Hi I have read many forums and websites that tell you how to upload an image to a server and I have managed to get this working, I can upload a file to a my server but storing the file name does work on the following example I found and I also need to create a form that allows more data to be entered to the database. I am stuck with this as a have done much PHP before. I have come to the end of trying different websites tutorials with no much success could anyone please help me! I need it done for a project I'm doing.
I am basically trying to make a CMS that allows users to upload a photo of a band member and have information stored about them so that it can be displayed on a webpage for the public to view.
My Table looks like this:
Field Type Null Default
id int(10) No
nameMember varchar(25) No
bandMember text No
photo varchar(30) No
aboutMember text No
otherBands text No
The form I want will look like this:
<h1>Adding a new Band Member or Affiliate</h1>
<form method="post" action="addMember.php" enctype="multipart/form-data">
<p>
Please Enter the Band Members Name.
</p>
<p>
Band Member or Affiliates Name:
</p>
<input type="text" name="nameMember"/>
<p>
Please Enter the Band Members Position. Example:Drums.
</p>
<p>
Member's Position:
</p>
<input type="text" name="bandMember"/>
<p>
Please Upload a Photo in gif or jpeg format. The file name should be named after the Members name. If the same file name is uploaded twice it will be overwritten!
</p>
<p>
Photo:
</p>
<input type="file" name="filep" size=35 />
<p>
Please Enter any other information about the band member here.
</p>
<p>
Other Member Information:
</p>
<textarea rows="10" cols="35" name="aboutMember">
</textarea>
<p>
Please Enter any other Bands the Member has been in.
</p>
<p>
Other Bands:
</p>
<input type="text" name="otherBands" size=30 />
<br/>
<br/>
<input TYPE="submit" title="Add data to the Database" value="Add Member"/>
</form>
The Example that uploads an Image to the server and only, that is this:
<?
if ($_POST["action"] == "Load")
{
$folder = "images/";
move_uploaded_file($_FILES["filep"]["tmp_name"] , "$folder".$_FILES["filep"]["name"]);
echo "
<p align=center>File ".$_FILES["filep"]["name"]."loaded...";
$result = mysql_connect("localhost", "******", "*****") or die ("Could not save image name
Error: " . mysql_error());
mysql_select_db("project") or die("Could not select database");
mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");
if($result) { echo "Image name saved into database
"; }
}
?>
And the Examples form I have to use is this:
<form action=addMember.php method=post enctype="multipart/form-data">
<table border="0" cellspacing="0" align=center cellpadding="3" bordercolor="#cccccc">
<tr>
<td>File:</td>
<td><input type="file" name="filep" size=45></td>
</tr>
<tr>
<td colspan=2><p align=center>
<input type=submit name=action value="Load">
</td>
</tr>
</table>
</form>
PS: Images file is open for writing to.
This question is related to
php
mysql
database
image-processing
file-upload
<form method="post" action="addMember.php" enctype="multipart/form-data">
<p>
Please Enter the Band Members Name.
</p>
<p>
Band Member or Affiliates Name:
</p>
<input type="text" name="nameMember"/>
<p>
Please Enter the Band Members Position. Example:Drums.
</p>
<p>
Band Position:
</p>
<input type="text" name="bandMember"/>
<p>
Please Upload a Photo of the Member in gif or jpeg format. The file name should be named after the Members name. If the same file name is uploaded twice it will be overwritten! Maxium size of File is 35kb.
</p>
<p>
Photo:
</p>
<input type="hidden" name="size" value="350000">
<input type="file" name="photo">
<p>
Please Enter any other information about the band member here.
</p>
<p>
Other Member Information:
</p>
<textarea rows="10" cols="35" name="aboutMember">
</textarea>
<p>
Please Enter any other Bands the Member has been in.
</p>
<p>
Other Bands:
</p>
<input type="text" name="otherBands" size=30 />
<br/>
<br/>
<input TYPE="submit" name="upload" title="Add data to the Database" value="Add Member"/>
</form>
save it as addMember.php
<?php
//This is the directory where images will be saved
$target = "your directory";
$target = $target . basename( $_FILES['photo']['name']);
//This gets all the other information from the form
$name=$_POST['nameMember'];
$bandMember=$_POST['bandMember'];
$pic=($_FILES['photo']['name']);
$about=$_POST['aboutMember'];
$bands=$_POST['otherBands'];
// Connects to your Database
mysql_connect("yourhost", "username", "password") or die(mysql_error()) ;
mysql_select_db("dbName") or die(mysql_error()) ;
//Writes the information to the database
mysql_query("INSERT INTO tableName (nameMember,bandMember,photo,aboutMember,otherBands)
VALUES ('$name', '$bandMember', '$pic', '$about', '$bands')") ;
//Writes the photo to the server
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{
//Tells you if its all ok
echo "The file ". basename( $_FILES['photo']['name']). " has been uploaded, and your information has been added to the directory";
}
else {
//Gives and error if its not
echo "Sorry, there was a problem uploading your file.";
}
?>
in the above code one little bug ,i fixed that bug.
Your part:
$result = mysql_connect("localhost", "******", "*****") or die ("Could not save image name
Error: " . mysql_error());
mysql_select_db("project") or die("Could not select database");
mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");
if($result) { echo "Image name saved into database
";
Doesn't make much sense, your connection shouldn't be named $result but that is a naming issue not a coding one.
What is a coding issue is if($result), your saying if you can connect to the database regardless of the insert query failing or succeeding you will output "Image saved into database".
Try adding do
$realresult = mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");
and change the if($result) to $realresult
I suspect your query is failing, perhaps you have additional columns or something?
Try copy/pasting your query, replacing the ".$_FILES['filep']['name']." with test and running it in your query browser and see if it goes in.
Adding the following avoids problems with quotes in file names, e.g.
"freddy's pic.jpg"
which are acceptable on some operating systems.
Before:
$pic=($_FILES['photo']['name']);
After:
$pic=(mysql_real_escape_string($_FILES['photo']['name']));
Your part:
$result = mysql_connect("localhost", "******", "*****") or die ("Could not save image name
Error: " . mysql_error());
mysql_select_db("project") or die("Could not select database");
mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");
if($result) { echo "Image name saved into database
";
Doesn't make much sense, your connection shouldn't be named $result but that is a naming issue not a coding one.
What is a coding issue is if($result), your saying if you can connect to the database regardless of the insert query failing or succeeding you will output "Image saved into database".
Try adding do
$realresult = mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");
and change the if($result) to $realresult
I suspect your query is failing, perhaps you have additional columns or something?
Try copy/pasting your query, replacing the ".$_FILES['filep']['name']." with test and running it in your query browser and see if it goes in.
Here is the answer for those of you looking like I did all over the web trying to find out how to do this task. Uploading a photo to a server with the file name stored in a mysql database and other form data you want in your Database. Please let me know if it helped.
Firstly the form you need:
<form method="post" action="addMember.php" enctype="multipart/form-data">
<p>
Please Enter the Band Members Name.
</p>
<p>
Band Member or Affiliates Name:
</p>
<input type="text" name="nameMember"/>
<p>
Please Enter the Band Members Position. Example:Drums.
</p>
<p>
Band Position:
</p>
<input type="text" name="bandMember"/>
<p>
Please Upload a Photo of the Member in gif or jpeg format. The file name should be named after the Members name. If the same file name is uploaded twice it will be overwritten! Maxium size of File is 35kb.
</p>
<p>
Photo:
</p>
<input type="hidden" name="size" value="350000">
<input type="file" name="photo">
<p>
Please Enter any other information about the band member here.
</p>
<p>
Other Member Information:
</p>
<textarea rows="10" cols="35" name="aboutMember">
</textarea>
<p>
Please Enter any other Bands the Member has been in.
</p>
<p>
Other Bands:
</p>
<input type="text" name="otherBands" size=30 />
<br/>
<br/>
<input TYPE="submit" name="upload" title="Add data to the Database" value="Add Member"/>
</form>
Then this code processes you data from the form:
<?php
// This is the directory where images will be saved
$target = "your directory";
$target = $target . basename( $_FILES['photo']['name']);
// This gets all the other information from the form
$name=$_POST['nameMember'];
$bandMember=$_POST['bandMember'];
$pic=($_FILES['photo']['name']);
$about=$_POST['aboutMember'];
$bands=$_POST['otherBands'];
// Connects to your Database
mysqli_connect("yourhost", "username", "password") or die(mysqli_error()) ;
mysqli_select_db("dbName") or die(mysqli_error()) ;
// Writes the information to the database
mysqli_query("INSERT INTO tableName (nameMember,bandMember,photo,aboutMember,otherBands)
VALUES ('$name', '$bandMember', '$pic', '$about', '$bands')") ;
// Writes the photo to the server
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{
// Tells you if its all ok
echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory";
}
else {
// Gives and error if its not
echo "Sorry, there was a problem uploading your file.";
}
?>
Code edited from www.about.com
You have an ID for each photo so my suggestion is you rename the photo. For example you rename it by the date
<?php
$date = getdate();
$name .= $date[hours];
$name .= $date[minutes];
$name .= $date[seconds];
$name .= $date[year];
$name .= $date[mon];
$name .= $date[mday];
?>
note: don't forget the file extension of your file or you can generate random string for the photo, but I would not recommend that. I would also recommend you to check the file extension before you upload it to your directory.
<?php
if ((($_FILES["photo"]["type"] == "image/jpeg")
|| ($_FILES["photo"]["type"] == "image/pjpg"))
&& ($_FILES["photo"]["size"] < 100000000))
{
move_uploaded_file($_FILES["photo"]["tmp_name"], $target.$name);
if(mysql_query("your query"))
{
//success handling
}
else
{
//failed handling
}
}
else
{
//error handling
}
?>
Hope this might help.
If you want to input more data into the form, you simply access the submitted data through $_POST.
If you have
<input type="text" name="firstname" />
you access it with
$firstname = $_POST["firstname"];
You could then update your query line to read
mysql_query("INSERT INTO dbProfiles (photo,firstname)
VALUES('{$filename}','{$firstname}')");
Note: Always filter and sanitize your data.
Your part:
$result = mysql_connect("localhost", "******", "*****") or die ("Could not save image name
Error: " . mysql_error());
mysql_select_db("project") or die("Could not select database");
mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");
if($result) { echo "Image name saved into database
";
Doesn't make much sense, your connection shouldn't be named $result but that is a naming issue not a coding one.
What is a coding issue is if($result), your saying if you can connect to the database regardless of the insert query failing or succeeding you will output "Image saved into database".
Try adding do
$realresult = mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");
and change the if($result) to $realresult
I suspect your query is failing, perhaps you have additional columns or something?
Try copy/pasting your query, replacing the ".$_FILES['filep']['name']." with test and running it in your query browser and see if it goes in.
If you want to input more data into the form, you simply access the submitted data through $_POST.
If you have
<input type="text" name="firstname" />
you access it with
$firstname = $_POST["firstname"];
You could then update your query line to read
mysql_query("INSERT INTO dbProfiles (photo,firstname)
VALUES('{$filename}','{$firstname}')");
Note: Always filter and sanitize your data.
Here is the answer for those of you looking like I did all over the web trying to find out how to do this task. Uploading a photo to a server with the file name stored in a mysql database and other form data you want in your Database. Please let me know if it helped.
Firstly the form you need:
<form method="post" action="addMember.php" enctype="multipart/form-data">
<p>
Please Enter the Band Members Name.
</p>
<p>
Band Member or Affiliates Name:
</p>
<input type="text" name="nameMember"/>
<p>
Please Enter the Band Members Position. Example:Drums.
</p>
<p>
Band Position:
</p>
<input type="text" name="bandMember"/>
<p>
Please Upload a Photo of the Member in gif or jpeg format. The file name should be named after the Members name. If the same file name is uploaded twice it will be overwritten! Maxium size of File is 35kb.
</p>
<p>
Photo:
</p>
<input type="hidden" name="size" value="350000">
<input type="file" name="photo">
<p>
Please Enter any other information about the band member here.
</p>
<p>
Other Member Information:
</p>
<textarea rows="10" cols="35" name="aboutMember">
</textarea>
<p>
Please Enter any other Bands the Member has been in.
</p>
<p>
Other Bands:
</p>
<input type="text" name="otherBands" size=30 />
<br/>
<br/>
<input TYPE="submit" name="upload" title="Add data to the Database" value="Add Member"/>
</form>
Then this code processes you data from the form:
<?php
// This is the directory where images will be saved
$target = "your directory";
$target = $target . basename( $_FILES['photo']['name']);
// This gets all the other information from the form
$name=$_POST['nameMember'];
$bandMember=$_POST['bandMember'];
$pic=($_FILES['photo']['name']);
$about=$_POST['aboutMember'];
$bands=$_POST['otherBands'];
// Connects to your Database
mysqli_connect("yourhost", "username", "password") or die(mysqli_error()) ;
mysqli_select_db("dbName") or die(mysqli_error()) ;
// Writes the information to the database
mysqli_query("INSERT INTO tableName (nameMember,bandMember,photo,aboutMember,otherBands)
VALUES ('$name', '$bandMember', '$pic', '$about', '$bands')") ;
// Writes the photo to the server
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{
// Tells you if its all ok
echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory";
}
else {
// Gives and error if its not
echo "Sorry, there was a problem uploading your file.";
}
?>
Code edited from www.about.com
Your part:
$result = mysql_connect("localhost", "******", "*****") or die ("Could not save image name
Error: " . mysql_error());
mysql_select_db("project") or die("Could not select database");
mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");
if($result) { echo "Image name saved into database
";
Doesn't make much sense, your connection shouldn't be named $result but that is a naming issue not a coding one.
