sfaisal Posted July 2, 2015 Share Posted July 2, 2015 Hello, Can anyone help me with my code, I am trying to retrieve a picture that I saved in a file on my server, the name was saved in MySQL and I want to retrieve it after it is uploaded and post it in a comment box. Here is the code: The problem I am having is te picture uploads perfect, but then it gives me a filename not found error when it is suppose to display the picture. Please help!!!!!!! <?php include('config3.php'); if($_POST) { // $_FILES["file"]["error"] is HTTP File Upload variables $_FILES["file"] "file" is the name of input field you have in form tag. if ($_FILES["file"]["error"] > 0) { // if there is error in file uploading echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; } else { // check if file already exit in "images" folder. if (file_exists("uploads/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { if(move_uploaded_file($_FILES["file"]["tmp_name"],"uploads/" . $_FILES["file"]["name"])) { // If file has uploaded successfully, store its name in data base $query_image = "INSERT into acc_images (image, status, acc_id) values ('".$_FILES['file']['name']."', 'display','')"; if(mysql_query($query_image)) { } } } } $query_image = "SELECT * FROM acc_images acc_ id='$id'"; $result = mysql_query($query_image); if(mysql_num_rows($result) > 0) { while($row = mysql_fetch_array($result)) { echo '<img alt="" src="uploads/'.$row["image"].'">'; } } echo 'File name not found in database'; } ?> <html> <body> <form action="upload.php" method="post"enctype="multipart/form-data"> <label for="file">Filename:</label> <input type="file" name="file" id="file" /> <br /> <input type="submit" name="submit" value="Submit" /> </form> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/297156-can-anyone-help-with-code-for-uploading-a-picture-from-mysql/ Share on other sites More sharing options...
scootstah Posted July 2, 2015 Share Posted July 2, 2015 (edited) Please use code tags and format your code properly. <?php include('config3.php'); if ($_POST) { // $_FILES["file"]["error"] is HTTP File Upload variables $_FILES["file"] "file" is the name of input field you have in form tag. if ($_FILES["file"]["error"] > 0) { // if there is error in file uploading echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; } else { // check if file already exit in "images" folder. if (file_exists("uploads/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { if (move_uploaded_file($_FILES["file"]["tmp_name"], "uploads/" . $_FILES["file"]["name"])) { // If file has uploaded successfully, store its name in data base $query_image = "INSERT into acc_images (image, status, acc_id) values ('" . $_FILES['file']['name'] . "', 'display','')"; if (mysql_query($query_image)) { } } } } $query_image = "SELECT * FROM acc_images acc_ id='$id'"; $result = mysql_query($query_image); if (mysql_num_rows($result) > 0) { while ($row = mysql_fetch_array($result)) { echo '<img alt="" src="uploads/' . $row["image"] . '">'; } } echo 'File name not found in database'; } ?> <html> <body> <form action="upload.php" method="post" enctype="multipart/form-data"> <label for="file">Filename:</label> <input type="file" name="file" id="file"/> <br/> <input type="submit" name="submit" value="Submit"/> </form> </body> </html>Your code doesn't work because you're not actually storing the file in the file system. Once a file is uploaded in PHP, it is stored in a temporary location until the script is done executing, where it is then deleted. You need to copy the file to a permanent location after it is uploaded. Take a look at move_uploaded_file() Edited July 2, 2015 by scootstah Quote Link to comment https://forums.phpfreaks.com/topic/297156-can-anyone-help-with-code-for-uploading-a-picture-from-mysql/#findComment-1515487 Share on other sites More sharing options...
jcbones Posted July 2, 2015 Share Posted July 2, 2015 You also have an error in your sql syntax. $query_image = "SELECT * FROM acc_images acc_ id='$id'"; Should be: $query_image = "SELECT * FROM acc_images WHERE acc_ id='$id'"; Quote Link to comment https://forums.phpfreaks.com/topic/297156-can-anyone-help-with-code-for-uploading-a-picture-from-mysql/#findComment-1515488 Share on other sites More sharing options...
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