Question

问题：

我有一个用JSON编码的变量，我试图弄清楚如何使用jQuery / AJAX将这个变量发送到PHP页面。

这是我到目前为止所尝试过的。

使用http://blueimp.github.io/jQuery-File-Upload/
根据以下JS代码由jQuery处理。

JS代码：

<script>    
    $(function () {
        'use strict';
        // Server-side upload handler:
        var url = 'process.php';

        $('#fileupload').fileupload({
            url: url,
            autoUpload: true,
            acceptFileTypes: /(\.|\/)(txt)$/i,
            maxFileSize: 5000000, // 5 MB               
            done: function (e, data) {

                setTimeout(function(){
                    $("#loading").html("Loading...");}, 1000);

                var formUrl = 'exec.php',

                // You'll have to re-encode to JSON, probably:
                formPerspective = JSON.stringify(data.result.formPerspective),

                // This is the newly added value ( Maybe this needs JSON aswell? )
                formTxtfile = JSON.stringify(data.result.formTxtfile),

                // Generate the form
                $form = $('<form action="'+ formUrl +'" method="post"></form>')
                        .append('<input type="hidden" name="Perspective" value="' + formPerspective + '">')
                        .append('<input type="hidden" name="Datafile" value="' + formTxtfile + '">')
                        .append('<input type="hidden" name="form_submitted">');

                // Submit the form
                $form.submit();
            },

            progressall: function (e, data) {
                var progress = parseInt(data.loaded / data.total * 100, 10);
                $('#progress .progress-bar').css(
                    'width',
                    progress + '%'
                );
            }
        }).prop('disabled', !$.support.fileInput)
            .parent().addClass($.support.fileInput ? undefined : 'disabled');
    });
</script>

如您所见，它根据以下代码调用PHP处理程序。

PHP代码：

<?php
    session_start();

    $folder      = 'upload';

    if (!empty($_FILES))
    {
        // Set temporary name
        $tmp    = $_FILES['files']['tmp_name'];

        // Set target path and file name
        $target = $folder . '/' . $_FILES['files']['name'];

        // Upload file to target folder
        $status = move_uploaded_file($tmp, $target);

        if ($status)
        {
            // Set session with txtfile name
            $_SESSION['txtfile']    = $_FILES['files']['name'];

            $text_file    = file('upload/' . $_SESSION['txtfile']);

            foreach ($text_file as $line_number => $line)
            {               
                if (strpos($line, "\t") === 0)
                {
                    // Remove commented column names and first \t
                    $dimensions = explode("\t", preg_replace("#\t/.+#", '', substr($line, 1)));
                }               
            }

            // Set dimensions
            $_SESSION['dimensions'] = str_replace('\n', '', $dimensions);

            $jsonReturn = array(
                'formPerspective'   => $_SESSION['dimensions'],
                'formTxtfile'       => $_SESSION['txtfile']
            );

            // Convert to JSON
            $encode = json_encode($jsonReturn);

            echo $encode;
        }
    }
?>

所需解决方案：

我想在$encode中获取信息并使用POST将其发送到“oe.lc”页面。在OE.lc上，使用以下方法选择变量：

$_POST["Perspective"]

此外，OE.lc使用以下方式检查表单是否已提交：

$_POST["form_submitted"]

JS应该在提交时重定向到OE.lc（现在它重定向到explorer.php）。

有什么想法吗？

Answer 1

尝试简单回显$encode，然后使用die();

退出脚本

编辑：您说要在表单中添加额外的值，首先需要更改确定JSON中返回Javascript的php文件：

$jsonReturn = array(
    'formPerspective' => $dimensions,
    'formTxtfile' => $_SESSION['txtfile'],
);

// Convert to JSON
$encode = str_replace('\n', '', json_encode($jsonReturn));

// The above will print out the following when echoed:
// {'formPerspective':["legs","skincover","weight","intelligence","speed"], 'formTxtfile': 'Not sure what this holds as data'}

die($encode);

然后，数据应位于您配置的data.result回调中的done。

您可能希望将浏览器转发给oe.lc，如何生成表单并提交？

编辑我们应该在这里添加新的准确值：

$('#fileupload').fileupload({
        // ...
        done: function (e, data) {
                // Determine the url:
            var formUrl = 'http://oe.lc',

                // You'll have to re-encode to JSON, probably:
                formPerspective = JSON.stringify(data.result.formPerspective),

                // This is the newly added value ( Maybe this needs JSON aswell? )
                formTxtfile = JSON.stringify(data.result.formTxtfile),


                // Generate the form
                $form = $("<form action='"+ formUrl +"' method='post'></form>")
                        .append("<input type='hidden' name='Perspective' value='" + formPerspective + "'>")
                        .append("<input type='hidden' name='Datafile' value='" + formTxtfile + "'>")
                        .append("<input type='hidden' name='form_submitted'>")
                        .appendTo('body');

                // Submit the form
                $form.submit();
        },
        // ...
    });

请告诉我这是否适合您！

Answer 2

这是完整的代码（虽然在Firefox中不起作用）：

<script>    
    $(function () {
        'use strict';

        // Server-side PHP upload handler
        var url = 'process.php';

        // File upload handler
        $('#fileupload').fileupload({
            url: url,
            dataType: 'json',
            autoUpload: true,
            acceptFileTypes: /(\.|\/)(txt)$/i,
            maxFileSize: 5000000, // 5 MB               
            done: function (e, data) {

                // Output alert message
                setTimeout(function(){
                    $("#loading").html("Loading...");}, 1000);

                // Form destination
                var formUrl = 'test.php',

                // Variables to be sent with the form
                formPerspective = data.result.formPerspective,
                formTxtfile = data.result.formTxtfile,

                // Generate the form
                $form = $("<form action='"+ formUrl +"' method='post'></form>")
                        .append("<input type='hidden' name='Perspective' value='" + formPerspective + "'>")
                        .append("<input type='hidden' name='Datafile' value='" + formTxtfile + "'>")
                        .append("<input type='hidden' name='form_submitted' value='true'>")
                        .appendTo('body');

                // Submit the form
                $form.submit();
            },

            // Progress indicator
            progressall: function (e, data) {
                var progress = parseInt(data.loaded / data.total * 100, 10);
                $('#progress .progress-bar').css(
                    'width',
                    progress + '%'
                );
            }
        }).prop('disabled', !$.support.fileInput)
            .parent().addClass($.support.fileInput ? undefined : 'disabled');
    });
</script>

在PHP中编码JSON并使用POST提交

2 个答案: