在PHP MYSQL中使用POST FORM的简短方法

Question

I am using angularjs php form post to insert or update row. Currently I have working example below and I want to know if this can be written in shorter way so that it can used in both UPDATE and INSERT scenario. I want to submit the whole FORM not just the defined value. Thanks for help.

function update($con){
$postdata = file_get_contents("php://input", true);
$request = json_decode($postdata);

$idx = $request->idx;
$images = $request->images;
$collection = $request->collection;
$title = $request->title;
$description = $request->description;
$height = $request->height;
$width = $request->width;
$length = $request->length;
$weight = $request->weight;
$price = $request->price;
$availability = $request->availability;
$active = $request->active;
$method = $request->method;
$searchkeys = $request->searchkeys;
$materialused = $request->materialused;
if(is_array($images)){
    $implodeImg = implode(',', $images);
}else{
    $implodeImg = $images;
}

$sqlIn = "UPDATE prodList_1 SET 
idx = '$idx',
images = '$implodeImg',
collection = '$collection',
title = '$title',
description = '$description',
height = '$height',
width = '$width',
length = '$length',
weight = '$weight',
price = '$price',
availability = '$availability',
active = '$active',
method = '$method',
searchkeys = '$searchkeys',
materialused = '$materialused'
WHERE idx = '$idx'";

if (mysqli_query($con, $sqlIn)) {
  echo "Record Updated successfully";
} else {
  echo "Error: " . $sqlIn . "<br>" . mysqli_error($con);
}

}

Answer 1

I've never used mysqli and probably will never after doing this but I believe it would be something like the following:

function update($con) {
    $postdata = file_get_contents("php://input", true);
    $request = json_decode($postdata);

    $idx = $request->idx;
    $images = $request->images;
    $collection = $request->collection;
    $title = $request->title;
    $description = $request->description;
    $height = $request->height;
    $width = $request->width;
    $length = $request->length;
    $weight = $request->weight;
    $price = $request->price;
    $availability = $request->availability;
    $active = $request->active;
    $method = $request->method;
    $searchkeys = $request->searchkeys;
    $materialused = $request->materialused;

    if (is_array($images)) {
        $implodeImg = implode(',', $images);
    }else{
        $implodeImg = $images;
    }

    // using INSERT ... ON DUPLICATE KEY UPDATE syntax
    // make sure idx is PRIMARY KEY
    $sql = "
        INSERT INTO table 
        (idx, images, collection, title, description, height, 
         width, length, weight, price, availability, active, 
         method, searchkets, materialused) 
        VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)
        ON DUPLICATE KEY UPDATE 
        images = VALUES(images), collection = VALUES(collection), 
        title = VALUES(title), description = VALUES(description), 
        height = VALUES(height), width = VALUES(width), 
        length = VALUES(length), weight = VALUES(weight), 
        price = VALUES(price), availability = VALUES(availability), 
        active = VALUES(active), method = VALUES(method), 
        searchkeys = VALUES(searchkeys), materialused = VALUES(materialused)
    ";

    // never trust user-submitted data
    // use prepared statement
    $stmt = mysqli_prepare($con, $sql);

    $stmt->bind_param(
        $stmt,
        'sssssssssssssss',
        $idx, $implodeImg, $collection, $title, $description, 
        $height, $width, $length, $weight, $price, $availability, $active,
        $method, $searchkeys, $materialused
    );

    mysqli_stmt_execute($stmt);

    if (mysqli_stmt_affected_rows($stmt)) {
        echo "Record Updated successfully";
    } else {
        echo "Error: " . $sqlIn . "<br>" . mysqli_error($con);
    }
}

So.. in terms of your question whether this can be written in a shorter way? Not really. However, you can kill two birds with one stone using INSERT ... ON DUPLICATE KEY UPDATE syntax.