rumon007
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need help to uninstall PHPTraid
rumon007 replied to rumon007's topic in PHP Installation and Configuration
:(sorry boss, no such file found for PHP but I have uninstalled Apache which had an installation file and please tell me how to uninstall mysql. -
I installed PHPTraid in my PC (XP) and now I want to uninstall it. But problem is it is not listed in Add or Remove Programs. How should I uninstall it from my PC
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I have a list box named select_country showing a list of countries. All the country names come from a mysql database table named call_rate. I also created a text field and a button in my form. I have written the following code to show all country names in the list box: CODE: $conn = mysql_connect("localhost", "Administrator", 7216708) or die(mysql_error()); mysql_select_db("test",$conn) or die(mysql_error()); $get_country="select id,country_name,rate from call_rate"; $get_country_result= mysql_query($get_country) or die(mysql_error()); <table width=50% hight=90% cellpadding=10 cellspacing=0 border=1 align=center> <tr> <td> <select name=\"select_country\">"; while($country_result= mysql_fetch_array($get_country_result)) { $country_id=$country_result[id]; $country_name=$country_result[country_name]; $call_rate=$country_result[rate]; $display_block.="<option value=\"$country_id\">$country_name</option>"; } $display_block.= "</select></td>"; $display_block.= " <td><input type = \"text\" name=\"edit_rate\" size=10></td> <td><input type = \"submit\" name = \"submit\" value=\"Update\"></td> <input type = \"hidden\" name = \"op\" value = \"edit\"> </tr> </table>"; And finally I printed the $display_block string. What I am trying to do is whenever I will select a country name from the list the associated call rate will be displayed in the text field. (‘rate’- field name in database). And when after any change in rate when I will click the button the new rate will replace the old rate in the database. Please help me in writing the code for doing this!!!
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I have a list box named select_country showing a list of countries. All the country names come from a mysql database table named call_rate. I also created a text field and a button in my form. I have written the following code to show all country names in the list box: CODE: <table width=50% hight=90% cellpadding=10 cellspacing=0 border=1 align=center> <tr> <td> <select name=\"select_country\">"; while($country_result= mysql_fetch_array($get_country_result)) { $country_id=$country_result[id]; $country_name=$country_result[country_name]; $call_rate=$country_result[rate]; $display_block.="<option value=\"$country_id\">$country_name</option>"; } $display_block.= "</select></td>"; $display_block.= " <td><input type = \"text\" name=\"edit_rate\" size=10></td> <td><input type = \"submit\" name = \"submit\" value=\"Update\"></td> <input type = \"hidden\" name = \"op\" value = \"edit\"> </tr> </table>"; And finally I printed the $display_block string. What I am trying to do is whenever I will select a country name from the list the associated call rate will be displayed in the text field. (‘rate’- field name in database). And when after any change in rate when I will click the button the new rate will replace the old rate in the database. Please help me in writing the code for doing this!!!
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both variables get all data from a database. for example - $id $country_name --------------------------- 1 USA 2 UK 3 India 4 Bangladesh
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I have created a list box writing a code as follows: <select name=\"select_to_edit\"> <option selected>- -Select a Country- -</option> </select> And I have to variables named $id and $country_name. Now I want to show the values of the $country_name variable in the List Box. Please someone help me to write the code.
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I have installed APACHE-PHP-MYSQL in my PC (localhost) successfully. I have wrote a script which is as follows: <?php setcookie("vegetable","artichoke",time()+3600,"/","localhost",0); ?> <html> <head><title></title></head> <body> <?php if (isset($_COOKIE[vegetable])) { echo "HELLO RUMON YOUR CHOSEN VEGETABLE IS $_COOKIE[artichoke]"; } else { echo "THIS MAY BE YOUR FIRST VISIT"; } ?> </body> </html> But whenever I run my script "THIS MAY BE YOUR FIRST VISIT" is shown. Please someone tell me what may be the problem!