vishalonne Posted November 8, 2012 Share Posted November 8, 2012 Hi I just wan to to pass the value of Input Type File html tag to a 2nd PHP page where I will insert the image in mysql but I am always getting a notice and isset() is not getting the $_FILE('IMAGE'). Here is the notice - Notice: Undefined index: IMAGE in C:\xampp\htdocs\billing\prodinsert.php on line 11 This is my HTML TAGS - <body> <hr /> <form id="form1" name="form1" method="post" action="prodinsert.php" enctype="multipart/form-data"> <input name="ICODE" type="text" size="10" maxlength="6" /> <input name="DESCR" type="text" size="50" maxlength="45" /> <input name="RATE" type="text" size="10" maxlength="9" /> <input type="hidden" name="MAX_FILE_SIZE" value="10000000" /> <input name="IMAGE" type="file" /> </form> And this is my PHP code- <?php $host="localhost"; $user="root"; $pass=""; $db="bill"; mysql_connect($host, $user, $pass) OR DIE (mysql_error()); mysql_select_db ($db) OR DIE ("Unable to select db".mysql_error()); $code=$_POST['ICODE']; $descp=$_POST['DESCR']; $rate=$_POST['RATE']; $image=$_POST['IMAGE']; if(!isset($_FILES[$image])) { echo '<p>Please select a file</p>'; echo $image; } else { echo "File Uploaded"; echo $image; } Where I am making mistake ??? Please guide me. Quote Link to comment Share on other sites More sharing options...
MDCode Posted November 8, 2012 Share Posted November 8, 2012 (edited) You cant simply post an image. You use $_FILES["IMAGE"]["name"] You also do not want to upload files to a database, use a directory Edited November 8, 2012 by ExtremeGaming Quote Link to comment Share on other sites More sharing options...
vishalonne Posted November 8, 2012 Author Share Posted November 8, 2012 You cant simply post an image. You use $_FILES["IMAGE"]["name"] You also do not want to upload files to a database. use a directory Thank you for Guiding me What is the 2nd paprameter ["name"] Quote Link to comment Share on other sites More sharing options...
MDCode Posted November 8, 2012 Share Posted November 8, 2012 The name of the file itself example: name --> picture.jpg <-- name Quote Link to comment Share on other sites More sharing options...
vishalonne Posted November 8, 2012 Author Share Posted November 8, 2012 The name of the file itself example: name --> picture.jpg <-- name Now I change my code to like this -> if(!isset($_FILES["IMAGE"])) { echo '<p>Please select a file</p>'; echo $image; } else { echo "File Uploaded"; It is now working showing File Uploaded but notice is still coming Notice: Undefined index: IMAGE in C:\xampp\htdocs\billing\prodinsert.php on line 11 Quote Link to comment Share on other sites More sharing options...
MDCode Posted November 8, 2012 Share Posted November 8, 2012 (edited) $image=$_POST['IMAGE']; Change that to $image = $_FILES["IMAGE"]["name"]; Edited November 8, 2012 by ExtremeGaming Quote Link to comment Share on other sites More sharing options...
vishalonne Posted November 8, 2012 Author Share Posted November 8, 2012 Change that to $image = $_FILES["IMAGE"]["name"]; Now it more problematic If I don't select any image then also it is showing File Uploaded Quote Link to comment Share on other sites More sharing options...
vishalonne Posted November 8, 2012 Author Share Posted November 8, 2012 (edited) Now it more problematic If I don't select any image then also it is showing File Uploaded YES YES ITS DONE Thank you Very much NOTICE GONE BUT I am getting Please select a File message and The name of the File Here is the modified code - $code=$_POST['ICODE']; $descp=$_POST['DESCR']; $rate=$_POST['RATE']; $image=$_FILES["IMAGE"]["name"]; if(!isset($_FILES[$image])) { echo '<p>Please select a file</p>'; echo $image; } else { echo "File Uploaded"; echo $image; ...... Edited November 8, 2012 by vishalonne Quote Link to comment Share on other sites More sharing options...
