我在 $dispatch
中获取数组值变量并在 JavaScript 中传递它 var dataSet3=<?php echo json_encode($dispatch); ?>;
但是当数组包含两行时,我的数据表只显示一行。
SQL查询:
$userSQL3 = "SELECT magazine.name_txt, dispatch.dispatch_id, dispatch.dispatch_dt, dispatch.notes_txt, dispatch_details.created_dt FROM dispatch_details left join dispatch on dispatch_id=fk_dispatch_id LEFT JOIN magazine ON magazine.magazine_id = dispatch.fk_magazine_id WHERE fk_subscriber_id = ". $subscriber_id;
$usersResult3 = mysqli_query($dbConn, $userSQL3);
while($userResult3 = mysqli_fetch_array($usersResult3))
{
$dispatch=array($userResult3) ;
echo"<pre>";
print_r($dispatch);
echo"</pre>";
}
JavaScript 代码:
var dataSet3=<?php echo json_encode($dispatch); ?>;
$(document).ready(function() {
$('#dispatch').DataTable( {
searching: false,
paging:false,
bLengthChange:false,
data: dataSet3,
columns: [
{ data: "fk_dispatch_id" ,"visible": false},
{ data: "fk_subscriber_id" ,"visible": false },
{ data: "created_dt" },
{ data: "updated_dt" ,"visible": false},
{ data: "dispatch_id" },
{ data: "dispatch_dt" },
{ data: "fk_publisher_id","visible": false },
{ data: "name_txt" },
{ data: "notes_txt" },
{ data: "created_dt" ,"visible": false },
{ data: "updated_dt" ,"visible": false}
],
} );
} );
最佳答案
使用下面更正的代码。
PHP
$userSQL3 = "SELECT magazine.name_txt, dispatch.dispatch_id, dispatch.dispatch_dt, dispatch.notes_txt, dispatch_details.created_dt FROM dispatch_details left join dispatch on dispatch_id=fk_dispatch_id LEFT JOIN magazine ON magazine.magazine_id = dispatch.fk_magazine_id WHERE fk_subscriber_id = ". $subscriber_id;
$usersResult3 = mysqli_query($dbConn, $userSQL3);
$dispatch = array();
while($userResult3 = mysqli_fetch_array($usersResult3, MYSQLI_ASSOC))
{
array_push($dispatch, $userResult3);
}
关于javascript - 如何在javascript中声明php数组变量并在数据表上查看,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39608417/