将图像上传到mysql第二张表

-1

我正在处理一个php表单，它将数据提交到一个表格，然后将图像提交到我的mysql数据库中的第二个表格。将图像上传到mysql第二张表

我坚持的位正在提交到第二个表为什么永远没有接缝工作的原因。

有人可以请指出我在正确的方向，我错了这个代码？

任何帮助都将不胜感激

<?php 
/* 
Attempt MySQL server connection. Assuming you are running MySQL 
server with default setting (user 'root' with no password) 
*/ 
$link = mysqli_connect("localhost", "***", "***", "***"); 

// Check connection 
if($link === false){ 
    die("ERROR: Could not connect. " . mysqli_connect_error()); 
} 

// Escape user inputs for security 
$id = mysqli_real_escape_string($link, $_POST['id']); 
$title = mysqli_real_escape_string($link, $_POST['title']); 
$price = mysqli_real_escape_string($link, $_POST['price']); 
$sqm = mysqli_real_escape_string($link, $_POST['sqm']); 
$sqm_land = mysqli_real_escape_string($link, $_POST['sqm_land']); 
$type = mysqli_real_escape_string($link, $_POST['type']); 
$area = mysqli_real_escape_string($link, $_POST['area']); 
$location = mysqli_real_escape_string($link, $_POST['location']); 
$bedroom = mysqli_real_escape_string($link, $_POST['bedroom']); 
$terrace = mysqli_real_escape_string($link, $_POST['terrace']); 
$orientation = mysqli_real_escape_string($link, $_POST['orientation']); 
$water = mysqli_real_escape_string($link, $_POST['water']); 
$seaview = mysqli_real_escape_string($link, $_POST['seaview']); 
$pool = mysqli_real_escape_string($link, $_POST['pool']); 
$ownerinfo = mysqli_real_escape_string($link, $_POST['ownerinfo']); 
$gaddress = mysqli_real_escape_string($link, $_POST['gaddress']); 
$description = mysqli_real_escape_string($link, $_POST['description']); 
$image = mysqli_real_escape_string($link, $_POST['image']); 
$lastid = mysqli_real_escape_string($link, $_POST['lastid']); 
$seq = mysqli_real_escape_string($link, $_POST['seq']); 



// attempt insert query execution 
$sql = "INSERT INTO property (title, price, sqm, sqm_land, type, area, location, bedroom, terrace, orientation, water, seaview, pool, ownerinfo, gaddress, description) VALUES 
('$title', '$price', '$sqm', '$sqm_land', '$type', '$area', '$location', '$bedroom', '$terrace', '$orientation', '$water', '$seaview', '$pool', '$ownerinfo', '$gaddress', '$description')"; 

function insertimages($image,$lastid,$seq){ 
     $query="insert into images(imagepath,property_id,imageorder) values('".$image."','".$lastid."','".$seq."')"; 
     $this->execQuery($query); 
    } 
if(mysqli_query($link, $sql)){ 
    echo "Records added successfully."; 
} else{ 
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link); 
} 




// close connection 
mysqli_close($link); 
?>

来源

2016-04-22 user2827404

你在哪里调用函数insertimages（）？ –

函数insertimages（$ image，$ lastid，$ seq）{ $ query =“insert into images（imagepath，property_id，imageorder）values（'”。$ image。“'，'”。$ lastid。“'，' ”。$起。 “'）”; $ this-> execQuery（$ query）; – user2827404

在这里，你需要声明insertimages功能，但没有要求。你可以这样调用：

if(mysqli_query($link, $sql)){ 
    insertimages($image,$lastid,$seq); 
    echo "Records added successfully."; 
} else{ 
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link); 
}

来源

2016-04-22 13:47:12

将图像上传到mysql第二张表

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