What is a coding issue is if($result), your saying if you can connect to the database regardless of the insert query failing or succeeding you will output "Image saved into database".
Try adding do
$realresult = mysql_query("INSERT into dbProfiles (photo) VALUES('".$_FILES['filep']['name']."')");
and change the if($result) to $realresult
I suspect your query is failing, perhaps you have additional columns or something?
Try copy/pasting your query, replacing the ".$_FILES['filep']['name']." with test and running it in your query browser and see if it goes in.
You have an ID for each photo so my suggestion is you rename the photo. For example you rename it by the date
<?php
$date = getdate();
$name .= $date[hours];
$name .= $date[minutes];
$name .= $date[seconds];
$name .= $date[year];
$name .= $date[mon];
$name .= $date[mday];
?>
note: don't forget the file extension of your file or you can generate random string for the photo, but I would not recommend that. I would also recommend you to check the file extension before you upload it to your directory.
<?php
if ((($_FILES["photo"]["type"] == "image/jpeg")
|| ($_FILES["photo"]["type"] == "image/pjpg"))
&& ($_FILES["photo"]["size"] < 100000000))
{
move_uploaded_file($_FILES["photo"]["tmp_name"], $target.$name);
if(mysql_query("your query"))
{
//success handling
}
else
{
//failed handling
}
}
else
{
//error handling
}
?>
Hope this might help.
Adding the following avoids problems with quotes in file names, e.g.
"freddy's pic.jpg"
which are acceptable on some operating systems.
Before:
$pic=($_FILES['photo']['name']);
After:
$pic=(mysql_real_escape_string($_FILES['photo']['name']));
<form method="post" action="addMember.php" enctype="multipart/form-data">
<p>
Please Enter the Band Members Name.
</p>
<p>
Band Member or Affiliates Name:
</p>
<input type="text" name="nameMember"/>
<p>
Please Enter the Band Members Position. Example:Drums.
</p>
<p>
Band Position:
</p>
<input type="text" name="bandMember"/>
<p>
Please Upload a Photo of the Member in gif or jpeg format. The file name should be named after the Members name. If the same file name is uploaded twice it will be overwritten! Maxium size of File is 35kb.
</p>
<p>
Photo:
</p>
<input type="hidden" name="size" value="350000">
<input type="file" name="photo">
<p>
Please Enter any other information about the band member here.
</p>
<p>
Other Member Information:
</p>
<textarea rows="10" cols="35" name="aboutMember">
</textarea>
<p>
Please Enter any other Bands the Member has been in.
</p>
<p>
Other Bands:
</p>
<input type="text" name="otherBands" size=30 />
<br/>
<br/>
<input TYPE="submit" name="upload" title="Add data to the Database" value="Add Member"/>
</form>
save it as addMember.php
<?php
//This is the directory where images will be saved
$target = "your directory";
$target = $target . basename( $_FILES['photo']['name']);
//This gets all the other information from the form
$name=$_POST['nameMember'];
$bandMember=$_POST['bandMember'];
$pic=($_FILES['photo']['name']);
$about=$_POST['aboutMember'];
$bands=$_POST['otherBands'];
// Connects to your Database
mysql_connect("yourhost", "username", "password") or die(mysql_error()) ;
mysql_select_db("dbName") or die(mysql_error()) ;
//Writes the information to the database
mysql_query("INSERT INTO tableName (nameMember,bandMember,photo,aboutMember,otherBands)
VALUES ('$name', '$bandMember', '$pic', '$about', '$bands')") ;
//Writes the photo to the server
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{
//Tells you if its all ok
echo "The file ". basename( $_FILES['photo']['name']). " has been uploaded, and your information has been added to the directory";
}
else {
//Gives and error if its not
echo "Sorry, there was a problem uploading your file.";
}
?>
in the above code one little bug ,i fixed that bug.
If you want to input more data into the form, you simply access the submitted data through $_POST.
If you have
<input type="text" name="firstname" />
you access it with
$firstname = $_POST["firstname"];
You could then update your query line to read
mysql_query("INSERT INTO dbProfiles (photo,firstname)
VALUES('{$filename}','{$firstname}')");
Note: Always filter and sanitize your data.
Source: Stackoverflow.com