我在Javascript/jQuery中設置了自定義模板,我需要從CodeIgniter控制器中提取數據並將返回的JSON插入到js/jQuery模板中。我相信我的邏輯是正確的,但由於某種原因,似乎沒有任何工作,我發現了以下錯誤在我的劇本的開頭:jQuery自定義模板無法正常工作
Uncaught SyntaxError: Unexpected end of input
我怎麼會去這樣做呢?到目前爲止,我寫的代碼如下所示:
$("#projects").click(function() {
jQuery.ajax({
type: "POST",
dataType: "JSON",
url: "<?=base_url()?>index.php/home/projectsSlider",
data: dataString,
json: {
returned: true
},
success: function (data) {
if (data.returned == true) {
$("#content").fadeOut(150, function() {
$(this).replaceWith(projectsSlider(), function() {
var html = projectsSlider(data.projectId, data.projectName, data.startDate, data.finishedDate, data.createdFor, data.contributors, data.screenshotURI, data.websiteURL);
jQuery(html).appendTo("#content").fadeIn();
});
});
}
}
});
});
這裏是我的PHP:
function projectsSlider() {
$query = $this->db->query("SELECT * FROM projects ORDER BY idprojects DESC");
foreach ($query->result() as $row) {
$projectId = $row->projectId;
$projectName = $row->projectName;
$startDate = $row->startDate;
$finishedDate = $row->finishedDate;
$createdFor = $row->createdFor;
$contributors = $row->contributors;
$projectDesc = $row->projectDesc;
}
$query1 = $this->db->query("SELECT * FROM screenshots s WHERE s.projectId = '{$projectId}' ORDER BY s.idscreenshot DESC");
foreach ($query1->result() as $row2) {
$screenshotURI = $row2->screenshotURI;
$websiteURL = $row->websiteURL;
}
echo json_encode(array('returned' => true,
'projectId' => $projectId,
'projectName' => $projectName,
'startDate' => $startDate,
'finishedDate' => $finishedDate,
'projectDesc' => $projectDesc,
'createdFor' => $createdFor,
'contributors' => $contributors,
'screenshotURI' => $screenshotURI,
'websiteURL' => $websiteURL));
}
}
任何想法,爲什麼發生這種情況?
我不知道那是怎麼回事,但尾部}不在我原來的代碼中。但jQuery似乎沒有正常工作。這是我的整個頁面上的代碼:http://pastebin.com/a31tsABt – 2012-07-10 00:19:35
沒關係,我想通了。我沒有結束我的document.ready語句 – 2012-07-10 00:25:58