我正在尝试将图像名称与数据库中的其他数据一起存储,但无法做到。数据库中没有插入任何内容。但是当我在数据库配置之前关闭大括号}时,图像会上传到目录中。但是当我在插入后关闭函数大括号}时,我调用了一个未定义的函数上传()。这是代码:
$imagesub = isset( $_FILES['image'] );
if ( $imagesub )
{
$output=upload();
}
return $output;
function upload()
{
include_once "class.php";
$uploader = new Uploader( "image" );
$uploader->saveIn("images");
$fileUploaded = $uploader->save();
$db = new mysqli("localhost", "root","","learndb");
if ($db->connect_error) {
die("Connection failed this is the error: " . $db->connect_error);
}
$stmt = $db->prepare("INSERT INTO studentrecords (Name, email, Phone, school,dob,father,feereceived,due,image) VALUES (?,?,?,?,?,?,?,?,?)");
if($stmt)
{
$stmt->bind_param("ssisssiis",$name,$email,$phone,$school,$dob,$father,$feereceived,$due,$fileUploaded);
$stmt->execute();
$out="<center>information entered.</center>";
echo "$out";
}
else
{
$out="DATABASE ERROR!!!";
echo "$out";
}
return $out;
}
这里是函数save()
public function save(){
$folderIsWriteAble = is_writable( $this->destination );
if( $folderIsWriteAble ){
$name = "$this->destination/$this->filename";
if($succes = move_uploaded_file( $this->fileData, $name ))
{
return $name;
}
}
}
最佳答案
@Micky 您在插入查询中只定义了 9 个字段,但您在 ´bind_param()` 中传递了 10 个参数。
$stmt->bind_param("ssisssiis",$name,$email,$phone,$school,$dob,$father,$feereived,$due,$fileUploaded);
/p>
应该是
$stmt->bind_param($name,$email,$phone,$school,$dob,$father,$feereceived,$due,$fileUploaded);
如果它不能解决您的问题,那么 var_dump 您的 $fileUploaded 变量并确保您为用于存储文件名的列定义了兼容的数据类型。
关于php - 图片已上传但未存储,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35447521/