ramyakumar

Hello Everyone, I have triple drop down menu. I want to display the contents of the table based on the third menu selection. The code is in the link http://php.pastebin.com/hgazULxi I know I need to include a onchange function to <select name="genus"> but as you can see I have a <div> already for it. I am confused how to create the function to display the table and also the how to include another div tag. Could some one help please??? Thank you Ramya

Hello everyone, I am trying triple drop down menu first and second menus are working but third one is not working and also the data is not being displayed on the webpage even though I have used "echo" command.. I am unsure what is the problem and how to correct it. So I am attaching my code in this email, could you please look into it and help me with it. I have 1st drop menu for State : Tennessee Alabama Georgia 2nd Drop down list for County: Anderson Bredford Benton 3rd Drop down list for Genus Acer Aristida Eg: Tennnesse -> Anderson, Bredford, Benton and Anderson -> Acer, Aristida and when I select Tennesse->Anderson->Acer it has to display all the information of the table based on these selected values. Could you please help me with this. This is my link for the complete code http://pastebin.com/vvLrpCcr The execution link is http://sozog.utc.edu/~tdv131/MYSQL/genus1c.htm

Hello, I am trying insert the values of csv file into a table. I want to concatenate the values of some columns of csv into one String column of table by delimiting them my commas. $im="INSERT into TB1 (col1, col2 ) values ('$a[2]','$a[3]')"; $b = ""; $b=".'$a[6]'."||".'$a[7]'."||".'$a[8]'."||".'$a[9]'."; $im1=$im+"UPDATE TB1 set latlng=.'$aline'."; mysql_query($im1) OR die(mysql_error()); I need to enter the value of $a[6] till $a[9] into a string column with delimiters and insert into the column col3 of TB1 with value as $a[4]. I am stuck up here could anyone help me with this issue.