通过AJAX使用JSON数据发布到php

191

巴克莱视觉

尝试通过json .getJSON将文本框值发布到数据库-此时，我只是想查看json是否发布到页面，我的UPDATE查询工作正常...

以下内容未按要求发布：

代码：

$(document).on("click", ".submit", function(event){
    alert($(this).text());

    var form_data = {
        FDID: $('.fdid-1').val(),
        CHOICE1: $('.choice-1').val(),
        CHOICE2: $(".choice-2").val()
        };      

    $.getJSON("modify.php",form_data,function(data){
        switch(data.retval){
            case 0: $("#status").html("Update successful!");
            break;
            case 1: $("#status").html("Unable to update!");
            break;
            default: $("#description").html("Database error, please try again.");
            break;
            }
    });
});

Modify.php：

<?php

header('content-type: application/json; charset=utf-8');
header("access-control-allow-origin: *");

$fdid = json_decode($_POST['FDID']);
$choice1 = json_decode($_POST['CHOICE1']);
var_dump($choice1);
// This is in the PHP file and sends a Javascript alert to the client
$message = $fdid;
echo "<script type='text/javascript'>alert('$message');</script>";
?>

代码的更多信息：

$.each( data, function ( i, val ) {

            ($('<div>')
            .attr({
                'data-role': 'collapsible',
                'data-content-theme': 'c',
                'data-collapsed': 'true',
                'id': 'cResults'
            })
            .html('<h4>' + this.LastName + ', ' + this.FirstName + '</h4>'
            + '<ul data-role="listview" data-filter="true" data-filter-placeholder="Search Choices..." data-inset="true" class="makecollapsibleul">'
            + '<li><form id="productForm" action="modify.php" method="post">'
            + '<label for="fdid-1">FDID:</label>'
            + '<input type="text" name="fdid-1" class="fdid-1" value=' + this.FDID + '>'
            + '</li><li>' 
            + '<label for="text-1">Choice 1:</label>'
            + '<input type="text" name="choice-1" class="choice-1" value=' + this.C1 + '>'
            + '</li><li>' 
            + '<label for="text-2">Choice 2:</label>'
            + '<input type="text" name="choice-2" class="choice-2" value=' + this.C2 + '>'
            + '</li><li>' 
            + 'IP: ' + this.IPADDRESS + '</li><input type="submit" class="submit" value="UPDATE" /></form><li>' 
            + 'Pick Date: ' + this.PICKDATE + '</li>'
            + '</ul>'))
            .appendTo('#primary');

                    //$(".title").append('<li>'+orderNum+' -- '+itemNum+'</li>');

            $('#makecollapsible').collapsibleset().trigger('create');
            $.mobile.hidePageLoadingMsg();

巴尔玛

您不应该调用json_encode()获取参数。它们使用www-form-urlencoded格式发送，PHP负责对其进行解码。

您需要调用json_encode以对要发送回的结果进行编码。

<?php

header('content-type: application/json; charset=utf-8');
header("access-control-allow-origin: *");

$fdid = $_POST['FDID'];
$choice1 = $_POST['CHOICE1'];
//var_dump($choice1);
// This is in the PHP file and sends a Javascript alert to the client
$message = $fdid;
$result = array('retval' => 0,
                'code' => "<script type='text/javascript'>alert('$message');</script>");
echo json_encode($result);
?>

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