Question

我正在处理一个php表单，它将数据提交到一个表，然后将图像映射到我的mysql数据库中的第二个表。

我坚持的一点就是提交到第二张桌子，因为它有什么理由它没有缝合工作。

有人可以指出我正确的方向指向我错误的代码吗？

任何帮助都将非常感谢

<?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);
?>

Answer 1

这里你已经声明了insertimages函数但没有调用。你可以这样打电话：

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);
}

将图像上传到mysql第二个表

1 个答案: