Selecting 4 Random MySQL DB Entries?
I have a table set up in my MySQL database called 'features'. This table has 10 entries, but I only want to select 4 random rows and display them out in tables.
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I have implemented a links bar on the side of a site, which I want to popuplate with links from a database randomly, but I feel the php workaround I used is inefficient. Is there a way to randomize the result in a mysql SELECT query? "SELECT * FROM links ORDER random" I know the code above is probably way off, but it illustrates what I want I think.
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i have a sql database with three fields: id, text, category how would i pull a random set of entries without repeated instances and print them in php? example: user has chosen two categories: art, design. user has also chosen to draw 25 entries. I need a script that will return 25 random entries that are in either category art or design without returning the same entry twice.
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I am trying to update some entries in a mysql database that will append the filename to the end of a thumbnail directory entry. So for example /images/thumb/ would become /images/thumb/12345.jpg. I am stumped because the below code gives no php or mysql errors but the entries are not updated in the mysql db! Code:
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I code some site and because Im a freak I made my own session-handling. When a user open up my site it will check if there is a ssid in the url if not generate one. this will be done by a function generate_ssid() and works fine. next step is to register the ssid in the table session this will be done by the following code: ****************************** function ssid_register($ssid,$login,$uid,$admin){ $connect = mysql_connect( $GLOBALS["db_host"], $GLOBALS["db_user"], $GLOBALS["db_password"]) or die($GLOBALS["db_error"]); mysql_select_db($GLOBALS["db_name"],$connect) or die($GLOBALS["db_error"].'<br>ErrorCode s.001-con'); $result = mysql_query("INSERT INTO session SET ssid='$season',ip='$rip'",$connect); echo ($rs); return; } ******************************** I redesigned the INSERT INTO part into all possible syntax but everytime I look into the db, there are two entries with two DIFFERENT ssids. The only way it works correct is to dont use a variable on the first place. Means if I use ssid='godhelpmetofindaworkingsolution' it works great but is not realy what I want. As told above a "INSERT INTO session (ssid) VALUES ($ssid)" also has the effect of two new db-entries.
Populate MySQL With Multiple Text Box Entries
I have about 300 rows of mysql data which I have chosen to diplay with a text box at the end of each row, that is available for adding comments. I would like to be able to add comments to entries and then have an update scripte that would take the text box, named "MC" and place the data in its row identified by a hidden field which would corrrespond to the Primary Key of the MySQL Database. I've done something similar using check box arrays, ie. Checkbox name="MC[]" My question is whether I can write a script that would test for a value in the corresponding text box, and update the MySQL data accordingly. I have also provided in my display script that if the MC column is already populated in MySQL it will not display in the form.
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I'm learning MySQL and how to use it through PHP. So I connect to the database without problems, but I can't select it, PHP Code:
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The topic is related to MySQL database. Suppose a table "address" contains the following records ------------------------------------------------------- | name | address | phone | ------------------------------------------------------- | mr x | 8th lane | 124364 | | mr x | 6th lane | 435783 | | mrs x | 6th lane | 435783 | | mr x | 8th lane | 124364 | ------------------------------------------------------- Quote:
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