You want to get the rank in MySQL. Here is one approach: SELECT id, name, @rn := @rn+1 as `rank` from `users` cross join (select @rn := 0) const ORDER BY points desc 1 solved How to get a ranking result for special record in mysql [closed]
You are misusing the case expression. There are two forms. The form you want is: (CASE WHEN userName IS NULL THEN ‘was null’ WHEN userName IS NOT NULL THEN ‘was not null’ END) AS caseExpressionTest Note: There is no userName after the CASE. This checks each condition stopping at the first. MySQL interprets booleans as … Read more
There are a few problems with your script. First off, you start by using PDO to connect to the database, then you use mysql_* functions (which are deprecated, stick to PDO !!!). Plus, you are not properly escaping your data, and your code is potentially vulnerable to SQL injection. Secondly, the query you are using … Read more
suggest that if your above SQL statement can query the result. maybe you can just skip the “order by” and simply just select * from (/* your SQL statement without “order by”*/) a order by a.message_id desc sorry that as it is difficult to see your screen capture and help you to resolve the SQL … Read more
select distinct Name as `Students Name` from Students where Age_Start between 15 and 27 SQL Fiddle demo 2 solved Select rows with in particular range sql query [closed]
select query will be something like below for selecting particular field $sql = “SELECT `coloumname1`,`coloumname2` from `tablename` where `someid`=’matchingvalue'”; for selecting all the field sql query will be like below $sql = “SELECT * from `tablename` where `someid`=’matchingvalue'”; hope you understand this, so from next time please google first and than come here if not … Read more
The actual problem I found is that windows 8.1 is not allowing some programs to change c:/programs.. folders. I re-installed on my D: drive by creating a folder say D:/PHP/xampp and its started working fine solved phpMyAdmin error: I am seeing errors in user accounts
Try – SELECT id, SUM( DISTINCT ss ) AS “Sum (ss)” FROM tblData GROUP BY id; Note : Without GROUP BY SUM() will try to add up all values of ss, even with the DISTINCT qualifier. If you use GROUP BY but not DISTINCT then SUM() will add up all the values of ss for … Read more
In this: UPDATE MATERIAL_MASTER SET MST_NAME=’XXX’ MAT_DESC=’YYY’ MAT_TYPE=’Raw Material’ MAT_GRP=’H’ UOM=’kg’ CURRENCY=’inr’ ENTITY_ASSEMBLED=’A’ where idMATERIAL_MASTER=3; You’re missing commas between fields in SET zone. The correct query is: UPDATE MATERIAL_MASTER SET MST_NAME=’XXX’, MAT_DESC=’YYY’, MAT_TYPE=’Raw Material’, MAT_GRP=’H’, UOM=’kg’, CURRENCY=’inr’, ENTITY_ASSEMBLED=’A’ where idMATERIAL_MASTER=3; 0 solved MySQL Error 1064 (42000) when running UPDATE query [closed]
I’ll give you a short example. Right now, you’r code will give you 1 option <select name =”Employee Name” style=”width: 160px” > <option value =””>Please select …</option></select> Let’s take an array like: $array = array(‘0’ => ‘test’, ‘1’ => ‘test1’); To populate your array as options, you can simply do <select> <?php foreach ($array as … Read more
Do you know what means “soft delete”? It means that every “delete” operation will be converted to SQL that only set some deleted flag to true. And if you will insert another row with the same value of unique field that have soft deleted row you will get this message. You have two ways to … Read more
i think at first you should read some of the basics in SQL syntax . As for what you asked i the syntax is like this: Insert into yourtable1 (column1, column2 ) select column1 ,column2 from yourtable2 where yourcondition solved selecting the table and some columns in another table [closed]
SELECT CName, BankName,YEAR(dateofopen) FROM bank where YEAR(dateofopen) = 1968; solved Mysql YEAR function not supported? [closed]
Do you want this? update emp set salary = salary + 1500 where joining_date < ‘2014-04-14’ 1 solved Need MySQL query for updating table [closed]
A good approach might be to use count(*) and GROUP BY. So you get a count of the number of answer rows per user per list in the completed table then join to a subquery which gets you a total count of items in the list. Something like this: select user_id, c.list_id, count(user_id) as list_count, … Read more