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masrawy

Problem Of Sending Email To User

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this is code to upload pic if (!isset($HTTP_POST_FILES['s_photo'])) exit(exit1);if (is_uploaded_file($HTTP_POST_FILES['s_photo']['tmp_name'])) {if ($HTTP_POST_FILES['s_photo']['size']>$max_size) { echo "حجم الملف كبير جدا\n";}if (file_exists($path . $Name)) {echo "اسم الصورة موجود مسبقا يرجي تغيير اسم الصورة وارسالها\n"; exit;}$res = copy($HTTP_POST_FILES['s_photo']['tmp_name'], $path . $Name);if (!$res) { echo "فشل تحميل الصورة<br>\n";}$s_photo=$Name;}if ($s_photo != ""){ $update=mysql_query("UPDATE ssubject SET s_name='$ss_name', s_disc='$s_disc', s_photo='$s_photo', replay='$rep' WHERE id='$hidden'"); if($update){echo "تم التعديل بنجاح";} else {echo "هناك خطأ في عملية التعديل";}}if ($s_photo == ""){ $update=mysql_query("UPDATE ssubject SET s_name='$ss_name', s_disc='$s_disc', replay='$rep' WHERE id='$hidden'"); if($update){echo "تم التعديل بنجاح";} else {echo "هناك خطأ في عملية التعديل";}}echo "<meta http-equiv=\"refresh\" content=\"2; URL=index.php?action=advs&advid=$hidden\">";}is it will be like that if (!isset($HTTP_POST_FILES['s_photo'])) exit(exit1);if (is_uploaded_file($HTTP_POST_FILES['s_photo']['tmp_name'])) {if ($HTTP_POST_FILES['s_photo']['size']>$max_size) { echo "حجم الملف كبير جدا\n";}if (file_exists($path . $Name)) {echo "اسم الصورة موجود مسبقا يرجي تغيير اسم الصورة وارسالها\n"; exit;}$res = move_uploaded_file($HTTP_POST_FILES['s_photo']['tmp_name'], $path . $Name);if (!$res) { echo "فشل تحميل الصورة<br>\n";}$s_photo=$Name;}if ($s_photo != ""){ $update=mysql_query("UPDATE ssubject SET s_name='$ss_name', s_disc='$s_disc', s_photo='$s_photo', replay='$rep' WHERE id='$hidden'"); if($update){echo "تم التعديل بنجاح";} else {echo "هناك خطأ في عملية التعديل";}}if ($s_photo == ""){ $update=mysql_query("UPDATE ssubject SET s_name='$ss_name', s_disc='$s_disc', replay='$rep' WHERE id='$hidden'"); if($update){echo "تم التعديل بنجاح";} else {echo "هناك خطأ في عملية التعديل";}}echo "<meta http-equiv=\"refresh\" content=\"2; URL=index.php?action=advs&advid=$hidden\">";}this code is wright ?yes or no?

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i did that and the script tell me that the picture was upload but i find (x) and no pici search for function fopen i didn't find it in the index.phpwhat is the problem sir ?

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You're trying to move the picture to "$path . $Name", so print that out to see where you're trying to move the picture to. You should probably check the permissions on that folder to make sure you're allowed to copy files to it.

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Print out the actual value that it's using for the path and name, not just the code for it.

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I mean add this code to the part that saves the file:echo $path . $Name;So you can actually see what name it's trying to save the file with.

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i did if (!isset($HTTP_POST_FILES['s_photo'])) exit(exit1);if (is_uploaded_file($HTTP_POST_FILES['s_photo']['tmp_name'])) {if ($HTTP_POST_FILES['s_photo']['size']>$max_size) { echo "حجم الملف كبير جدا\n";}if (file_exists($path . $Name)) {echo "اسم الصورة موجود مسبقا يرجي تغيير اسم الصورة وارسالها\n"; exit;}$res = move_uploaded_file($HTTP_POST_FILES['s_photo']['tmp_name'], $path . $Name);if (!$res) { echo "فشل تحميل الصورة<br>\n";}$s_photo=$Name;}if ($s_photo != ""){$update=mysql_query("UPDATE ssubject SET s_name='$ss_name', s_disc='$s_disc', s_photo='$s_photo', replay='$rep' WHERE id='$hidden'");if($update){echo "تم التعديل بنجاح";} else {echo "هناك خطأ في عملية التعديل";}}if ($s_photo == ""){$update=mysql_query("UPDATE ssubject SET s_name='$ss_name', s_disc='$s_disc', replay='$rep' WHERE id='$hidden'");if($update){echo "تم التعديل بنجاح";} else {echo "هناك خطأ في عملية التعديل";}}echo "<meta http-equiv=\"refresh\" content=\"2; URL=index.php?action=advs&advid=$hidden\">";}but no pic uploaded sir is ther any reasone for that

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if (!isset($HTTP_POST_FILES['s_photo'])) exit(exit1);if (is_uploaded_file($HTTP_POST_FILES['s_photo']['tmp_name'])) {if ($HTTP_POST_FILES['s_photo']['size']>$max_size) { echo "حجم الملف كبير جدا\n";}if (file_exists($path . $Name)) {echo "اسم الصورة موجود مسبقا يرجي تغيير اسم الصورة وارسالها\n"; exit;}$res = move_uploaded_file($HTTP_POST_FILES['s_photo']['tmp_name'], $path . $Name);echo $path . $Name;if (!$res) { echo "فشل تحميل الصورة<br>\n";}$s_photo=$Name;}if ($s_photo != ""){$update=mysql_query("UPDATE ssubject SET s_name='$ss_name', s_disc='$s_disc', s_photo='$s_photo', replay='$rep' WHERE id='$hidden'");if($update){echo "تم التعديل بنجاح";} else {echo "هناك خطأ في عملية التعديل";}}if ($s_photo == ""){$update=mysql_query("UPDATE ssubject SET s_name='$ss_name', s_disc='$s_disc', replay='$rep' WHERE id='$hidden'");if($update){echo "تم التعديل بنجاح";} else {echo "هناك خطأ في عملية التعديل";}}echo "<meta http-equiv=\"refresh\" content=\"2; URL=index.php?action=advs&advid=$hidden\">";}like that ?

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First, everywhere you see $HTTP_POST_FILES, change it to $_FILES. The $HTTP_POST vars are old and have been deprecated. This code is not doing any error checking on the uploaded file. There could be an error happening and you wouldn't know what it is (which might be the problem here). Add this near the top of the page:print_r($_FILES);

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i did that <?print_r($_FILES); include("system/config.php");now i see in the top of the page this Array ()i try to upload pictre ths what i see Array ( [s_photo] => Array ( [name] => 140x80-20090221-213043.gif [type] => image/gif [tmp_name] => /tmp/phpQ4Zk6F [error] => 0 => 23939 ) ) do the function fopen can make this error ?

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