Question

我试图让.js文件填充MySQL数据库中的内容。

我有一个.php文件可以完成所有MySQL的工作：

$result = mysql_query("SELECT *,
DATE_FORMAT(gig_dateTime, '%Y') as gigYear,
DATE_FORMAT(gig_dateTime, '%m') as gigMonth,
DATE_FORMAT(gig_dateTime, '%d') as gigDay,
DATE_FORMAT(gig_dateTime, '%H') as gigHour,
DATE_FORMAT(gig_dateTime, '%i') as gigMinute
FROM live
WHERE gig_dateTime > NOW()
ORDER BY gig_dateTime DESC");
$to_encode = array();
while($row = mysql_fetch_assoc($result)) {
    $to_encode[] = $row;
}
echo json_encode($to_encode);

然后在.js文件中我有以下内容：

$.getJSON('live-dates.php', function(data) {
    var myConcerts = [];
    $.each(data, function(index, d){            
        myConcerts.push(
            "{",
            "year:", d.gigYear, ",",
            "month:", d.gigYear, ",",
            "day:", d.gigYear, ",",
            "hour:", d,gigHour, ",",
            "minute:", d.gigMinute, ",",
            "latitude:", d.gig_lat, ",",
            "longitude:", d.gig_long, ",",
            "location:", d.gig_location, ",",
            "infoWindow:", d.gig_info, "",
            "}"
        );
    });
});

但在加载页面时，我得到未捕获的ReferenceError：$未定义

在显示.js文件的index.php文件中，我包含以下内容：

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script src="js/myconcerts.js" type="text/javascript"></script>

我正在尝试做什么，或者我是否采取了错误的方式？

由于

Answer 1

也许你需要用$.getJSON(...)包裹$(function(){ /* $.getJSON() HERE */ });，这与$(document).ready(...);相同。

js文件从mysql获取内容

1 个答案: