ehsan_cs

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who can find my error this code for update does not work i think the error in query i get this message Notice: Undefined variable: detail in C:\xampp\htdocs\database\mange.php on line 156 /////////////////////////////////////////////////////////////////////////////////////////////// //update <? if(isset($pid)){ $uquery= mysql_query("update product SET product_name=`$product_name`,price=`$price`,category=`$category`,detail=`$detail`,picture=`$picture` WHERE id = `$pid`"); while($row = mysql_fetch_array($query)){ $product_name=$row[1]; $price=$row[2]; $category=$row[3]; $productdetail_name=$row[4]; $picture=$row[5]; } ?> <form id="form2" name="form2" method="get" action="mange.php"> <table width="593" border="1" align="right"> <tr> <td width="92">product name</td> <td width="369"> <input type="text" name="product_name"value="<? echo $price;?>" id="textfield2" /></td> </tr> <tr> <td>product price</td> <td><input type="text" name="price"value="<? echo $price;?>" id="textfield2" /></td> </tr> <tr> <td>category</td> <td><select name ="category"> <option>"<? echo $detail;?>"</option> <option>book</option> <option>electorinc </option> <option>clothing</option> </select> </td> </tr> <tr> <td>product datails</td> <td><textarea name="detail"rows=5 value="<? echo $detail;?>"></textarea></td> </tr> <tr> <td>product image </td> <td><input type="file" name="picture" /><input type="hidden" name= "upload" ></td> </tr> <tr> <td></td> <td><input type="submit" name="button" value="update " /></td> </tr> <tr> <td height="128"> </td> <td> </td> </tr> </table> </form> <? } ?>

Who can help me quickly

THANK U FOR REPLY I want to activate the edit link to update info When I click on the link showing all the information stored in databases in EACH input text then when click on submit button save change in DB I WONT TO MAKE LIKE THIS VIDEO

good morning how i can active edit link when i click on it dispay all row information by pid virible and wen cilck on button save the change .. see my code in attachment <? extract($_GET); //connect to mysql if (!($dbc=mysql_connect("localhost","root",""))) die("could not connect to database "); //select database $db_selected = mysql_select_db("ehsan",$dbc); if(!$db_selected) { die("cannt connect "); } $query= mysql_query("select `id`,`product_name`,`price`,`category`,`detail`,`picture`,`date_addd` FROM `product` "); if(isset($id)) { $queryid= mysql_query("select id, product_name,price,category,detail,picture,date_addd FROM product WHERE id = '$id'"); $rowid =mysql_fetch_row($queryid); } else{ $productlist="no items"; } ?> <? extract($_GET); // delete from DB if (!($dbc=mysql_connect("localhost","root",""))) die("could not connect to database "); //select database $db_selected = mysql_select_db("ehsan",$dbc); if(!$db_selected) { die("cannt connect "); } if(isset($deleteid)){ $dquery=mysql_query("DELETE FROM `product` WHERE `id`='$deleteid'") ; if(!($result=mysql_query($dquery))) { print"the item was deleted seccsessfly "; die (mysql_error()); } } ?> <!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>add items </title> <style type="text/css"> .title { color: #0C6; background-color: #666; border: thin dotted #000; position: relative; text-align: right; } </style> </head> <body> <div> <p> </p> <p> </p> <p> </p> </div> <div><h1>store list</h1> <p> <?php //get info from title from DB and print it; while($row =mysql_fetch_row($query)) { print(" $row[0] $row[1]- <a href='mange.php?pid=$row[0]'>edit</a>•<a href='mange.php?deleteid=$row[0]'>delete</a><br/>"); ?> <? } ?> </p> </div> <form id="form1" name="form1" method="post" action="conection.php"> <table width="513" border="1"> <tr> <td width="92">product name</td> <td width="405"> <input type="text" name="product_name" id="textfield2" /></td> </tr> <tr> <td>product price</td> <td><input type="text" name="price" id="textfield2" /></td> </tr> <tr> <td>category</td> <td><select name ="category"> <option>book</option> <option>electorinc </option> <option>clothing</option> </select> </td> </tr> <tr> <td>product datails</td> <td><textarea name="detail"rows=5 ></textarea></td> </tr> <tr> <td>product image </td> <td><input type="file" name="picture" /><input type="hidden" name= "upload" ></td> </tr> <tr> <td></td> <td><input type="submit" name="button" value="add this item " /></td> </tr> <tr> <td height="128"> </td> <td> </td> </tr> </table> </form> </body> </html>

how can solve my problem :clapping:

does not work $fg have constant can apply your in the code shown below <textarea name="cod" id="cod2" cols="38" rows="25"> <?php print "$fg" ; ?> </textarea> <? [code] [/code] thank u so much

how i can display the output of script when click to run button

please lock to attachment give my the result by php thank u

MR SAMUS CODE ID RIGTH THANK U MR dddougal To participate

thank u MR Samu s

any one give my html or php code :clapping: or modify to my code

using dreamwer S/W or netbeans S/W with xammpS/W

there are some error

i have error in $query ="insert ..... i don't know what it's extract($_POST); //connect to mysql if (!($dbc=mysql_connect("localhost","root",""))) die("could not connect to database "); //select database $db_selected = mysql_select_db("ehsan",$dbc); if(!$db_selected) { die("cannt connect "); } $query= "INSERT INTO tutorial (code,titile,video) VALUES ('$code','$titile','$video')"; if (! mysql_query($query)) die (mysql_error()); mysql_close(); ?> *********************************************************************************** <form id="form1" name="form1" method="post" action=""> <p> <label for="textfield"></label> <input type="text" name="code" id="textfield" /> code </p> <p> <label for="textfield2"></label> <input type="text" name="titile" id="textfield2" /> title</p> <p> <label for="textfield3"></label> <input type="text" name="video" id="textfield3" /> video</p> <p> <input type="submit" name="button" id="button" value="Submit" /> </p> </form>

thank u can u make the code with php and html code becouse i cant understand ajax code

how i can disply the video on div #video by dynamic linking in navigation menu the video store in database <?php while($row =mysql_fetch_row($query)) { $titile =$row[2]; $ids= $row[1]; ?> <p> <a href="index.php?id.=<?php echo $ids ;?>".><? print "$titile<br>"?> </a></p> <? } ?> </div>

thank u so much

just using html5 ,cs3 and php with out javascript

how i can ply video by click on navigation form but display the video in same page please look at the attachment