
cruxaders
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hirs the code...my proble is when i upload a image yes its upload but the problem is the previous image that he got...that still the image i see after i click submit..then i press f5 to see the updated image i uploaded.... what i want is after i press submit button i want the previous image will be automatic replaced by current/updated images no need to press F5 $id = "sir"; $sql = "SELECT * FROM `users` WHERE `username` = '".$id."'"; $res = mysql_query($sql) or die (mysql_error()); $row = mysql_fetch_assoc($res); $location = $row ['imagelocation']; echo "avatar <img src=\"$location\" width=\"100\" height=\"100\">"; echo "<form action=\"\" name=\"form1\" enctype=\"multipart/form-data\" method=\"post\" >"; echo "<input type=\"file\" name=\"image\" />"; echo "<input type=\"submit\" name=\"submit\" value=\"submit\" action =\"\">"; echo "</form>"; if($_POST['submit']) { $name = $_FILES['image']['name']; $tmp_name = $_FILES['image']['tmp_name']; $location = "avatar/".$name; move_uploaded_file($tmp_name,$location); $sql = mysql_query("UPDATE `users` SET `imagelocation` = '".$location."' WHERE `username` = '".$id."'") or die (mysql_error());
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i got a submit button... if user click the submit button iwant to page will auto referesh... can someone teach me
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another code..got a problem...can someone help me again
cruxaders replied to cruxaders's topic in PHP Coding Help
still ddnt see anything no error..all white what should i do... -
When i try to run the page...i ddnt see anything...its just a plane white background.. is there a problem in my code <?php $con = mysql_connect ("localhost","root",""); if(!$con) { die ('could not connect:' . mysql_error()); } mysql_select_db ("users", $con); $username = "sample"; $sql = "SELECT * FROM `users` WHERE `id`='".username."'"; $res = mysql_query($sql) or die(mysql_error()); while($row = mysql_fetch_assoc($res)) { $imagename = $_FILES['image']['imagename']; $tmp_name = $_FILES['image']['tmp_name']; $location = "avatar/".$imagename; move_uploaded_file($tmp_name,$location); $avatar = mysql_query("SELECT imagelocation FROM `users` WHERE `id`='".$username."'")or die(mysql_error()); $row3 = mysql_fetch_assoc($avatar); $location = $row3['imagelocation']; echo"<table width=\"686\" height=\"178\" border=\"0\">"; echo"<tr> <td width=\"160\" height=\"30\">First Name</td> <td width=\"321\">" . $row['name'] . "</td> </tr>"; echo"<tr> <td height=\"31\">Email Address</td> <td>" . $row['email'] ."</td> </tr>"; echo"<tr> <td height=\"28\">Username</td> <td>" . $row['username'] ."</td> </tr>"; echo"<tr> <td height=\"30\">ProfilePic</td> <td> </td> </tr>"; echo"<tr> <td><table width=\"100\" border=\"0\" align=\"center\"> <tr> <td><img src =\"$location\" width=\"100\" hieght=\"100\"></td> </tr>"; echo"</table></td>"; echo"<td><form id=\"form1\" name=\"form1\" enctype=\"multipart/form-data\" method=\"post\" action=\"\"> <input type=\"file\" name=\"image\" id=\"image\" />"; echo"</form></td> </tr> </table>"; } mysql_close($con); ?>
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Can some please check my code if what is wrong
cruxaders replied to cruxaders's topic in PHP Coding Help
ill already figure out the image... tnx for the help and time -
Can some please check my code if what is wrong
cruxaders replied to cruxaders's topic in PHP Coding Help
still the image ddnt appear -
Can some please check my code if what is wrong
cruxaders replied to cruxaders's topic in PHP Coding Help
$avatar2 = mysql_query("SELECT users.imagelocation FROM users , forum_replies WHERE users.id=forum_replies.uid ")or die(mysql_error()); $row4 = mysql_fetch_assoc($avatar); $location2 = $row4['imagelocation']; <table width=\"100\" border=\"0\" align=\"center\"> <tr> <td><img src =\"$location2\" width=\"100\" hieght=\"100\"></td> </tr> </table> -
Can some please check my code if what is wrong
cruxaders replied to cruxaders's topic in PHP Coding Help
this the other code with fetch $avatar2 = mysql_query("SELECT users.username, forum_replies.message FROM users,forum_replies WHERE users.id= forum_replies.uid ")or die(mysql_error()); $row3 = mysql_fetch_assoc($avatar2); $location2 = $row4['imagelocation']; <table width=\"100\" border=\"0\" align=\"center\"> <tr> <td><img src =\"$location2\" width=\"100\" hieght=\"100\"></td> </tr> </table> -
Can some please check my code if what is wrong
cruxaders replied to cruxaders's topic in PHP Coding Help
if thats the correct code why the image not yet shawn..what will i do to make the image visible -
Can some please check my code if what is wrong
cruxaders replied to cruxaders's topic in PHP Coding Help
btw i forgot to tell i got 2 tables 1st table is users, with fields -id -username -imagelocation 2nd table is forum_replies with fields -id -uid -message what i want to do in sql is to get the image location.. but when i echo the $avatar2 it appears Resource id #3 idk what the problem on my code -
can someone check if what the problem with my code $avatar2 = mysql_query("SELECT `users`.`imagelocation' FROM `users` `forum_replies` WHERE `users`.`id`= `forum_replies`.`uid`")or die(mysql_error());
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my problem is if my local date is february 1 or march 30... if i press previous month i cant see the month of february, but if i press the change the local date to february 2 or march 29 i can see the month of february in my event calendar... what should i do to see the month of february even i set the local date in february 1 or march 30.... heres the code if (isset($_GET['day'])){ $day = $_GET['day']; } else { $day = date("j"); } if(isset($_GET['month'])){ $month = $_GET['month']; } else { $month = date("n"); } if(isset($_GET['year'])){ $year = $_GET['year']; }else{ $year = date("Y"); } $currentTimeStamp = strtotime( "$day-$month-$year"); $monthName = date("F", $currentTimeStamp); $numDays = date("t", $currentTimeStamp);
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i got a problem on undefined index...but if i use Xampp i can run in nicely with our error..but in wamp it say undefined index... what i supposed to do.... so that i can run it in both xampp and wamp nicely
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oh got it now..tnx