Question

我有这个ajax

$.ajax({
                        type: "GET",
                        url: '../connect.php',
                        data: "OrB=" + ajaxsend+"&&IOr="+i,
                        success: function(data)
                        { 
                             var x = $.parseJSON(data);
                             var el='<div class="CommentsAw Comment_Hs">\
                                      <img src="../users/'+x[0]+'">\
                                      <span>'+x[1]+'</span>\
                                      <span class="s2">'+x[2]+'</span>\
                                    </div>'
                                    $(".F_W_comments").html().remove();
                                    $(".F_W_comments").html(el);
                        }
        });

和php

if (isset($_GET['OrB'])) {
        $OB=$_GET['OrB'];
        $I=$_GET['IOr'];
        if ($OB=='OO') {
            $OB='`Date` ASC';
        }else if ($OB=='No') {
            $OB='`Date` DESC';
        }
        $query=$con->query("SELECT id,comment FROM uploads WHERE Rand='$I'");
        $row=$query->fetch_row();
        $Commentsq=$con->query("SELECT * FROM (SELECT * FROM comments WHERE Post_id='$row[0]' ORDER BY $OB LIMIT 4) AS sub ORDER BY `DATE` ASC") or die($con->error);
        while ($CommentRow=$Commentsq->fetch_row()) {
                $CommenterPp=$con->query("SELECT Profile_pic FROM user_opt WHERE Username='$CommentRow[3]'");        
                $CommenterPicture=$CommenterPp->fetch_row();
                $CommenterPp=$con->query("SELECT Username FROM users WHERE Id='$CommentRow[3]'");        
                $CommenterName=$CommenterPp->fetch_row();
                echo json_encode(array($CommenterPicture,$CommenterName,$CommentRow));             
        }
    }

但它在控制台中给我一个错误，就像这个

VM654:2 Uncaught SyntaxError: Unexpected token [ in JSON at position 107
[["5734919677561.jpg"],["Murad"],["1842","3","21","1","2016-05-08 21:56:52"]]                    
                [["5734919677561.jpg"],["Murad"],["1843","GOodm","21","1","2016-05-08 21:56:54"]]                    
                [["5734919677561.jpg"],["Murad"],["1845","re","21","1","2016-05-08 21:56:54"]]                    
                [["5734919677561.jpg"],["Murad"],["1844","re","21","1","2016-05-08 21:56:54"]]

我想要的是connect.php从数据库中获取数据然后将其传递给ajax。但结果是没有什么可能因为我的代码中存在一些错误

Answer 1

回显几个json编码的字符串并不意味着有效 json。

您需要做的是json_encode您的数据，echo 一次。

$commenters = array();    // result array
while ($CommentRow=$Commentsq->fetch_row()) {
    $CommenterPp=$con->query("SELECT Profile_pic FROM user_opt WHERE Username='$CommentRow[3]'");        
    $CommenterPicture=$CommenterPp->fetch_row();
    $CommenterPp=$con->query("SELECT Username FROM users WHERE Id='$CommentRow[3]'");        
    $CommenterName=$CommenterPp->fetch_row();
    $commenters[] = array($CommenterPicture,$CommenterName,$CommentRow);             
}
// while loop over
echo json_encode($commenters);

在你的js中你应该迭代一个对象数组，而不是一个简单的对象，例如：

success: function(data) {
    var x = $.parseJSON(data);
    for (var k in x) {
        console.log(x[k]);
    }
}

Answer 2

暂时删除数组中的数据，生成的JSON结构如下：

[][][][]

那是无效的JSON。解析器需要一个对象或数组，而不是像这样连接在一起的几个数组。

不是在循环中输出响应，而是在循环之后构建总响应并将其输出一次。基本上在PHP中的while循环之前创建一个空数组，然后将元素推送到循环内的数组上，然后在循环之后回显JSON编码的数组。

我的PHP非常生疏，但在PHP-ish伪代码中它的结构如下：

$result = array();
while ($CommentRow=$Commentsq->fetch_row()) {
    // other code...
    $result[] = array($CommenterPicture,$CommenterName,$CommentRow);
}
echo json_encode($result);

通过PHP返回带有while循环的ajax

2 个答案: