implode(',', $idArray)); Is that possible?

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  • My simple PHP is outputting wrong things.

    - by Sergio Tapia
    EDIT: I forgot to add semi colons. Now there is another problems. I'm getting a error: Parse error: syntax error, unexpected T_STRING in C:\xampp\htdocs\useraccess.php on line 12 It outputs: 0){ echo 'si'; } ?> When it should only output 'si' in the body. Here's the code: <html> <head> </head> <body> <? $user = mysql_real_escape_string($_GET["u"]) $pass = mysql_real_escape_string($_GET["p"]) $query = "SELECT * FROM usario WHERE username = '$user' AND password = '$pass'" mysql_connect(localhost, "sergio", "123"); @mysql_select_db("multas") or die( "Unable to select database"); $result=mysql_query($query); if(mysql_numrows($result) > 0){ echo 'si'; } ?> </body> </html>

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  • Problems with Date, preparedStatement, JDBC and PostgreSQL

    - by GuidoMB
    I Have to get a movie from a PostgreSQL database that matches a given title and release date. title is a character(75) and releaseDate is a date. I Have this code: String query = "SELECT * FROM \"Movie\" WHERE title = ? AND \"releaseDate\" = ?)"; Connection conn = connectionManager.getConnection(); PreparedStatement stmt = conn.prepareStatement(query); java.sql.Date date = new java.sql.Date(releaseDate.getTime()); stmt.setString(1, title); stmt.setDate(2, date); ResultSet result = stmt.executeQuery(); but it's not working because the releaseDate is not matching when it should. The query SELECT * FROM "Movie" WHERE title = A_MOVIE AND "releaseDate" = A_DATE works perfectly on a command shell using psql

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  • In LINQ-SQL, wrap the DataContext is an using statement - pros cons

    - by hIpPy
    Can someone pitch in their opinion about pros/cons between wrapping the DataContext in an using statement or not in LINQ-SQL in terms of factors as performance, memory usage, ease of coding, right thing to do etc. Update: In one particular application, I experienced that, without wrapping the DataContext in using block, the amount of memory usage kept on increasing as the live objects were not released for GC. As in, in below example, if I hold the reference to List of q object and access entities of q, I create an object graph that is not released for GC. DataContext with using using (DBDataContext db = new DBDataContext()) { var q = from x in db.Tables where x.Id == someId select x; return q.toList(); } DataContext without using and kept alive DBDataContext db = new DBDataContext() var q = from x in db.Tables where x.Id == someId select x; return q.toList(); Thanks.

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  • SQL: Need to SUM on results that meet a HAVING statement

    - by Wasauce
    I have a table where we record per user values like money_spent, money_spent_on_candy and the date. So the columns in this table (let's call it MoneyTable) would be: UserId Money_Spent Money_Spent_On_Candy Date My goal is to SUM the total amount of money_spent -- but only for those users where they have spent more than 10% of their total money spent for the date range on candy. What would that query be? I know how to select the Users that have this -- and then I can output the data and sum that by hand but I would like to do this in one single query. Here would be the query to pull the sum of Spend per user for only the users that have spent 10% of their money on candy. SELECT UserId, SUM(Money_Spent), SUM(Money_Spent_On_Candy) / SUM(Money_Spent) AS PercentCandySpend FROM MoneyTable WHERE DATE >= '2010-01-01' HAVING PercentCandySpend > 0.1;

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  • Count Distinct With IF in MySQL?

    - by user1600801
    I need to do a query with count distinct and IF, but the results always are 0. What I need to do, is count the different users from a table in different months, using IF. My individual query, for one month is this: SELECT COUNT(DISTINCT(idUsers)) AS num_usuarios FROM table01 WHERE date1='201207' But I need to get the results by different months in the same query. What I'm trying to do is this: SELECT IF(date1=(201207), count(distinct(idUsers)), 0) as user30, IF(fecha1=(201206), count(distinct(idUsers)), 0) as user60, IF(fecha1=(201205), count(distinct(idUsers)), 0) as user90, IF(fecha1=(201204), count(distinct(idUsers)), 0) as user120, IF(fecha1=(201203), count(distinct(idUsers)), 0) as user150 FROM table01 But the all the results are always 0.

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  • Problem with date_select when using :discard option. (Rails)

    - by MikeH
    I'm using a date_select with the option :discard_year => true If a user selects a date in the date select, and then he comes back and returns the select to the prompt values of Month and Day, Rails automatically sets the select values to January 1. I know this is the intended functionality if a month is selected and a day is left blank, but that's not the case here. In my example, the user sets both the month and day back to the prompt. By Rails forcing January 1, I'm getting bad results. I've tried every parameter available in the api. :default = nil, :include_blank = true. None of those change the behavior I'm describing. I've isolated the root of the problem, which is this: Because I'm discarding the :year parameter, when the user tries to return the month and day to the prompt values, Rails doesn't see an empty prompt select. It perhaps sees a year selected with empty month and day, which it then sets to January 1. This is the case because the :discard_year parameter does in fact set a date in the database, it just removes it from the view. How can I code around this problem?

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  • Yahoo Query Language Problem

    - by Damiano
    Hello everybody! Today, I've started with Yahoo Query Language. I would use it to retrive stocks details, so I'm talking about Yahoo Finance. I think there is a bug on this language. This is my query: select * from yahoo.finance.quoteslist where symbol='@^GSPC' I ALWAYS get 51 results! it's impossible, take a look at: http://it.finance.yahoo.com/q/cp?s=^GSPC There are 500 results! I also tried some paging parameters. select * from yahoo.finance.quoteslist(50,30) where symbol='@^GSPC' (to get from 50 to 80) select * from yahoo.finance.quoteslist(100) where symbol='@^GSPC' (to get the first 100 results) select * from yahoo.finance.quoteslist where symbol='@^GSPC' limit 30 offset 50 but ALWAYS the last stock is: <quote symbol="BBY"> <Symbol>BBY</Symbol> <LastTradePriceOnly>41.03</LastTradePriceOnly> <LastTradeDate>5/7/2010</LastTradeDate> <LastTradeTime>4:00pm</LastTradeTime> <Change>-0.48</Change> <Open>41.35</Open> <DaysHigh>42.35</DaysHigh> <DaysLow>39.60</DaysLow> <Volume>14129531</Volume> </quote> Why do I have this kind of problem? Thank you so much for your support! (P.S. I've tested it on Yahoo YQL console)

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  • MySQLi Prepared Statement Query Issue

    - by Benjamin Flak
    I'm relatively new to MySQLi prepared statements, and running into an error. Take this code: $user = 'admin'; $pass = 'admin'; if ($stmt = $mysqli->query("SELECT * FROM members WHERE username='$user' AND password='$pass'")) { echo $stmt->num_rows; } This will display "1", as it should. This next piece of code though, returns "0": $user = 'admin'; $pass = 'admin'; if ($stmt = $mysqli->prepare("SELECT * FROM members WHERE username=? AND password=?")) { $stmt->bind_param("ss", $user, $pass); $stmt->execute(); echo $stmt->num_rows; } Any ideas why?

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  • YAHOO QUERY LANGUAGE BUG!

    - by Damiano
    Hello everybody! Today, I've started with Yahoo Query Language. I would use it to retrive stocks details, so I'm talking about Yahoo Finance. I think there is a bug on this language. This is my query: select * from yahoo.finance.quoteslist where symbol='@^GSPC' I ALWAYS get 51 results! it's impossible, take a look at: http://it.finance.yahoo.com/q/cp?s=^GSPC There are 500 results! I also tried some paging parameters. select * from yahoo.finance.quoteslist(50,30) where symbol='@^GSPC' (to get from 50 to 80) select * from yahoo.finance.quoteslist(100) where symbol='@^GSPC' (to get the first 100 results) select * from yahoo.finance.quoteslist where symbol='@^GSPC' limit 30 offset 50 but ALWAYS the last stock is: <quote symbol="BBY"> <Symbol>BBY</Symbol> <LastTradePriceOnly>41.03</LastTradePriceOnly> <LastTradeDate>5/7/2010</LastTradeDate> <LastTradeTime>4:00pm</LastTradeTime> <Change>-0.48</Change> <Open>41.35</Open> <DaysHigh>42.35</DaysHigh> <DaysLow>39.60</DaysLow> <Volume>14129531</Volume> </quote> Why do I have this kind of problem? Thank you so much for your support! (P.S. I've tested it on Yahoo YQL console)

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  • Jquery .Show() .Hide() not working as expected

    - by fizgig07
    I'm trying to use the show and hide to display a different set of select options when a certain report type is selected. I have a couple problems with this: The .show .hide only execute properly if I pass params, slow fast, in the the first result of my conditional statement. If I take out the params or pass params in both results, only one select shows and it never changes.. here's is the code that currently kind of works. if ($('#ReportType').val() == 'PbuseExport') { $('#PbuseServices').show('fast'); $('#ReportServiceDropdown').hide('fast'); } else { $('#PbuseServices').hide(); $('#ReportServiceDropdown').show(); } After i've used this control I am taken to a differnt page. When I use the control again, it reatins the original search values and repopulates the control. Then again I only want to show one select option if a certain report is chosen.. This works correctly if the report type I originally searched on is not the "PbuseExport". If I searched on the report type "PbuseExport", then both selects show on the screen, and only until I change the report type does it show only one select. I know this probably isn't very clear.. Here is the code that handles the change event on the report type drop down. var serviceValue = $("#ReportType").val(); switch (serviceValue) { case 'PbuseExport': $('#PbuseServices').show('fast'); $('#ReportServiceDropdown').hide('fast'); default: $('#PbuseServices').hide(); $('#ReportServiceDropdown').show(); break; }

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