我在update.php中有这个表单html代码。为了更新,需要使用mysql更新脚本链接到另一个页面save_seeker.php,以便执行它。有没有办法在提交表单的同一页面上执行脚本,以便在查询执行后它保留在同一页面上?
<form action= "save_seeker.php" method = "post">
Update details !<br><br>
First Name
<input type = "text" name = "fname" value = "<?php echo $disp['fname'];?>"><br><br>
Last Name
<input type = "text" name = "lname" value = "<?php echo $disp['lname'];?>"><br><br>
Contact number
<input type = "text" name = "contact" value = "<?php echo $disp['contact'];?>"><br><br>
Email-id
<input type = "email" name = "email" value = "<?php echo $disp['email'];?>"><br><br>
Address
<input type = "text" name = "address" value = "<?php echo $disp['address'];?>"><br><br>
Experience
<input type = "number" name = "experience" value = "<?php echo $disp['experience'];?> "><br><br>
Qualification
<input type = "text" name = "qualification" value = "<?php echo $disp['qualification'];?>"><br><br>
<input type = "Submit" value = "Update">
</form>
答案 0 :(得分:2)
您可以通过将其添加到表单操作中来在同一文件中进行处理。
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<!-- Input fields in here-->
<input type="submit" name="form_submit" value="Submit">
</form>
现在通过检查是否在post变量中设置了
来检查提交操作<?php
if ( isset( $_POST['form_submit'] ) ) {
// Do processing here.
}
?>
答案 1 :(得分:-1)
您可以使用include('page-with-update.php')
,调用更新函数并使用$ _SERVER [“PHP_SELF”]。
我希望有所帮助!!