我有一个用 php-pear 创建的表，以及一个处理单击事件以将单元格信息提交到新表单页面的 jquery 函数。jquery是这样的：
var row = $this.parent('tr').contents('th:eq(0)').html();
var departmentID = $(".deptSelect").val();
var headerObj = $(this).parents('.main').find('th').eq(colIndex);
var toPass = $.trim(headerObj.text());
var picked = $("#picked").val();
var testDate = new Date(picked + " " + row);
var today = new Date();
if (testDate < today)
{
if (roleID > 2)
{
alert("You Cannot Schedule a New Job in the Past!");
}
return;
}
var thisForm = '';
if (roleID == 5)
{
thisForm = '../forms/tentativeJobForm.php';
}
else
{
thisForm ='../forms/newJobForm.php';
}
var f = document.createElement("form");
f.setAttribute('class','jobTime');
f.setAttribute('method','post');
f.setAttribute('action',thisForm);
var iii = document.createElement('input');
iii.setAttribute('type','hidden');
iii.setAttribute('name','departmentID');
iii.setAttribute('value',departmentID);
f.appendChild(iii);
var i = document.createElement('input');
i.setAttribute('type','hidden');
i.setAttribute('name','sTime');
i.setAttribute('value',picked + " " + row);
f.appendChild(i);
var ii = document.createElement('input');
ii.setAttribute('type','hidden');
ii.setAttribute('name','ast');
ii.setAttribute('value',toPass);
f.appendChild(ii);
document.getElementsByTagName('body')[0].appendChild(f);
if (roleID > 2)
$('.jobTime').submit;
else
return;
});
这在本质上是有效的，但是我的用户抱怨在他们安排新工作时看不到日历页面，并且它与已经安排的另一个工作冲突。我的新要求是生成的冲突页面作为弹出窗口打开。不是存在冲突的警报，而是具有冲突信息的原始表单。上面的表单提交到我的php页面，下面的函数处理冲突：
function checkRows($stmt, $msg=NULL, $params=NULL)//, $updatedInfo=NULL)
{
if ($stmt != NULL)
{
$rows_affected = sqlsrv_rows_affected($stmt);
if ($rows_affected > 0)
{
$starting = new DateTime($params[5]);
$ending = new DateTime($params[6]);
$starting = date_format($starting,'m/d/Y h:i a');
$ending = date_format($ending,'m/d/Y h:i a');
echo "Job:" . $params[0] . " was Successfully Scheduled for $starting to $ending<br>";
}
}
else
{
$userSTime = new DateTime($params[0]);
$userSTime = date_format($userSTime,'m/d/Y h:i a');
$userETime = new DateTime($params[1]);
$userETime = date_format($userETime,'m/d/Y h:i a');
$_SESSION['jbNum'] = $params[2];
$_SESSION['asset'] = $params[4];
$_SESSION['userSTime'] = $userSTime;
$_SESSION['userETime'] = $userETime;
$_SESSION['userDesc'] = trim($params[3]);
$_SESSION['conJbNum'] = $msg['JobNum'];
$_SESSION['conSTime'] = date_format($msg['StartTime'], 'm/d/Y h:i a');
$_SESSION['conETime'] = date_format($msg['EndTime'], 'm/d/Y h:i a');
$_SESSION['dueDate'] = $params[5];
$_SESSION['comment'] = $params[6];
$_SESSION['destination'] = $params[7];
$_SESSION['jStat'] = $params[8];
if ($_SESSION['recurring'] == 'n')
{
echo "<meta http-equiv='refresh' content='0;URL=../forms/newJobForm.php' target = 'mainFrame'/>";
exit;
}
else
{
echo "Scheduling for Job:" . $params[2] ." failed for $userSTime to $userETime<br>";
}
}
}
当遇到冲突时，原始表单将使用来自会话变量的新信息打开。如何在弹出窗口中打开表单？我知道这应该不会太难，但我无法让它发挥作用。仅供参考 target='_blank' 已被建议但仅打开一个新选项卡。那是行不通的。