Javascript 使用 php 调用引导模式显示事件
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Call bootstrap modal show event with php
提问by pckmn
I want to call modal('show')event with php code. How can I do this? I tried the following:
我想modal('show')用php代码调用事件。我怎样才能做到这一点？我尝试了以下方法：
if(isset($_POST['Edit'])){
echo "<script> $('#Modal').modal('show') </script>";}
I run this code, but echo "<script> $('#Modal').modal('show') </script>";doesnt work. How can I show modal when image input is clicked? I don't want to call with image onlick.
我运行此代码，但echo "<script> $('#Modal').modal('show') </script>";不起作用。单击图像输入时如何显示模态？我不想用图像 onlick 打电话。
Input:
输入：
<input type="image" src="image.png" name="Edit" value="Edit" alt="Edit" />
Modal:
模态：
<div class="modal hide fade" id="Modal" >
<form method="post">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" style="margin-top:-10px">×</button>
</div>
<div class="modal-body">
<textarea id="text" name="text">Test</textarea>
</div>
<div class="modal-footer">
</div>
</form>
</div>
回答by user3133586
Set the document.ready function as shown below:
如下设置 document.ready 函数：
echo "<script type='text/javascript'>
$(document).ready(function(){
$('#Modal').modal('show');
});
</script>";
On a side note, this work around is also great if you have a modal login form and need to alert the user on submit (or $_POST) for any submission errors in a seperate modal.
附带说明一下，如果您有一个模态登录表单并且需要在提交（或 $_POST）时提醒用户在单独的模态中出现任何提交错误，则此解决方法也很棒。
回答by Devang Rathod
You can use jquery fadeIn function to show model
您可以使用 jquery 的淡入功能来显示模型
$('#Modal').modal('show');
Change to
改成
$('#Modal').fadeIn('show');
Hope it will help you.
希望它会帮助你。
