我在将值输入到 textarea
. 其他一切都很好,你知道怎么做吗?
html格式:
<form id="formData2" action="artistuploader.php" method="post"
enctype="multipart/form-data">
<input type="hidden" name="size" value="1000000"></input>
<br/>
<input id="inputField" type="text" name="actname" placeholder="Act Name" >
<br>
<input id="inputField" type="text" name="fullname" placeholder="Full Name" >
<br>
<input id="inputField" type="text" name="genre" placeholder="Genre" >
<br>
<textarea id="inputField" name="biography" form="formData2" placeholder="Biography"<?php echo $biography; ?>></textarea>
<br>
<input id="inputField" type="file" name="artistImage" placeholder="Artwork" >
<br>
<input id="inputField" type="text" name="imagepath" placeholder="Image path URL" >
<br>
<input id="submitButton" type="submit" name="uploadArtist" value="Register Artist">
</form>
PHP
<?php
$msg = "";
//if Upload button is pressed
if (isset($_POST['uploadArtist'])){
$target = "uploads/artistPics".basename($_FILES['artistImage']['name']);
//connecting to our database
$db = mysqli_connect("127.0.0.1", "user", "pass", "tablename");
$tmp_name = $_FILES['artistImage']['tmp_name'];
$name = $_FILES['artistImage']['name'];
//getting the submitted form data
$ActName = $_POST['actname'];
$FullName = $_POST['fullname'];
$Genre = $_POST['genre'];
$ArtistPhoto = $_FILES['artistImage']['name'];
$imageURLpath = $_POST['imagepath'];
$Biography = $_POST['biography'];//having problem with this line here
//saving submitted data into database table songsDB
$sql = "INSERT INTO artistsdb (ActName,FullName,Genre,ArtistPhoto,Biography,imageURLpath) VALUES ('$ActName','$FullName','$Genre','$ArtistPhoto','$Biography','$imageURLpath')";
mysqli_query($db, $sql); //stores the submitted data into table
//now moving the uploaded image to uploads folder
if(move_uploaded_file($_FILES['artistImage']['tmp_name'], $target)){
$msg = "Uploaded Successful";
}else{
$msg = "There was a problem uploading Data";
}
}
//header("refresh:1; url=index.php"); ?>
1条答案
按热度按时间wpx232ag1#
更换您的
textarea
具有