javascript 在mysql中成功插入数据后如何显示警报框？
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How to show alert box after successful data insertion in mysql?
提问by Sankalp Nigam
I want to show JavaScript alert after successful data insertion in MySQL. How to do this? I have written this code but it shows JavaScript alert everytime I open this page and as soon as i click on OK of JavaScript alert it redirects me to finalmem.php, without the action of taking values from users!
我想在 MySQL 中成功插入数据后显示 JavaScript 警报。这个怎么做？我已经编写了这段代码，但每次打开此页面时它都会显示 JavaScript 警报，一旦我单击 JavaScript 警报的“确定”，它就会将我重定向到 finalmem.php，而不会从用户那里获取值！
$sql="INSERT INTO members VALUES ('$name', '$email', '$ybr', '$ach')";
if(!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
else
{
echo '<script language="javascript">';
echo 'alert("Successfully Registered"); location.href="finalmem.php"';
echo '</script>';
}
Thanks in advance.
提前致谢。
回答by Neel Ion
Use is set isset($_POST['submit'])to check whether user submits the form or not
用于设置isset($_POST['submit'])检查用户是否提交表单
<?php
include 'SQLIDB.php';
if(isset($_POST['submit'])){
$name=$_POST['name'];
$email=$_POST['email'];
$ybr=$_POST['ybr'];
$ach=$_POST['ach'];
$sql="INSERT INTO members VALUES ('$name', '$email', '$ybr', '$ach')";
if(!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
else
{
echo '<script language="javascript">';
echo 'alert("Successfully Registered"); location.href="finalmem.php"';
echo '</script>';
}
}
?>
<form action="" method="post">
<input type="text" name="name">
<input type="text" name="email">
<input type="text" name="ybr">
<input type="text" name="ach">
<input type="submit" name="submit" value="Submit">
</form>
回答by Shehary
You can try like this
你可以这样试试
<?php session_start();
//Include database detail here
if(isset($_POST['name'])){
$name = mysqli_real_escape_string($_POST["name"]);
$ybr = mysqli_real_escape_string($_POST["ybr"]);
email = mysqli_real_escape_string($_POST["email"]);
$ach = mysqli_real_escape_string($_POST["ach"]);
//Do your data validation here
$_SESSION['sname'] = $name;
$_SESSION['sybr'] = $ybr;
$_SESSION['semail'] = email;
$_SESSION['sach']= $ach;
$sql="INSERT INTO members VALUES ('$name', '$email', '$ybr', '$ach')";
if(!mysqli_query($con,$sql)) {
die('Error: ' . mysqli_error($con));
} else {
echo '<script language="javascript">';
echo 'alert("Successfully Registered"); location.href="finalmem.php"';
echo '</script>';
}
}
?>
回答by Met El Idrissi
You need to know two things, one of them it's to confirm if your data it's saved succefully.
您需要知道两件事，其中之一是确认您的数据是否已成功保存。
For this you can play with
为此，您可以玩
mysql_affected_rows()
and this function will return the number of rows affected by your query. If zero, it was not inserted. If >= 1, it was.
此函数将返回受查询影响的行数。如果为零，则未插入。如果 >= 1，则是。
So:
所以：
if ( mysql_affected_rows() >= 1 ){ /* inserted! now do something... */ }
If you are using an auto-incrementing column for row ID, you can use mysql_insert_id() as well:
如果您为行 ID 使用自动递增列，您也可以使用 mysql_insert_id() ：
if ( mysql_insert_id() > 0 ) { /* inserted! now do something... */ }
Then you can work with jQuery UI for show a dialog like this:
然后你可以使用 jQuery UI 来显示这样的对话框：
[https://jqueryui.com/dialog/][1]
You need tot load the .css and .js files to run jQuery and inside your code put this:
您需要加载 .css 和 .js 文件来运行 jQuery，并在您的代码中输入：
<script>
$(function() {
$( "#dialog" ).dialog();
});
</script>
And this in your view:
在你看来：
<div id="dialog" title="Basic dialog">
<p>This is the default dialog which is useful for displaying information. The dialog window can be moved, resized and closed with the 'x' icon.</p>
</div>
Normally this it's super ugly to do for me, because the best way it's doing this by AJAX.
通常这对我来说是非常丑陋的，因为最好的方法是通过 AJAX 来做这件事。
