techker
-
Posts
812 -
Joined
-
Last visited
Everything posted by techker
-
ok i got it going i remove the " " cause it was giving me a white space error.. <? // Connects to your Database mysql_connect("localhost", "l_l", "l") or die(mysql_error()); mysql_select_db("techker_l") or die(mysql_error()); $id = mysql_escape_string($_GET['id']); $q = "SELECT * FROM cardio_select WHERE id = '$id'"; $res = mysql_query($q); $row = mysql_fetch_assoc($res); $plan =$row['simple']; $plan2 =$row['combo']; if ( $plan == "on" ) { echo "<meta http-equiv=Refresh content=2;url=client_cardio.php>";; } elseif($plan2 == "on"){ echo "<meta http-equiv=Refresh content=2;url=emplacement.php>";; } ?> is this the write way to do it?
-
sorry fergot to post it..lol ok so i have 2 questions in this: <? // Connects to your Database mysql_connect("localhost", "l", "techker") or die(mysql_error()); mysql_select_db("techker_llls") or die(mysql_error()); $id = mysql_escape_string($_GET['id']); $q = "SELECT * FROM cardio_select WHERE id = '$id'"; $res = mysql_query($q); $row = mysql_fetch_assoc($res); $plan = "$row['simple']"; //is this ok?will it get table simple? $plan2 = "$row['combo']"; if ( $plan == "on" ) {; //if that table's value is on then it should get page x } elseif($plan2 == "on"){; //if this one is on then it gets page b } ?> now it's the getting that page the part were i freeze.. i was looking at fetch but i think it is only for arays. then i looked at another page i did and the echo echo "<meta http-equiv=Refresh content=4;url=emplacement.php>"; but was woundering if 2 would the script get confused?
-
hey guys i have a script that selects client from a databse.now that client is either in a programme simple or combo. there is a link on that page that gets a grid with exercises on it. now i need a script between that process thats like if the client is simple fetch page...if client is combo fetch page x..
-
cool.i don't want to get each pic.i usaly put when echo that shows gets all the pics one by one.(cause of the query)
-
hey guys i want to to a picture gallery for my cuz.now i made an admin section to upload the pics and it inserts the name and description in a database.. now when i use a php echo witha select all pics from.. it shows all the pics put in one straight line.. is there a way to make it show like 4 images per row.. pic pic pic pic pic pic pic pic http://www.humanflytattoo.com/tattoo.php (like this)
-
ok i setled the prob by putting a hidden filed that echo the name of the pic. so in my add pic i just added the post for that field..and voila!thx.. but im still wondering why it does not work..
-
so what would be the value?i need the named of the image that is ticked. whats the [0] added?
-
hey guys im trying to do a select pic to insert the name in a database. the insert works but it does not insert the pic name only an id.. i can't seem to figure it out.. <? $dbh = mysql_connect("localhost","user","pass") or die("There was a problem with the database connection."); $dbs = mysql_select_db("db", $dbh) or die("There was a problem selecting the categories."); $type=$_POST['type']; $sql = "SELECT * FROM `gymball` WHERE `type` = '$type' "; $fileLIST=mysql_query($sql);?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>GymGraph</title> </head> <body> <p align="center"><u>Sélectionner UNE photo</u></p> <table width="346" border="0" align="center" cellpadding="0" cellspacing="0"> <tr><? while($row = mysql_fetch_array($fileLIST)) { ?> <td width="173"><?php echo '<a href="/Gymgraph/Gymgraph/gymball/'. $row['name'] .'" target="_blank"> <img src="/Gymgraph/Gymgraph/gymball/'. $row['name'] .'" border="0" alt="" width=115/> </a> <br />'; ?></td> <td width="173"><label> <input type="checkbox" name="select" id="pic" value='<? $row['name'] ?>'/> Selection de photo</label></td> </tr> <? } ?> </table> <p> <p> <label> <div align="center"> <select name="location" id="location"> <option value="pic1">1</option> <option value="pic2">2</option> <option value="pic3">3</option> <option value="pic4">4</option> <option value="pic5">5</option> <option value="pic6">6</option> <option value="pic7">7</option> <option value="pic8">8</option> </select> Emplacement de la photo </div> </label> </p> <label> <div align="center"> <input type="submit" name="submit" id="submit" value="Submit" /> </div> </label> <div align="center"> Envoyer la sélection </div> </p> </form> </body> </html> and this is the insert pic <?php $pic=$_POST['pic']; $location=$_POST['location']; mysql_connect("localhost", "user", "t") or die(mysql_error()) ; mysql_select_db("techker_gymgraphpics") or die(mysql_error()) ; //Writes the information to the database mysql_query("INSERT INTO $location (name) ". "VALUES ('$pic')"); //Tells you if its all ok $id= mysql_insert_id(); echo "<p>This file has the following Database ID: <b>$id</b>"; echo "You'll be redirected to Home Page after (4) Seconds"; echo "<meta http-equiv=Refresh content=4;url=gymball_select.php>"; ?>
-
All good thx!!
