
gathos
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Everything posted by gathos
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ok i like this place, thats why i'm posting this here. and the people seem nice. so i was wondering if anyone could direct me to a tutorial or some advice about putting startup messages on teh computer. what i want is after the post message (I think thats what there called) my computer sits for a while, with a blinking white line, then windows starts to boot up. well looking at this white line blinking reminds me of the matrix, when neo is sleeping and his computer says hello neo. and tells hiom to follow the white rabbit. i want a message to appear then wait like a couple of seconds then go into the windows boot. thanks guys,
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hi, so i'm getting an errorwhen i view the page. Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';' in /home/user/public_html/site/bobdisplay.php on line 45 now line 45 is the end of the while echo and all it is, is "; so it's not that line that is causing the problems. however, i know which line is causing the problem. it's the <a href='...'></a> as soon as i remove that line it all works, so what i need is another set of eyes to look over the line to see what i'm missing. while($row = mysql_fetch_assoc($result)){ echo "<table width= '800 ' border= '1 '>"; echo "<tr> <td>Publisher:</td> <td>".$row['publisher']."</td> </tr> <tr> <td>Comic Title:</td> <td>".$row['title']."</td> </tr> <tr> <td>Issue:</td> <td>".$row['issue']."</td> </tr> <tr> <td>Story arc:</td> <td>".$row['arc']."</td> </tr> <tr> <td>Release date:</td> <td>".$row['date']."</td> </tr> <tr> <td></td> <td> <a href='../Editcomic.php?publisher='".$row['publisher']."&title=".$row['title']."&issue=".$row['issue']."&arc=".$row['arc']."&date=".$row['date']."&price=".$row['price']."&id=".$row['id']"'>Edit comic</a> </td> </tr> "; } echo "</table>"; ?> </body> </html> oh and by the way, yes it is for a comic database. I SWEAR IT"S NOT FOR ME BUT MY BROTHER!!! (who i mooch off of to read the comics hehehe) thanks ahead of time. gathos
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please help followed tutorial but DB not working
gathos replied to gathos's topic in PHP Coding Help
thanks barand, it works now it was named table- i wasn't thinking. i appreciate your help. -
Ok here is my problem every time i load my page database.php I see Connection to the server was successful! Database was selected! Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/user/public_html/typh.ca/database.php on line 6 this is my code <?php mysql_connect("localhost", "****", "****") or exit(mysql_error()); echo "Connection to the server was successful!<br/>"; mysql_select_db("database") or die(mysql_error()); echo "Database was selected!<br/>"; $result = mysql_query("SELECT * FROM table"); while ($row = mysql_fetch_assoc($result)) { echo $row["id"]; echo $row["name"]; echo $row["city"]; echo $row["age"]; } ?> please help me, this database thing is driving me bonkers, i cannot find the error.
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ok thanks darkwater, i guess i'll just head back to the basics and start over, with more tutorials. as a side note, i applaude, how much you help the community, thank you very much.
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thanks for the wickedfast reply dark water. lol i did but i guess i though the as $design was unessecary. well i put it in, now i get Warning: Invalid argument supplied for foreach() in /folder/cookie_table.php on line 7 however below that i do get a thin line which i think is supposed to be my table, but nothing is in it.
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ok so i was wondering if someone could help me? this is what i want to do, click on a picture it add's it to a cookie then on another page load that image and others that were added from the cookie and display them all at once i made two pages cookie_setup.php looks like this <? //load values from previous page $design = $_GET['$id']; //load values currently in cookie $count = $_COOKIE['$count']; $designarray = $_COOKIE['designarray[$count]']; //create new array variable $count = $count + 1; //set new values $designarray['$count'] = $design; //apply new values to cookie setcookie('designarray[$count]','$design', time() +7200); setcookie('count', '$count', time() +7200); ?> and teh other page to display whats in the cookie <html> <body> <table width="100%" border="1"> <? $count = $_COOKIE['count']; $designarray = $_COOKIE['designarray[$count]']; foreach ($designarray[$count]) { echo " <tr> /n"; echo "<td>$design</td> /n"; echo "<td><img src='../images/cabinet_glass_lg_$design.jpg' height='100' alt='image'></td> /n"; echo "</tr> /n"; } ?> </table> </body> </html> when i try to open cookie_table i get teh error message: Parse error: syntax error, unexpected ')' in /folders*/cookie_table.php on line 7 *change actual file path for security reasons hope you guys can help me, thanks ahead of time.
