[英]how to display a hidden table when submit button is clicked and data is inserted in database table in php?
I have one from through which I am entering project name and sheet name which will be saved in project table. 我有一个从中输入项目名称和工作表名称,它将保存在项目表中。 Based on this I am inserting empty values on another table 'Data'.
基于此,我在另一个表'Data'上插入空值。 I am using editable table to update these empty values in database.
我正在使用可编辑的表来更新数据库中的这些空值。 I want to hide this table till project name and sheet name is submitted and once value is submitted then only table will be displayed.
我希望隐藏此表,直到提交项目名称和工作表名称,并且一旦提交了值,则只显示表格。 I have tried a lot but after clicking submit button my table gets displayed till page is reloaded and disappers when loading is completed.
我已经尝试了很多但是在点击提交按钮后,我的表格会一直显示,直到页面重新加载,并且在加载完成后会显示消息。 My code parts are like below.
我的代码部分如下所示。
<script>
$(function(){
$('button#showit').on('click',function(){
$('#myform').show();
});
});
</script>
<form class="form-horizontal" action="db.php" id="#form1"
onSubmit="function();" method="post">
<input type="text" class="form-control1" name="projectname"
id="projectname" required >
<input type="text" class="form-control1" name="sheetname" id="sheetname"
required >
<button type="submit" name="submit" id="button" class="btn-
success btn" >Submit</button>
</form>
<div class="tab-content" id="myform" style="display:none">
<div class="tab-pane active" id="horizontal-form">
<?php
$s4 = "SELECT * FROM data ORDER BY data_id DESC LIMIT 1";
$res = mysqli_query($bd, $s4);
while ($row = mysqli_fetch_array($res))
{
?>
<table class="table table-bordered" style="width:50%;">
<thead>
<tr>
<th>Sr.no</th>
<th>Column Name</th>
</tr>
</thead>
<tbody>
<tr class="odd">
<td>1</td>
<td id="d1:<?php echo $row['data_id']; ?>" contenteditable="true"><?php
echo $row['d1']; ?></td>
</tr>
.
.
.
</tbody>
</table>
<?php } ?>
</div>
</div>
on db.php I have used to get back on my previous page
<script>
alert('Sheet Created Successfully');
window.location.href='index.php?success';
</script>
I cant help you much in your code, maybe you should readjust some things, but, here´s some help: 我在你的代码中无法帮助你,也许你应该重新调整一些东西,但是,这里有一些帮助:
$('#firstform').submit(function () {
//handle here a function to send data to db.php, then:
sendData();
$('#firstform').hide();
$('#hidedform').show();
return false;
});
If you control the submit button not to refresh the page, you can hide first form and display second one. 如果您控制提交按钮不刷新页面,则可以隐藏第一个表单并显示第二个表单。 If you post more code I´ll try to be clearer.
如果你发布更多的代码我会尝试更清楚。
("return false" makes submit function not to refresh the page) (“return false”使提交功能不刷新页面)
here are another post about not refreshing...maybe it helps 这是另一篇关于不爽快的帖子......也许有帮助
<!-- Example of jQuery Reference --> <script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"></script> <!-- Your Hiding Script --> <script type='text/javascript'> $(document).ready(function(){ $("#AddNewRecord").hide(); }); </script> <!-- Your Markup --> <table border="1" cellpadding="2" id="AddNewRecord"> <tr style="width:25px;"> <td>Site</td> </tr> </table>
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