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freddixon

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  1. Hi, The easiest way is to create an SQL script file and then run it in one of the mysql clients such as mysqladministrator. The following example is very simple but should give you the idea. This part creates the database table DROP TABLE IF EXISTS `database_name`.`table_name`; CREATE TABLE `table_name`.`Types` ( `Type_ID` INT(10) NOT NULL, `Type_Name` VARCHAR(25) NULL, PRIMARY KEY (`Type_ID`); ) This second part inserts the data into the database table, it is this bit that you have created from excel. INSERT INTO `database_name`.`table_name`(`Type_ID`, `Type_Name`) VALUES (1, "Dogs"), (2, "Cats"), (3, "Pigs"), (4, "Donkeys"); Hope this helps, Fred
  2. [!--quoteo(post=352442:date=Mar 7 2006, 10:31 AM:name=Whaih)--][div class=\'quotetop\']QUOTE(Whaih @ Mar 7 2006, 10:31 AM) [snapback]352442[/snapback][/div][div class=\'quotemain\'][!--quotec--] Hi there, I'm new to php and i have trouble exporting data from Excel to MySql. There are available software application in the market that is able to do this but i would like to try it by coding in PHP. I know that i need to export the data to a csv file first before i put it in MySql. However i absolutely no idea where and how to start. Are there any kind soul to guide me along or are there any available sample coding for me to reference? [/quote]
  3. [!--quoteo(post=352119:date=Mar 6 2006, 05:46 PM:name=kenrbnsn)--][div class=\'quotetop\']QUOTE(kenrbnsn @ Mar 6 2006, 05:46 PM) [snapback]352119[/snapback][/div][div class=\'quotemain\'][!--quotec--] Please post your code. This sounds more like something that could be accomplished using Javascript. Ken [/quote]
  4. Hi, Quick question for anone who knows more about PHP / Javascript than me. I have a line of code as below in one of my PHP files: echo "<select name='left' id='left' size='1' onChange='get-flag_middle(left.value)'>"; Can I add a second / third action to this line so that I have got onChange='xxxx' onChange='yyy' etc? If not how can I do it? Thanks a lot.
  5. Hi, I have got a site that needs to have three dynamic menus and several text boxes to be to visble throughoput the select process. The application is a database select and display system with all the data residing in mySql tables. I have a main page with three functions defined on it. These functions call PHP files, action the results using "onchange" and display the results on the main page using <span ids>. This process works fine and recalls data from the database etc fine. The problem is that anyone visiting this page would just see the opening menu displayed, the rest of the page is blank. As each menu is selected the next menu pops up and so on. When all selections have been made the screen displays as it should and you can backtrack and change any menu settingas you want and it does dynamically change everything as required. The problem is anyone using this WEB page would see a page building in front of thier eyes as they make selections. I need the opening page to display the page at it would look when all selections are made but with blank boxes with associated captions etc. I have trie to put holding textboxes on the page but these seem to be over written when one of the functions are selected. Any help or ideas would be most welcome.
  6. Hi, I'm relatively new to HTML and PHP, but have been programming PLSQL for some years, and I am aware that this question might be simple, but if you don't know the answer... I am trying to produce a site that has two <select type> lists on one of the pages that are populated from the database. The first list works fine but to get the second one to work I need to send the result value from the first select to the PHP script that controls the second. I don't know if this can be done easily as I realise that PHP is server based and HTML / Java is local. Example of first select follows: <select name="first" id="first" onChange="price_up(flag_form.first.value)"> <?php do { ?> <option value="<?php echo $row_rs_first_list['Category_ID']?>"><?php echo $row_rs_first_list['Category_Type']?></option> <?php } while ($row_rs_first_list = mysql_fetch_assoc($rs_first_list)); $rows = mysql_num_rows($rs_first_list); if($rows > 0) { mysql_data_seek($rs_first_list, 0); $row_rs_first_list = mysql_fetch_assoc($rs_first_list); } ?> </select> The function "price_up" just displays the value of the select. Anybody got any advice? Thanks
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