Php 编辑内爆字符串
我有一个应用程序,我可以在其中分配任务,如清洁浴室、洗车等。我可以分配一个任务给3到4个人,使用下面的代码Php 编辑内爆字符串,php,foreach,while-loop,odbc,explode,Php,Foreach,While Loop,Odbc,Explode,我有一个应用程序,我可以在其中分配任务,如清洁浴室、洗车等。我可以分配一个任务给3到4个人,使用下面的代码 if (isset($_POST["btnassign"])){ $proj = $_REQUEST['projhid']; $analyst = $_POST['analyst']; $commaList = implode('| ', $analyst); $queupass = "UPDATE projects set assignedto='$commaList', assigne
if (isset($_POST["btnassign"])){
$proj = $_REQUEST['projhid'];
$analyst = $_POST['analyst'];
$commaList = implode('| ', $analyst);
$queupass = "UPDATE projects set assignedto='$commaList', assignedby = '$uname' where projectname='$proj'";
$queresupass = odbc_exec($conn,$queupass);
$notifassign = "New Project Assignment";
$queprojgn = "INSERT INTO notification (notification,datetime,isread,createdby,createddate)
values('$notifassign',GETDATE(),0,'$uname',GETDATE())";
$queprojresnot = odbc_exec($conn,$queprojgn);
echo "<script type='text/javascript'>";
echo "window.close();";
echo "</script>";
这是最终的工作,但我需要的是爆炸它在表中有一个删除锚标签,如下面的一个地方,任何帮助将不胜感激
这可以通过以下方式实现:
,如下所示
$que=“从projectname='$projname'的项目中选择assignedto”代码>
$analysis=explode(“|”,$res)代码>
$new\u analysis=array\u diff($analysis,$remove\u worker\u id)代码>
$result=内爆(“|”,$new|)代码>
$que = "SELECT assignedto FROM PROJECTS where projectname = '$projname'";
$queres = odbc_exec($conn,$que);
$res = odbc_result($queres, 1);
$analysts_arr = explode("| ", $res);
$analysts = implode(",", $analysts_arr);
$que2 = "SELECT [user_id, first name, last name] FROM [user table] where [user_id] IN '$analysts'";
$query = odbc_exec($conn,$que2);
$result = odbc_result_all($query);
foreach($result as $val){
$user_id = $val['user_id'];
$users[$user_id] = $val;
}
在此之后,您所要做的就是迭代$analysts\u arr
,并创建表所需的HTML
希望这有帮助
干杯 据我所知,一旦用户删除一个worker,您需要在数据库中编辑分配给的值。正确吗?是的。因此,如果最初分配给的是1 | 2 | 3 | 4 | 5,我想删除姓氏,Juan假设他的id是3,那么分配给的将成为1 | 2 | 4 | 5,但我将如何在表格中显示它,就像我原来帖子上的最后一张照片一样?
$que = "SELECT assignedto FROM PROJECTS where projectname = '$projname'";
$queres = odbc_exec($conn,$que);
$res = odbc_result($queres, 1);
$analysts_arr = explode("| ", $res);
$analysts = implode(",", $analysts_arr);
$que2 = "SELECT [user_id, first name, last name] FROM [user table] where [user_id] IN '$analysts'";
$query = odbc_exec($conn,$que2);
$result = odbc_result_all($query);
foreach($result as $val){
$user_id = $val['user_id'];
$users[$user_id] = $val;
}