[英]PHP database SQL insert
我試圖將數據從我的應用程序寫入外部數據庫。 我的PHP頁面沒有任何響應。 當我查看發送到PHP頁面的變量時,它們收到的效果很好,並且此時沒有任何問題。 但是,當我使用SQL進行INSERT時,它將出錯。 (我認為)。 當我進入我的PHPadmin頁面並執行下一個SQL命令時,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')
所以我有下一個數據庫;
ID(PRIMARY KEY AUTOINCREMENT),
FBid (varchar(255)),
Datum (datetime),
Lat (Double),
Longi(Double),
Image(Blob).
這是我的php頁面:
<?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";
}
感謝大伙們!
此致Stijn
查看數據庫連接,您正在使用mysqli進行錯誤的准備 。 在INSERT語句中,它看起來像是PDO版本。 如果要使用PDO版本,請查看此鏈接 。 您不能混合使用PDO和mysqli。 mysqli_prepare的過程樣式如下:
$stmt = mysqli_prepare($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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