Aqui está a resposta para aqueles que parecem como eu fiz em toda a web tentando descobrir como fazer essa tarefa. Carregar uma foto para um servidor com o nome do arquivo armazenado em um banco de dados mysql e outros dados de formulário que você deseja em seu banco de dados. Por favor, deixe-me saber se ajudou.
Em primeiro lugar, o formulário que você precisa:
<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>
Em seguida, este código processa seus dados do formulário:
<?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.";
}
?>
Código editado de www.about.com
