[英]PHP database SQL insert
I am trying to write data form mine app to a external database. 我试图将数据从我的应用程序写入外部数据库。 I just get no response form my PHP page. 我的PHP页面没有任何响应。 When I look at the variables that I send to the PHP page, they are received good and nothing goes wrong at that moment. 当我查看发送到PHP页面的变量时,它们收到的效果很好,并且此时没有任何问题。 But when I do an INSERT with SQL it goes wrong. 但是,当我使用SQL进行INSERT时,它将出错。 (I think). (我认为)。 When I go to mine PHPadmin page and I do next SQL command, it works: 当我进入我的PHPadmin页面并执行下一个SQL命令时,它可以正常工作:
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')
So i have next database; 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).
And this is my php page: 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";
}
Thank you guys! 感谢大伙们!
Regards, Stijn 此致Stijn
Looking at the database connection, you are using mysqli prepare wrongly. 查看数据库连接,您正在使用mysqli进行错误的准备 。 In the INSERT statement, it looks like a PDO version. 在INSERT语句中,它看起来像是PDO版本。 If you want to use PDO version, have a look at this link . 如果要使用PDO版本,请查看此链接 。 You can't mix PDO and mysqli. 您不能混合使用PDO和mysqli。 The procedural style for mysqli_prepare is like below: mysqli_prepare的过程样式如下:
$stmt = mysqli_prepare($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);
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