PHP 数据库 SQL 插入
PHP database SQL insert
我正在尝试将我的应用程序中的数据写入外部数据库。
我的 PHP 页面没有任何回应。当我查看发送到 PHP 页面的变量时,它们收到的很好,当时没有任何问题。
但是当我使用 SQL 执行 INSERT 时,它出错了。 (我认为)。
当我进入我的 PHP 管理页面并执行下一个 SQL 命令时,它有效:
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";
}
谢谢大家!
此致,
斯蒂恩
查看数据库连接,您正在使用 mysqli prepare wrongly. In the INSERT statement, it looks like a PDO version. If you want to use PDO version, have a look at this link。你不能混合使用 PDO 和 mysqli。 mysqli_prepare 的程序风格如下:
$stmt = mysqli_prepare($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);
我正在尝试将我的应用程序中的数据写入外部数据库。 我的 PHP 页面没有任何回应。当我查看发送到 PHP 页面的变量时,它们收到的很好,当时没有任何问题。 但是当我使用 SQL 执行 INSERT 时,它出错了。 (我认为)。 当我进入我的 PHP 管理页面并执行下一个 SQL 命令时,它有效:
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";
}
谢谢大家!
此致, 斯蒂恩
查看数据库连接,您正在使用 mysqli prepare wrongly. In the INSERT statement, it looks like a PDO version. If you want to use PDO version, have a look at this link。你不能混合使用 PDO 和 mysqli。 mysqli_prepare 的程序风格如下:
$stmt = mysqli_prepare($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);