The blank table column comes from a blank MySQL database field...(I was initially going to blob images from there to show status change). If you know of a way to eliminate the database field and still show the status of the values in a web table, please point me in the right direction. I am one of those people who have to look up everything for JavaScript. (My main scripting beast was/is ASP...this PHP is a new project for a company on a Solaris platform!)
-----Original Message----- From: Tony S. Wu [mailto:[EMAIL PROTECTED]] Sent: Monday, September 30, 2002 8:04 PM To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Changing colors in table? I don't see where you put your blank table column. I am guessing it might be somewhere within this for loop: for ($i=0; $i < $num_tables; $i++) { echo("<center><h3> $arr_tablenames[$i] </h3></center>"); echo('<table align="center" border="1"><tr>'); $result = mysql_db_query($dbname, "select * from $arr_tablenames[$i]", $id_link); for ($ii=0; $ii < $arr_num_fields[$i]; $ii++) { echo("<th>"); echo $hash_field_names[$i][$ii]; echo("</th>"); } echo("</tr><tr>"); $number_of_rows = @mysql_num_rows($result); for ($iii = 0; $iii < $number_of_rows; $iii++) { $record = @mysql_fetch_row($result); for ($ii=0; $ii < $arr_num_fields[$i]; $ii++) { echo("<td>"); echo $record[$ii]; echo("</td>"); } echo("</tr>"); } echo("</table>"); } If that's the case, all you need to do is, let's pretend $table_value holds the table value you need to determine the background color: echo "<td"; if ($table_value < 5) { echo " bgcolor = \"#999999\""; // whatever color you want } else if ($table_value >= 5 && $table_value < 10) { echo " bgcolor = \"#888888\""; // another color } else // if no value matched { echo ">"; // close <td> tag } Tony S. Wu [EMAIL PROTECTED] "Nope, this world ain't perfect. But at least I know it's not because of me." Blue Tiger at [EMAIL PROTECTED] wrote: > Can I force it to display the data sorted by the primary key? > > Here's the code: > // store table names in an array > $arr_tablenames[] = ''; > > // store number of fields per table(index 0,1,2..) in an array > $arr_num_fields[] = ''; > for ($i=0; $i < $num_tables; $i++) { > $arr_tablenames[$i] = mysql_tablename($tables, $i); > $arr_num_fields[$i] = mysql_num_fields(mysql_db_query($dbname, > "select * from $arr_tablenames[$i]", $id_link)); > } > > // store field names in a multidimensional array: > // [i] == table number, [ii] == field number for that table > for ($i=0; $i < $num_tables; $i++) { > for ($ii=0; $ii < $arr_num_fields[$i]; $ii++) { > $result = mysql_db_query($dbname, "select * from > $arr_tablenames[$i]", $id_link); > $hash_field_names[$i][$ii] = mysql_field_name($result, $ii); > > } > } > > for ($i=0; $i < $num_tables; $i++) { > echo("<center><h3> $arr_tablenames[$i] </h3></center>"); > echo('<table align="center" border="1"><tr>'); > $result = mysql_db_query($dbname, "select * from > $arr_tablenames[$i]", $id_link); > for ($ii=0; $ii < $arr_num_fields[$i]; $ii++) { > echo("<th>"); > echo $hash_field_names[$i][$ii]; > echo("</th>"); > } > echo("</tr><tr>"); > $number_of_rows = @mysql_num_rows($result); > for ($iii = 0; $iii < $number_of_rows; $iii++) { > $record = @mysql_fetch_row($result); > for ($ii=0; $ii < $arr_num_fields[$i]; $ii++) { > echo("<td>"); > echo $record[$ii]; > echo("</td>"); > } > echo("</tr>"); > } > echo("</table>"); > } > > Is that too much to sort through??? > > -----Original Message----- > From: Tony S. Wu [mailto:[EMAIL PROTECTED]] > Sent: Monday, September 30, 2002 7:31 PM > To: [EMAIL PROTECTED] > Subject: Re: [PHP-DB] Changing colors in table? > > If the table's background color will be set when the page is first > loaded, > you can just use the bgcolor attribute of <td> tag. > If you want to dynamically change the background color, you'll need to > combine your page with javascript. > > Tony S. Wu > [EMAIL PROTECTED] > > "Nope, this world ain't perfect. But at least I know it's not because of > me." > > Blue Tiger at [EMAIL PROTECTED] wrote: > >> Okay..... >> Disclaimer: I am very much the newbie, so please bear with me. >> >> I have a piece of code that displays an array from a MySQL table. One >> field in the table is blank (and is displayed as an empty column of >> cells) at the moment. I want to have the blank cells display an image, >> value, or change its background color based on whether the numerical >> values in the previous 4 fields meet certain criteria or not. Make >> sense? Maybe? >> >> An example of what I want to show: >> If table 2 has a value that is less than 5, then the last cell > displayed >> changes color to reflect that. >> If table 2 has a value that is between 5 and 10, then the last cell >> displayed changes color to reflect that as well. >> >> Please help if you can....many thanks in advance. (I can show my code >> thus far as well if that helps..) >> >> Thanks, >> Ryan -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
