通过PHP将数据插入SQL表的更正

Question

I've been using php and mysql for my computing project and I've just run into this small problem. I've tried countless numbers of variations to this line of code but I always seem to receive an SQL Error.

Background Information:

• HTML Document that contains a form, allowing the Admin to upload an image path, a name for the photo and a comment as well
• PHP Document containing SQL that runs the code for this. I've worked out how to insert either an image path, a name or a comment but not all three at once...

Here's the line of code that causes the problem, specifically inserting the data into the database.

$sql="INSERT INTO image_tbl (image, name, comment) VALUES ('{$_POST{[ VALUES GO HERE, WHAT SYNTAX/HOW??? ]}')";

The HTML Form is here (for anyone interested):

<form action="insertTest.php" method="POST" enctype="multipart/form-data">

Image: <input type="text" name="image" /><br>

Name: <input type="text" name="name" /><br>

Comment: <input type="text" name="comment" /><br>

<input type="submit">

</form>

PHP Doc (whole upload file, quite small, changed password for security xD):

<?php

$servername = "localhost";
$username = "root";
$password = "password";
$dbname = "admin_db";


// Create connection
$conn = new mysqli($servername = "localhost", $username = "root", $password = "password", $dbname = "admin_db");
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 


$sql="INSERT INTO image_tbl (image, name, comment) VALUES ('{$_POST{[]}')";

if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?>

Answer 1

$image = $_POST['image'];
$name = $_POST['name'];
$comment = $_POST['comment'];


$sql="INSERT INTO image_tbl (image, name, comment) VALUES ('$image', '$name', '$comment')";

Answer 2

you can set variables to the associative array (with name from html form) and include each in the sql query. Something like this:

$image = $_POST['image'];
$name = $_POST['name'];
$comment = $_POST['comment'];
$sql="INSERT INTO image_tbl (image, name, comment) VALUES ('$image','$name','$comment')";

Answer 3

Separate your variables $imgpath=$_POST["path"]; $imgname=$_POST["name"]; $imgcomment=$_POST["comment"]; now insert into table.... INSERT INTO imgtable(path,name,comment) VALUES('$imgpath','$imgname','$imgcomment')