-
<? mysql_connect("localhost", "user_techker", "pass"); mysql_select_db("soul3438_gymgraphtrainers"); $showtablequery = "SHOW TABLES FROM [soul3438_gymgraphtrainers]"; $showtablequery_result = mysql_query($showtablequery); while($showtablerow = mysql_fetch_array($showtablequery_result)) { echo $showtablerow[0]."<br />"; } ?>
-
Warning: mysql_query() [function.mysql-query]: Access denied for user 'soul3438'@'localhost' (using password: NO) in /home/soul3438/public_html/Gymgraph/show.php on line 4 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /home/soul3438/public_html/Gymgraph/show.php on line 4 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/soul3438/public_html/Gymgraph/show.php on line 5
-
it dosnt work it gives me an access denied all the time..i have changed all the info to my server. i removed to test $mysql_access and still nothing
-
hey guys can i select all in a databases? i want to show all tables created..
-
szo i just inverted the id with training and it works fine now!lol thx for the quick reply!i can always count on this forum!!
-
so with & works if i remove the space.. this is my link to the file <a href="grapht.php?trainer=<? echo $row['trainer']?>&id=<? echo $row['id'] ?>" target="_blank"> in the browser it gives http://site.ca/admin/GymGraph/grapht.php?trainer=mike%20&id=4 the %20
-
hey guys is it possible to do this? like(cause i can't explain the srting if it is the real name..lol) link :http://site.ca/admin/GymGraph/grapht.php?id=4,trainer=mike $trainer= mysql_escape_string($_GET['trainer']); $id = mysql_escape_string($_GET['id']); $sql = "SELECT * FROM `$trainer` WHERE `id`= $id" ; $res= mysql_query($sql); $row= mysql_fetch_assoc($res);?> i changed it to mysql_real_escape_string and still nothing. it is as soon as i pass 2 infor in the link it does not work..
-
ya cause i will upload images to my database.(small images)and i wnat to make a liste box that will show the images name and when selected show the image in the corresponding box. look at this grid it will give you an idea. it hase images in the boxes for training.. http://soulfitness.ca/scan0001.jpg
-
hey guys i was wondering if somebody can point me in the write direction for this. im looking for a script or tutorial that has a list box that i will populate with mysql data that conatins images.when you select the image name the script shows the image.. so list box to select an image that will show it.lol
-
hmm..it is true that the members can share the rss links but just like there username and password of there account.but anywyas other people can't take advantage of there promotions. si it is interesting cause he can even broadcast if h eis closed or holidays.. how do i get this set up?i have googled rss feed tutorials but i get about 50 pages of sites.. i juste need a script that he can write the news and the scirpt that the clients use and a widget??
-
Hey guys i have a quick question concerning what is RSS i have my budy that has a members only promotion section on his site.now if it's what i think it is(RSS)it can send information to rss feed about anything(promotions) the client justs need too subscribe to it or download a widget.. is that write?or is this public meaning that it will broadcast to all ... and if it's what i think were do i start with this..lol
-
it was ?gim=1
-
why is that?
-
it does not work the code is wrong.?
-
doesn't work..very good idea do.. and i can't view the source..lol http://tech-design.info/pics.php?pid=28
-
maybe it does work but i can't seem to figure it out in my script.. if it try ".144." it gives me an error and the other nothing?
