Question

嘿，伙计们可以帮助我。使用jquery文件，无需在线访问Googleapis，这将使我的引导程序在完全脱机时正常工作...我试图激活通知如果成功消息，以便用户可以取消bt取消b＆＃34; X＆＃34;不工作....下面是我的项目文件的链接，请看看 Bootstrap project files

Answer 1

你拼错了，这就是为什么它没有关闭'ALERT'

下面：

<?php


$page = $_GET['page']; 
$limit = $_GET['rows']; 
$sidx = $_GET['sidx']; 
$sord = $_GET['sord']; 


$result = $mysqli->query("SELECT COUNT(*) AS count FROM users"); 
$row = mysqli_fetch_array($result,MYSQLI_ASSOC); 

$count = $row['count']; 
if( $count > 0 && $limit > 0) { 
    $total_pages = ceil($count/$limit); 
} else { 
    $total_pages = 0; 
} 
if ($page > $total_pages) $page=$total_pages;
$start = $limit*$page - $limit;
if($start <0) $start = 0; 

$SQL = "SELECT * FROM users ORDER BY $sidx $sord LIMIT $start , $limit"; 
$result = $mysqli->query( $SQL ); 

$i=0;
$responce = new stdClass();
$responce->page = $page; 
$responce->total = $total_pages; 
$responce->records = $count;

while($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) {
    $responce->rows[$i]['id']=$row['id_user'];
    $responce->rows[$i]['cell']=array($row['id_user'],$row['firstname'],$row['lastname'],$row['username'],$row['level']);
    $i++;
}

echo json_encode($responce);
?>

错误：

$(function () {
    "use strict";
    jQuery("#list2").jqGrid({
        url:'users_grid_load_data.php?q=2', 
        dataType: "Json",
        mtype: 'Get',       
        colNames:['Id','First Name', 'Last Name', 'Username','Level'], 
        colModel:[ 
             {name:'id_user',index:'id_user', width:55}, 
             {name:'firstname',index:'firstname', width:90}, 
             {name:'lastname',index:'lastname', width:90}, 
             {name:'username',index:'username', width:90},
             {name:'level',index:'level', width:80}          
        ], 
        rowNum:10, rowList:[10,20,30], 
        pager: '#pager2', 
        sortname: 'id_user', 
        viewrecords: true, 
        sortorder: "asc", 
        height:"auto",
        width:"auto",
        caption:"LIST OF USERS" 
    }); 
    jQuery("#list2").jqGrid('navGrid','#pager2',{edit:false,add:false,del:false});
});

不是<div class="alert alert-danger fade in"> <a href="#" class="close" data-dissmiss="alert" aria-label="close" >×</a>Don't forget Ramsey's Birthday </div>

使用此：

data-dissmiss="alert" // DISMISS

尝试在bootsrap中使用Wells时出现脱机Bootstrap jquery错误

1 个答案: