如何在javascript中声明php数组变量并查看数据表答案

【问题标题】：how to declare php array variable in javascript and view on datatable如何在javascript中声明php数组变量并查看数据表
【发布时间】：2017-01-29 05:22:15
【问题描述】：

我在 $dispatch 变量中获取数组值，并在 JavaScript var dataSet3=<?php echo json_encode($dispatch); ?>; 中传递它，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}
        ],
    } );
} );

【问题讨论】：

@Anant 在数组视图中获取值但数据表未显示值！
var dataset3 是否包含所需的 json？
@Anant 使用 ajax 的链接，但我直接将值 js 传递给数据表：- datatables.net/examples/data_sources/js_array.html
@ArunKumaresh 对不起，我是初学者，所以更了解这样的 json
@Anant 是的，我收到以下错误：-*sub.php:111 GET https:cdn.datatables.net/1.10.12/css/jquery.dataTa%3Chead%3Ebles.min.css sub .php 是我的文件名

标签： javascript php mysqli datatables

【解决方案1】：

使用下面的更正代码。

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); 
}

【讨论】：