我在Javascript/jQuery中有一个自定义模板设置，我需要从CodeIgniter控制器中提取数据，并将返回的JSON插入到js/jQuery模板中。我确实相信我的逻辑是正确的，但由于某种原因，似乎什么都不起作用，我在脚本的开头出现了以下错误：
未捕获的SyntaxError：输入意外结束
我该怎么做？到目前为止，我写的代码如下所示：
$("#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();
});
});
}
}
});
});
这是我的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));
}
}
有什么关于为什么会发生这种情况的想法吗？
问题看起来像是有一个尾随的
编辑：当我重构时，其他人回答了，但我还是提供了这个版本：
function projectsSlider() {
$query = $this->db->query("SELECT * FROM projects ORDER BY idprojects DESC");
$project = $query->fetch(PDO::FETCH_OBJECT);
$project->screenshots = array();
$query = $this->db->query("SELECT * FROM screenshots WHERE projectId = '$projectId' ORDER BY idscreenshot DESC");
foreach ($query->result() as $screenshot) {
$project->screenshots[] = $screenshot;
}
echo json_encode(array('returned' => true,'project'=>$project));
}
由于您可以获取一个对象，因此不必进行所有的循环和转换。
}。您应该查看PHP linter以避免这些问题。