Ajax-在PHP中使用Json时出错

I'm trying to insert JSON data into MySQL using Ajax. But I don't get anything, The query isn't working.

I tried everything, and I don't know if the problem is in Ajax, PHP or in the JSON.

Where should I start looking?

JS CODE

$("#btnprueba").click(function () {
            var array1 = [];
            $("#tabla .DataRow").each(function () {

                var firstTableData = {};
                var Aidi = $(this).find('td').eq(0).text().trim();
                    firstTableData.ID_Articulo = $(this).find('td').eq(0).text().trim();
                    firstTableData.Descripcion = $(this).find('td').eq(1).text().trim();
                    firstTableData.Valor_Venta = $(this).find('td').eq(2).text().trim();
                    firstTableData.Cantidad = $(this).find('td').eq(3).text().trim();
                    firstTableData.Subtotal = $(this).find('td').eq(4).text().trim();
                    array1.push(firstTableData);

            });

            var JsonValues = JSON.stringify(array1);
                alert(JsonValues);
            $.ajax({
                type: "POST",
                url: "GuardarDatosFactura2.php",
                data: "J="+JsonValues,
                success: function(response){
                    alert(response);
                },
                error: function(e){
                    console.log(e.message);
                },

            });

        });

Json in the variable JsonValues

[{"ID_Articulo":"001","Descripcion":"Caramelos","Valor_Venta":"6500","Cantidad":"2","Subtotal":"13000"}]

PHP CODE

<?php 
header("Content-Type: application/json; charset=UTF-8");

$CON = mysqli_connect("localhost","root","","BDfactura") or die ("error");

$data = json_decode($_POST['J'],false);

    $ID=$data->ID_Articulo;
    $Canti=$data->Cantidad;   
    $Vlr=$data->Valor_Venta;

    $cadena2 = "INSERT INTO ItemXVenta (IdArticulo, Cantidad, ValorVenta) VALUES ('$ID','$Canti','$Vlr')";
    $create2 = mysqli_query($CON,$cadena2);

    if($cadena2){
        $MSG= "Se guardaron los datos";
    }
    else{
        $MSG= "No se guardaron los datos";
    }
    echo($MSG);



?>

You have to get the array values from json decoded data like this :

<?php 
header("Content-Type: application/json; charset=UTF-8");

$CON = mysqli_connect("localhost","root","","BDfactura") or die ("error");

$data = json_decode($_POST['J'],false);

    $ID=$data[0]->ID_Articulo;
    $Canti=$data[0]->Cantidad;   
    $Vlr=$data[0]->Valor_Venta;

    $cadena2 = "INSERT INTO ItemXVenta (IdArticulo, Cantidad, ValorVenta) VALUES ('$ID','$Canti','$Vlr')";
    $create2 = mysqli_query($CON,$cadena2);

    if($cadena2){
        $MSG= "Se guardaron los datos";
    }
    else{
        $MSG= "No se guardaron los datos";
    }
    echo($MSG);



?>

This will help you, check I have loop over your received json in post request and generated insert query using for loop to create a single insert statement

<?php 
header("Content-Type: application/json; charset=UTF-8");

$CON = mysqli_connect("localhost","root","","BDfactura") or die ("error");

$data = json_decode($_POST['J'],false);

$comma = "";
$query = "INSERT INTO ItemXVenta (IdArticulo, Cantidad, ValorVenta) VALUES ";
foreach ($data as $key => $value) {
    $ID = mysqli_real_escape_string($CON, $value->ID_Articulo);
    $Canti = mysqli_real_escape_string($CON, $value->Cantidad);
    $Vlr = mysqli_real_escape_string($CON, $value->Valor_Venta);
    $query .= $comma . "('$ID','$Canti','$Vlr')";
    $comma = ",";
}

$create2 = mysqli_query($CON, $query);

if($cadena2){
    $MSG= "Se guardaron los datos";
}
else{
    $MSG= "No se guardaron los datos";
}
echo($MSG);
?>

Op, don't json encode your data yourself. Let jQuery handle that. Or at least pass the JSON encoded data in an object.

What you need to do is pass the data as a javascript object to jQuery. This way you can let jQuery and PHP do all the heavy lifting.

     $.ajax({
            type: "POST",
            url: "GuardarDatosFactura2.php",
            data: {J: array1},
            //    ^^ Javascript object, passing the array as raw info.
            //    This way it will be passed as an object to your php.
            success: function(response){
                alert(response);
            },
            error: function(e){
                console.log(e.message);
            },

        });

Then in PHP you can do

$data = $_POST['J'];

PHP should have transformed it already to a native php object/array.

Just do a var_dump($data); die(); to see what it has become.

If your object is very deeply nested php might not autodecode it. Then encode it and pass the encoded var.

data: {J: JSON.stringify(array1)},

and in php

$data = json_decode($_POST['J'], false);

Notice, not asked, but need to know!

You are using unprepared variables in your sql.

"INSERT INTO ItemXVenta (IdArticulo, Cantidad, ValorVenta) VALUES ('$ID','$Canti','$Vlr')";

this is bad. If your website will become front facing, Anyone can then dump your database, read out your database and wreak havoc on your database because you're providing them direct access to your internals.

Please use prepared statements and never, ever use a query without prepared statements. Also you're not counting for duplicate keys. If a product already exists it should be updated, not re-inserted.

if ($stmt = $mysqli->prepare("INSERT INTO ItemXVenta (IdArticulo, Cantidad, ValorVenta) VALUES (?,?,?) ON DUPLICATE KEY UPDATE Cantidad=?, ValorVenta=?")) {
      $stmt->bind_param("i", $ID);
      $stmt->bind_param("s", $Canti);
      $stmt->bind_param("s", $Vlr);
      $stmt->bind_param("s", $Canti);
      $stmt->bind_param("s", $Vlr);

      mysqli_stmt_execute($stmt);
}