MDCode Posted November 8, 2012 Share Posted November 8, 2012 You're recalling the $image variable inside the $_FILES. Change if(!isset($_FILES[$image])) to if(!isset($_FILES["IMAGE"])) Quote Link to comment Share on other sites More sharing options...
vishalonne Posted November 8, 2012 Author Share Posted November 8, 2012 You're recalling the $image variable inside the $_FILES. Change if(!isset($_FILES[$image])) to if(!isset($_FILES["IMAGE"])) CAn I just check the this by this way if(!isset($image)) Quote Link to comment Share on other sites More sharing options...
MDCode Posted November 8, 2012 Share Posted November 8, 2012 (edited) Not sure. $image will always be set because you're giving it a value. You can first try checking if the form was submitted. Then give it a value and it should work Edited November 8, 2012 by ExtremeGaming Quote Link to comment Share on other sites More sharing options...
vishalonne Posted November 8, 2012 Author Share Posted November 8, 2012 CAn I just check the this by this way if(!isset($image)) As you suggested me to change if(!isset($_FILES[$image])) TO if(!isset($_FILES["IMAGE"])) now if submit the form without selecting any image file it is showing File Uploaded Quote Link to comment Share on other sites More sharing options...
vishalonne Posted November 8, 2012 Author Share Posted November 8, 2012 I just want to save the image of an item in mysql database Quote Link to comment Share on other sites More sharing options...
MDCode Posted November 8, 2012 Share Posted November 8, 2012 (edited) You said that was the problem before *facepalm* Ok try this... $code=$_POST['ICODE']; $descp=$_POST['DESCR']; $rate=$_POST['RATE']; $image=$_FILES["IMAGE"]["name"]; if($_FILES["IMAGE"]["error"] > 0) { echo "Error: " . $_FILES["IMAGE"]["error"] . "<br />"; } else { if(!isset($_FILES["IMAGE"]["name"])) { echo '<p>Please select a file</p>'; } else { echo "File Uploaded"; echo $image; } Edited November 8, 2012 by ExtremeGaming Quote Link to comment Share on other sites More sharing options...
vishalonne Posted November 8, 2012 Author Share Posted November 8, 2012 You said that was the problem before *facepalm* Ok try this... $code=$_POST['ICODE']; $descp=$_POST['DESCR']; $rate=$_POST['RATE']; $image=$_FILES["IMAGE"]["name"]; if($_FILES["IMAGE"]["error"] > 0) { echo "Error: " . $_FILES["IMAGE"]["error"] . "<br />"; } else { if(!isset($_FILES["IMAGE"]["name"])) { echo '<p>Please select a file</p>'; } else { echo "File Uploaded"; echo $image; } Thank you for your co operation After using the code you gave If I select image it it shows "File Uploaded" with the name of the image file And if I don't select image file it show Error: 4 Can't understand why 4? why not it shows "Please select a file" Quote Link to comment Share on other sites More sharing options...
MDCode Posted November 8, 2012 Share Posted November 8, 2012 (edited) Error code 4 means that there was no file uploaded A full list can be found at http://php.net/manua...load.errors.php If you wish to change it simply do: $code=$_POST['ICODE']; $descp=$_POST['DESCR']; $rate=$_POST['RATE']; $image=$_FILES["IMAGE"]["name"]; if($_FILES["IMAGE"]["error"] > 0) { // create a list of errors here if($_FILES["IMAGE"]["error"] == "4") { echo "Please select a file."; } } else { echo "File Uploaded"; echo $image; } Edited November 8, 2012 by ExtremeGaming Quote Link to comment Share on other sites More sharing options...
vishalonne Posted November 8, 2012 Author Share Posted November 8, 2012 Error code 4 means that there was no file uploaded A full list can be found at http://php.net/manua...load.errors.php If you wish to change it simply do: $code=$_POST['ICODE']; $descp=$_POST['DESCR']; $rate=$_POST['RATE']; $image=$_FILES["IMAGE"]["name"]; if($_FILES["IMAGE"]["error"] > 0) { // create a list of errors here if($_FILES["IMAGE"]["error"] == "4") { echo "Please select a file."; } } else { echo "File Uploaded"; echo $image; } Great You are Genius Thank you very much Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.