rjlowe
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You've got too many close braces "}" I always label my close braces, otherwise I end up with errors like you have. See my //comments in your code snippett [!--quoteo(post=353423:date=Mar 9 2006, 03:55 PM:name=PcGamerz13)--][div class=\'quotetop\']QUOTE(PcGamerz13 @ Mar 9 2006, 03:55 PM) [snapback]353423[/snapback][/div][div class=\'quotemain\'][!--quotec--] 1. I get this error Parse error: parse error, unexpected '}' in Not Telling on line 142 2. I got four field in table mail: Sent_By,Sent_To,Message,Date. I need this to show From: Sent_By, Message: Message,Date: Date. Please help fix this. [code]<? $sql = "SELECT * FROM `mail` WHERE Sent_To='$Username'"; $result = mysql_query($sql); if (mysql_num_rows($result) > 0) { // Opens the IF Clause while ($row = mysql_fetch_row($result)) { //Opens The While Loop echo '<div>'; echo '<p>----<br /><font face="verdana" size="2" color="red">From: </font>' . $row[1]; echo '<br /><font face="verdana" size="2" color="red">Message: </font>' . $row[2]; echo '<br /><font face="verdana" size="2" color="red">Date: </font>' . $row[3]; echo '<br />----</p>' . '</div>'; } // CLOSES the While Loop }else{ // Continue the IF Clause echo ("<font face='verdana' size='2'>You have no messages</font>"); } // CLOSES the IF Clause } // Dunno what this is closing? } // Dunno what this is closing? mysql_free_result($result); ?>[/code] [/quote]
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Can't you just do a: INSERT ....... ON DUPLICATE KEY UPDATE ....... ??? check out [a href=\"http://dev.mysql.com/doc/refman/5.0/en/insert-on-duplicate.html\" target=\"_blank\"]http://dev.mysql.com/doc/refman/5.0/en/ins...-duplicate.html[/a] for more syntax info on how that might work for you.. --RobBob [!--quoteo(post=353407:date=Mar 9 2006, 02:57 PM:name=mouli)--][div class=\'quotetop\']QUOTE(mouli @ Mar 9 2006, 02:57 PM) [snapback]353407[/snapback][/div][div class=\'quotemain\'][!--quotec--] Nearly true... I want to upload images to a database with a code number for the record. When I upload an image I specify the code number of the record I want the image added to. For instance I want to upload 45.jpg to field 'webpic1' in record with 'id' = L203 If 'id'= L203 exists then I want it to simply update 'webpic1' with 45.jpg If 'id'=L203 doesn't exist then I want it to insert a new record with 'id'=L203 with 'webpic1'=45.jpg I can do it by querying the database for the existance of 'id'=L203, then using an IF statement to decide whether to insert or update but I was thinking there must be a quicker, slicker way to do it. Any ideas much appreciated. mouli [/quote]
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SWEET! That worked like a charm! Thx! :D [!--quoteo(post=350021:date=Feb 27 2006, 03:30 PM:name=AndyB)--][div class=\'quotetop\']QUOTE(AndyB @ Feb 27 2006, 03:30 PM) [snapback]350021[/snapback][/div][div class=\'quotemain\'][!--quotec--] You can 'trick' browsers into a fresh download if they think it's a different image. Easy way to do that is to give the browser an image name like sameoldimage.jpg?xyz=randomnumber. [/quote]
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OK here's the deal, I have a script that I use to upload images and then provide the user with <form> buttons to rotate the image if needed. It works great in FireFox and Opera, but IE refuses to reload the image when the <form> button is pressed. If I use the "Refresh" button, however, then the image updates (and rotates one more iteration - ugh). Is there a way to FORCE IE to reload an image instead of using it's cache? I've tried: <HEAD> <META HTTP-EQUIV="Pragma" CONTENT="no-cache"> <META HTTP-EQUIV="Expires" CONTENT="-1"> </HEAD> and that didn't do anything... I've also tried various tags for the img itself such as just plain HTML, or throwing some php in ( like echo "<img src=".$src.">"; ) and that didn't do anything either.... My main userbase is going to be IE, so just using FF or O isn't really an option either... Another (bad) option would be to change the name of the image in order to force/trick IE into thinking it's a new image file... I would really prefer NOT to do this. Thanks y'all :P