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thanks sasa i appreciate your help it works now
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sorry your right it wasn't nice, i'm just under alot of stress so again i'm sorry.
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thanks for taking time out of your day to help here is the code - it neded sixteenth added so i did, but it doesn't work. <html> <head> <title>Price calculator</title> </head> <body> <?php $width = $_POST["width"]; $width = $_POST["width16"]; $height = $_POST["height"]; $height16 = $_POST["height16"]; $price = "0.25"; $totalheight = $_POST["totalheight"]; $totalwidth = $_POST["totalwidth"]; $total = $_POST["total"]; ?> <p><img src="Phil/images/b_spiderman.gif" height="100"></p> <form method="post"> <p> <input type="text" name="height"> <select name="height16" id="height16"> <option value="0">Choose sixteenth</option> <option value="0.0625">1/16</option> <option value="0.125">1/8</option> <option value="0.1875">3/16</option> <option value="0.250">1/4</option> <option value="0.3125">5/16</option> <option value="0.375">3/8</option> <option value="0.4375">7/16</option> <option value="0.500">1/2</option> <option value="0.5625">9/16</option> <option value="0.625">5/8</option> <option value="0.6875">11/16</option> <option value="0.750">3/4</option> <option value="0.8125">13/16</option> <option value="0.875">7/8</option> <option value="0.9375">15/16</option> </select> <br> <input type="text" name="width"> <select name="width16" id="width16"> <option value="0">Choose sixteenth</option> <option value="0.0625">1/16</option> <option value="0.125">1/8</option> <option value="0.1875">3/16</option> <option value="0.250">1/4</option> <option value="0.3125">5/16</option> <option value="0.375">3/8</option> <option value="0.4375">7/16</option> <option value="0.500">1/2</option> <option value="0.5625">9/16</option> <option value="0.625">5/8</option> <option value="0.6875">11/16</option> <option value="0.750">3/4</option> <option value="0.8125">13/16</option> <option value="0.875">7/8</option> <option value="0.9375">15/16</option> </select> </p> <p><br> <input type="submit" name="submit" value="Calculate"> </p> </form> <? if(!empty($_POST)){ // set the PRICE to the square footage cost of the inserts $price = 0.25; $totalwidth = $width + $width16; $totallength = $length + $length16; $total = $totalwidth * $totalheight * $price; echo "Your inserts are an estimated cost of $ $total"; } ?> </body> so i get a result of $0 when it echo's price and FYI i would take classes but i can't because i work 2 jobs and help my wife take care of my baby, since she's to sick to look after him all the time, and i can't afford to spend any money on classes. but you didn't know that. and THIS isn't my job i was joking, i work on stained glass, i told my boss i can do html - hell everyone does html, but he want's it all. asp java php mysql css ecommerce, and is willing to pay for me to learn on my own but not willing to pay for classes. so thank you I really apreaciate your time, but if you feel that I should go to classes instead of asking for help, fine don't answer.
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sorry I'm a super noob, I have no idea what to use instead if i remove the fist section of php code: <?php $length = $_POST["length"]; $height = $_POST["height"]; $price = $_POST["price"]; $total = $_POST["total"]; ?> the calculator stops working, i'm not sure what else to use instead. as for the global variable again i have no idea what to use instead. I basically thought myself a freaking genious to stumble upon this. (and i demostrate of how much of a genious by my spelling)
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THANK YOU SO MUCH!!!! now i can keep my job pretending i know what i'm doing !!!!! lol thanks again
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ok so here is the page i need help with http://www.typh.ca/phpcalctest.php now what i need is to HIDE (or something) the sentence that contains the estimate until the user clicks submit. i have one condition - it has to be simple (cause i'm an idiot.) any ways heres my code: <html> <head> <title>Price calculator</title> </head> <body> <?php $length = $_POST["length"]; $height = $_POST["height"]; $price = $_POST["price"]; $total = $_POST["total"]; ?> <p><img src="Phil/images/b_spiderman.gif" height="100"></p> <form method="post" action="<?php echo $PHP_SELF;?>"> <p>Length: <input type="text" name="length" id="length"> <br> Height: <input type="text" name="height" id="height"> </p> <p> <input type="submit" value="submit" name="submit"> </p> </form> <? // set the PRICE to the square footage cost of the inserts $price = 0.25; $total = $length * $height * $price; echo "Your inserts are an estimated cost of $ $total"; ?> </body> Please help me, i'm hitting a brick wall and my brain is starting to detach.