PHP数据库SQL insert

Question

我正在尝试将数据从我的应用程序写到一个外部数据库。我只是没有从PHP页面中得到任何响应。当我查看发送到PHP页面的变量时，它们会被很好地接收，并且没有什么问题。但是，当我使用SQL进行插入时，就会出错。(我想)。当我转到我的PHPadmin页面并执行next SQL命令时，它可以工作：

INSERT INTO images (FBid,Datum,Lat,Longi,Image) 
VALUES ('1846465164',
'2016-08-25 14:14:15',10.5,5.69,'/9j/
 4AAQSkZJRgABAQAAAQABAAD/2wBDAAEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQE
 BAQEBQBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQH/2wBDAQEBAQEBAQEBAQEBAQEBAQEBAQEB')

所以我有下一个数据库；

ID(PRIMARY KEY AUTOINCREMENT), 
FBid (varchar(255)), 
Datum (datetime), 
Lat (Double), 
Longi(Double), 
Image(Blob).

这是我的php页面：

<?php



 if($_SERVER['REQUEST_METHOD']=='POST'){

   define('HOST','localhost');
 define('USER','XXXXXXXXX');
 define('PASS','XXXXXXXXX');
 define('DB','database2');

 $con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect');

 $image = $_POST['image'];
 $FBid = $_POST['FBid'];
 $date = $_POST['Date'];
 $long = $_POST['long'];
 $lat = $_POST['lat'];

 $stmt = $con->prepare(
                "INSERT INTO images (FBid,Datum,Lat,Longi,Image) 
                 VALUES (:Fbid,:date,:lat,:long,:image)"); 
        $stmt->bindParam(":Fbid",$FBid);
        $stmt->bindParam(":date", $date);
        $stmt->bindParam(":lat", $lat);
        $stmt->bindParam(":long", $long);
        $stmt->bindParam(":image","s",$image);
        $stmt->execute();

$check = mysqli_stmt_affected_rows($stmt);

 if($check == 1){
 echo "Image Uploaded Successfully";
 }else{
 echo "Error Uploading Image";
 }
 mysqli_close($con);
 }else{
 echo "Error";
 }

谢谢你们！

你好，斯蒂恩

Answer 1

查看数据库连接，您错误地使用了mysqli准备。在INSERT语句中，它看起来像PDO版本。如果您想使用PDO版本，请看一下这个链接。你不能把PDO和mysqli混在一起。mysqli_prepare的过程样式如下所示：

$stmt = mysqli_prepare($con, "INSERT INTO images VALUES (?, ?, ?, ?, ?)");
if ( !$stmt ) {
   die('mysqli error: '.mysqli_error($con);
}
mysqli_stmt_bind_param($stmt, 'ssddb', $FBid,$date,$lat,$long,$image);
if ( !mysqli_stmt_execute($stmt)) {
   die( 'stmt error: '.mysqli_stmt_error($stmt) );
}

$check = mysqli_stmt_affected_rows($stmt);
if($check == 1){
 echo 'Image successfully uploaded';
}else{
 echo 'Error uploading image';
}
mysqli_stmt_close($stmt);