从PHP脚本返回JSON和HTML

Question

您在这里搜索了问题,但找不到任何东西.我是编写PHP和jQuery的新手,所以请耐心等待.

我正在尝试做的是使用jQuery向我的脚本发送ajax请求,该脚本对我的数据库中的数据运行mysql查询,并使用php的json_encode将其序列化为JSON格式.然后使用可用的json2.js脚本解析响应.所有这一切都很好,但我也希望从这个脚本中返回除JSON之外的更多数据.

主要是,我想在json_encode之前回显以下行:

echo "<h1 style='margin-left: 25px;'>$num_rows Comments for $mysql_table</h1>";

但是,我的jQuery正在评估ajax成功期间的整个响应,使得json.parse函数由于脚本的返回格式无效而失败.

        success: function(data) {
            //retrieve comments to display on page by parsing them to a JSON object
            var obj = JSON.parse(data);
                    //loop through all items in the JSON array
                    for (var x = 0; x < obj.length; x++) {
                        //Create a container for the new element
                        var div = $("<div>").addClass("bubble").appendTo("#comments");
                        //Add author name and comment to container
                        var blockquote = $("<blockquote>").appendTo(div);
                            $("<p>").text(obj[x].comment).appendTo(blockquote);
                        var cite = $("<cite>").appendTo(div);
                            $("<strong>").text(obj[x].name).appendTo(cite);
                            $("<i>").text(obj[x].datetime).appendTo(cite);
                    }
                $("#db").attr("value", '' + initialComments + '');
    }   

有谁知道我怎么能返回html行以及json_encode使用这个脚本不仅仅是json人口？

谢谢你,这个网站在回答我的菜鸟问题时非常精彩.

我的PHP:`

    for ($x = 0, $numrows = mysql_num_rows($result); $x < $numrows; $x++) {
$row = mysql_fetch_assoc($result);
    $comments[$x] = array("name" => stripslashes($row["name"]), "comment" => stripslashes($row["comment"]), "datetime" => date("m/d/Y g:i A", strtotime($comment['datetime'])));        
}

//echo "<h1 style='margin-left: 25px;'>$num_rows Comments for $mysql_table</h1>";

$response = json_encode($comments);
echo $response;`

Answer 1

不要echo行,将其保存在变量中.构造一个简单的数组 $response = array( 'html' => $the_line_you_wanted_to_echo, 'jsobject' => $the_object_you_were_going_to_send_back );并将其发送回(via json_encode).

此外,您不需要json2.js,jQuery有一个优秀的JSON解析器.

你可以像这样加载 $.get( 'your/url', { params : here }, success, 'JSON' );

已更改以匹配新引入的迭代.

for ($x = 0, $num_rows = mysql_num_rows($result); $x < $num_rows; $x++) {
    $row = mysql_fetch_assoc($result);
    $comments[$x] = array(
        "name" => stripslashes($row["name"]), 
        "comment" => stripslashes($row["comment"]), 
        "datetime" => date("m/d/Y g:i A", strtotime($comment['datetime']))
    );        
}

$html = "<h1 style='margin-left: 25px;'>$num_rows Comments for $mysql_table</h1>";

echo json_encode(array( 'comments' => $comments, 'html' => $html ));

那么,在你的javascript中,你有

function success( parsedObject ){
    parsedObject.html; // "<h1 style..."
    parsedObject.comments; // an array of objects
    parsedObject.comments[0].name 
    + " on " + parsedObject.comments[0].datetime 
    + " said \n" + parsedObject.comments[0].comment; // for example
}

Answer 2

如上所述,只需将您想要的所有数据放回到数组中并对其进行编码.

<?php

echo json_encode(array(
    'html' => $html,
    'foo' => $bar,
    'bar' => $baz
));

?>

如上所述,你不需要json2.js.您可以通过将数据类型指定为json来使用jQuery的任何ajax函数解析JSON数据.

$.ajax({
    type: 'POST',
    url: 'path/to/php/script.php',
    dataType: 'json',
    data: 'foo=bar&baz=whatever',
    success: function($data) {
        var html = $data.html;
        var foo = $data.foo;
        var bar = $data.bar;

        // Do whatever.
    }
});

编辑几乎是Horia所说的.我能看到的唯一其他变体是你想要同一个数组中的所有内容.

例如:

PHP:

<?php

// You have your comment array sent up as you want as $comments
// Then just prepend the HTML string onto the beginning of your comments array.
// So now $comments[0] is your HTML string and everything past that is your comments.
$comments = array_unshift($comments, $your_html_string);

echo json_encode($comments);

?>

jQuery的:

$.ajax({
    type: 'POST',
    url: 'path/to/php/script.php',
    dataType: 'json',
    data: 'foo=bar&baz=whatever',
    success: function($comments) {
        // Here's your html string.
        var html = $comments[0];

        // Make sure to start at 1 or you're going to get your HTML string twice.
        // You could also skip storing it above, start at 0, and add a bit to the for loop:
        // if x == 0 then print the HTML string else print comments.
        for (var x = 1; x < $comments.length; x++) {
            // Do what you want with your comments.
            // Accessed like so:
            var name = $comments[x].name;
            var comment = $comments[x].comment;
            var datetime = $comments[x].datetime;
        }
    }
});