Question

我使用Bootstrap模式为用户显示一些内容＆＃39;错误，这是我的代码：

<div class="modal fade" id="modal1">
    <div class="modal-dialog">
        <div class="modal-content">
            <div class="modal-header">
                <button type="button" class="close" data-dismiss="modal"><span aria-hidden="true">&times;</span><span class="sr-only">Close</span></button>
                <h4 class="modal-title text-warning">ERROR</h4>
            </div>
            <div class="modal-body text-warning" id="a">You need to login to leave your comments.
            </div>
            <div class="modal-footer">
                <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
                <button type="button" class="btn btn-primary">Login</button>
            </div>
        </div>
    </div>
</div>

<!-- second modal -->
<div class="modal fade" id="modal2">
    <div class="modal-dialog">
        <div class="modal-content">
            <div class="modal-header">
                <button type="button" class="close" data-dismiss="modal"><span aria-hidden="true">&times;</span><span class="sr-only">Close</span></button>
                <h4 class="modal-title text-warning"></h4>
            </div>
            <div class="modal-body text-warning">Please fill all these fields for more accurate results
            </div>
            <div class="modal-footer">
                <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
                <button type="button" class="btn btn-primary">Login</button>
            </div>
        </div>
    </div>
</div>

然后我使用2个变量来确定将显示哪个模态。问题是我的代码仅在我使用1模态时有效，当我添加另一个模态时，它不会显示任何内容。有人可以指出我的错误吗？

感谢您的时间

EDITED 这是我的完整代码

<?php 
$error1 = "";
$error2 = "";
if ((isset($_POST["MM_insert_comment"])) && ($_POST["MM_insert_comment"] == "form")) {
if($autologin){
    $fac = $_POST["facilities"]; $fee = $_POST["fee"];
    $qua = $_POST["quality"]; $act = $_POST["activities"];
    //$name  = $_SESSION['MM_Username'];
    $name = $real_username;
    if ( ($fac!=NULL)&&($fee!=NULL)&&($qua!=NULL)&&($act!=NULL) ){
        $insertSQL = sprintf("INSERT INTO comment (facilities,fee,quality,activities,UserName) 
                            VALUES ('$fac','$fee','$qua','$act','$name')");
        mysqli_query($localhost,$insertSQL) or die(mysqli_error($localhost));
        header('Location: '.$thispage);
    }
    else {
        $error1 = true;
    }
} else {
    $error2= true; 
}
}?>

和Js代码显示模态：

<script type="text/javascript">
$(document).ready(function(){
    var error1 = <?php echo $error1; ?>;
    if(error1){
        $('#modal1').modal('show');
    }
    var error2 = <?php echo $error1; ?>;
    if(error2){
        $('#modal2').modal('show');
    }
});
</script>

//我不认为我的问题是关于这段代码的，所以我之前没有发布过它：）

Answer 1

您可以使用Stackable

选项尝试Bootstrap-Modal

不能有2个或更多Bootstrap模式一起工作

感谢您的时间

1 个答案: