Question

这段代码工作正常我关心的是它的输出

Javascript方

        function AjaxRetrieve()
        {
            var rid = document.getElementById('trg').value;
            $.get('includes/getChat.php?chat='+uid + '&rid=' + rid + '&name=' + user,function(data)
            {

               $("#clog").html(data);



            });
        }

PHP方

$sql6="SELECT msgid FROM thread WHERE combination1=:msgids OR combination2=:submsgids LIMIT 1";
          $msg_id = $con4->prepare($sql6);
          $msg_id->bindParam(':msgids', $comb, PDO::PARAM_STR);
          $msg_id->bindParam(':submsgids', $comb, PDO::PARAM_STR);
          $msg_id->execute();
          $msgd = $msg_id->fetchColumn();
          $tbpre = $msgd;
        $sql7 = "SELECT * FROM ".$tbpre."chat_conversation WHERE msgid=:chat";

        $stmt7=$con3->prepare($sql7);
        $stmt7->bindValue( 'chat', $msgd, PDO::PARAM_STR);
        $stmt7->execute();
        $rows = $stmt7->fetchAll(PDO::FETCH_ASSOC);

        echo json_encode($rows);

输出如下：

[{"msgid":"1","message_content":"asd","username":"ab","message_time":"2014-04-02 13:58:03","recipient":"cd"}]

如何显示与此类似的JSON结果：

1
asd
ab
2014-04-02
cs

Answer 1

您可以在jQuery中使用json dataType，然后使用JSON.stringify使用缩进对对象进行字符串化，并将其输出到pre标记以保留空格和换行符

function AjaxRetrieve(){
    var rid  = document.getElementById('trg').value,
        data = {chat : uid, rid : rid, name : user};

    $.get('includes/getChat.php', data, function(result) {

        var pre = $('<pre />', {text : JSON.stringify(result, undefined, 4)})
        $("#clog").html(pre);

    }, 'json');
}

FIDDLE

或只输出值列表

function AjaxRetrieve() {
    var rid = document.getElementById('trg').value,
        data = {chat: uid, rid: rid, name: user};

    $.get('includes/getChat.php', data, function (result) {
        var res = $([]);

        $.each(result[0], function(key, value) {
            res = res.add($('<div />', {text : value}));
        });

        $("#clog").html(res);

    }, 'json');
}

Answer 2

在客户端解析响应并显示它。

$.get('includes/getChat.php?chat='+uid + '&rid=' + rid + '&name=' + user,function(data){
    var obj = JSON.parse(data);

    var temp = obj[0].msgid + "<br>"+ obj[0].message_content + "<br>"+ obj[0].username + "<br>"+ obj[0].message_time + "<br>"+ obj[0].recipient;
        $("#clog").html(temp);
});

Answer 3

PHP 5.4提供了JSON_PRETTY_PRINT选项，可用于json_encode()调用。

以下是链接：

http://php.net/manual/en/function.json-encode.php

<?php

$json_string = json_encode($data, JSON_PRETTY_PRINT);

输出我的JSON结果

3 个答案: