Comments on: Creating ajaxy Chained Selects with jQuery http://www.script-tutorials.com/creating-ajaxy-chained-selects-with-jquery/ Tutorials for Web Developers Wed, 15 May 2013 09:10:19 +0000 hourly 1 By: Peter http://www.script-tutorials.com/creating-ajaxy-chained-selects-with-jquery/comment-page-1/#comment-63846 Peter Fri, 08 Feb 2013 01:29:49 +0000 http://www.script-tutorials.com/?p=685#comment-63846 Hey, great code!! after a good week or more of searching, code that finally works as advertised... I also had problems with the second dropdown working when I realized something. I was copy and pasting the database table info query from the page, and not using the included sql.sql file. As stated at the top, it is limited info. Use the sql.sql file to get larger and more functional tables. Before I did that, it would fill in the first dropdown, but none of the others. well of course. No data in the tables. That solved my problems. Again, thanks a lot, you have helped me over a very large hurdle in my coding. Hey, great code!! after a good week or more of searching, code that finally works as advertised… I also had problems with the second dropdown working when I realized something. I was copy and pasting the database table info query from the page, and not using the included sql.sql file. As stated at the top, it is limited info. Use the sql.sql file to get larger and more functional tables. Before I did that, it would fill in the first dropdown, but none of the others. well of course. No data in the tables. That solved my problems. Again, thanks a lot, you have helped me over a very large hurdle in my coding.

]]> By: admin http://www.script-tutorials.com/creating-ajaxy-chained-selects-with-jquery/comment-page-1/#comment-56673 admin Sun, 20 Jan 2013 14:25:00 +0000 http://www.script-tutorials.com/?p=685#comment-56673 Hi Sasa, In this case you can easily add a trigger to 'change' event for our first element: $('select[name=country]').trigger("change"); Hi Sasa,
In this case you can easily add a trigger to ‘change’ event for our first element:
$(‘select[name=country]‘).trigger(“change”);

]]> By: admin http://www.script-tutorials.com/creating-ajaxy-chained-selects-with-jquery/comment-page-1/#comment-56671 admin Sun, 20 Jan 2013 14:12:32 +0000 http://www.script-tutorials.com/?p=685#comment-56671 Hi Prince, And, what's a problem? You always can keep add your info in a single SQL table. Also you can have multiple chained fields (selectors). Once you have selected some value from one level - you should perform a necessary search to find suitable values for linked field. Hi Prince,
And, what’s a problem? You always can keep add your info in a single SQL table. Also you can have multiple chained fields (selectors). Once you have selected some value from one level – you should perform a necessary search to find suitable values for linked field.

]]> By: Sasa http://www.script-tutorials.com/creating-ajaxy-chained-selects-with-jquery/comment-page-1/#comment-55050 Sasa Wed, 16 Jan 2013 21:43:02 +0000 http://www.script-tutorials.com/?p=685#comment-55050 In your live demo, first "Country" is "Argentina" without states in "State". Try to add states for Argentina. I added states to Argentina, but they don't showed under "State" on first access to the page. After that, if I select other country and get back to Argentina I can select states under "State". Try to add states for Argentina in your live demo and see what happens. Maybe I'm wrong. In your live demo, first “Country” is “Argentina” without states in “State”. Try to add states for Argentina. I added states to Argentina, but they don’t showed under “State” on first access to the page. After that, if I select other country and get back to Argentina I can select states under “State”. Try to add states for Argentina in your live demo and see what happens. Maybe I’m wrong.

]]> By: prince http://www.script-tutorials.com/creating-ajaxy-chained-selects-with-jquery/comment-page-1/#comment-44993 prince Mon, 17 Dec 2012 10:24:37 +0000 http://www.script-tutorials.com/?p=685#comment-44993 Also dont want to create more then 1 table . every thing should be in one table. Also dont want to create more then 1 table . every thing should be in one table.

]]> By: prince http://www.script-tutorials.com/creating-ajaxy-chained-selects-with-jquery/comment-page-1/#comment-44992 prince Mon, 17 Dec 2012 10:23:00 +0000 http://www.script-tutorials.com/?p=685#comment-44992 thnx for this tutorial but now I want to create Chained Selects with 6 or 7 select menu. have any idea? thnx for this tutorial

but now I want to create Chained Selects with 6 or 7 select menu.

have any idea?

]]> By: admin http://www.script-tutorials.com/creating-ajaxy-chained-selects-with-jquery/comment-page-1/#comment-36905 admin Wed, 21 Nov 2012 17:40:11 +0000 http://www.script-tutorials.com/?p=685#comment-36905 Hi Sasa, Can you please give us an example? What do you mean - default country? Hi Sasa, Can you please give us an example? What do you mean – default country?

]]> By: Sasa http://www.script-tutorials.com/creating-ajaxy-chained-selects-with-jquery/comment-page-1/#comment-23009 Sasa Sun, 14 Oct 2012 12:51:17 +0000 http://www.script-tutorials.com/?p=685#comment-23009 Hello, cannot select STATE for default COUNTRY. I think that is necessary to fulfill array of states with default country. Hello,
cannot select STATE for default COUNTRY. I think that is necessary to fulfill array of states with default country.

]]> By: admin http://www.script-tutorials.com/creating-ajaxy-chained-selects-with-jquery/comment-page-1/#comment-17959 admin Sun, 20 May 2012 13:46:23 +0000 http://www.script-tutorials.com/?p=685#comment-17959 Hello Allen, Yes, I mentioned json_encode. Basically, JSON is PHP library, and as of PHP 5.2.0, the JSON extension is bundled and compiled into PHP by default. But if your PHP version is below 5.2.0, you can install that JSON separately. As I know, JSON is a part of PECL. Try to read about installation of PECL extensions. Hello Allen,
Yes, I mentioned json_encode. Basically, JSON is PHP library, and as of PHP 5.2.0, the JSON extension is bundled and compiled into PHP by default. But if your PHP version is below 5.2.0, you can install that JSON separately. As I know, JSON is a part of PECL. Try to read about installation of PECL extensions.

]]> By: allen http://www.script-tutorials.com/creating-ajaxy-chained-selects-with-jquery/comment-page-1/#comment-17919 allen Thu, 17 May 2012 13:35:22 +0000 http://www.script-tutorials.com/?p=685#comment-17919 Hi, I'm new to coding, and I would really like to get this to work for a form I'm creating. I'm have issues with the second drop down. You said it might have to deal with the json_encode function. How do i fix that? Hi, I’m new to coding, and I would really like to get this to work for a form I’m creating. I’m have issues with the second drop down. You said it might have to deal with the json_encode function. How do i fix that?

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