how to call jquery function after php if condition?

Question

I want to display jquery sweet alert after php insert query successfully executed.. but whenever i tried to call jquery function after if condition, didn't work..or alert appear for 1-2 sec before page reload for form submit and disappear immediately..it only works with button onclick method..

<html>
<head>
<script src="lib/sweet-alert.js"></script>
<link rel="stylesheet" href="lib/sweet-alert.css">
</head>

<body>
<form method="post">
  <label for="name">name</label>
  <input type="text" name="name" id="name" />
  <input type="submit" value="submit" name="submit" class="sub" id="sub" />

<?php
   $a=mysql_connect("localhost","root","");
   $a1=mysql_select_db("test",$a);

  if(isset($_POST['submit']))
  {
         $name=$_POST['name'];
         $i=mysql_query("insert into student (name) values ('$name')");
    if($i)
    {

        echo "<script>a();</script>";
    }
  }
 ?>

 <script type="text/javascript">
 function a(){
 swal("Here's a message!");
 };
</script>

</body>
</html>

it is jquery plugin function..tristanedwards.me/sweetalert

makjal
– makjal

2014-12-19 12:18:45 +00:00
Commented Dec 19, 2014 at 12:18 — makjal
– makjal, Commented Dec 19, 2014 at 12:18

Varedis · Accepted Answer · 2014-12-19 12:05:57Z

4

You are trying to mix two incompatiable languages, PHP is Server side, JQuery is Client Side. The PHP script is already finished processing once your Browser displays the page so you can not then decide to interact with the browser from a PHP script.

You need to use Ajax from a javascript page if you want this kind of functionality.

answered Dec 19, 2014 at 12:05

Varedis

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2 Comments

Sayed Over a year ago

That's rude! and not true as well. You can always echo responses through JS via PHP, given that the response is being echoed after either a page load, a redirection or an AJAX request!

makjal Over a year ago

can u tell me..how to use ajax??.i don't know ajax

Kamran Eyyubov · Accepted Answer · 2014-12-19 12:09:08Z

0

Put javascript function declaration before your php script.

<script type="text/javascript">
 function a(){
 swal("Here's a message!");
 };
</script>
<?php
   $a=mysql_connect("localhost","root","");
   $a1=mysql_select_db("test",$a);

  if(isset($_POST['submit']))
  {
         $name=$_POST['name'];
         $i=mysql_query("insert into student (name) values ('$name')");
    if($i)
    {

        "<script>a();</script>";
    }
  }
 ?>

answered Dec 19, 2014 at 12:09

Kamran Eyyubov

3004 silver badges12 bronze badges

3 Comments

makjal Over a year ago

it didn't work..swal function for plugin can only be placed before closing body tag..

Kamran Eyyubov Over a year ago

Put your full php file here. There is no plugin included in your codes here

Kamran Eyyubov Over a year ago

And as @Ghost said, put echo before <script>a();</script> like this echo "<script>a();</script>"

Harsh Soni · Accepted Answer · 2014-12-19 12:19:19Z

0

<?php 
if($_REQUEST['submit']){
    $i = "yes";
    if($i == "yes"){
    echo "<script>alert('hiii');</script>";
    }
}
?>
<form action="" method="post">
<input type="submit" name="submit" value="submit">
</form>

answered Dec 19, 2014 at 12:19

Harsh Soni

801 gold badge1 silver badge9 bronze badges

4 Comments

makjal Over a year ago

I know..alert is working perfectly but jQuery plugin doesn't working. I want to use sweetalert jQuery plugin

Harsh Soni Over a year ago

I think something conflicting with your plugin try this: in Google chrome and Go to page which you want to see effect now press Ctrl+shift+j and see console check any error is there ? or give me live link

makjal Over a year ago

error: Uncaught TypeError: Cannot read property 'querySelector' of null

makjal Over a year ago

check out jquery plugin link.. tristanedwards.me/sweetalert.it is working with button but i want to display alert after php insert query ..with php it's not working.

makjal · Accepted Answer · 2015-03-27 04:47:59Z

I have 1 demo.php file for form and alert. And demo1.php file for php query. I have to use Ajax for this. demo.php:

<html>
<head>
<script src="lib/sweet-alert.js"></script>
<link rel="stylesheet" href="lib/sweet-alert.css">
<script src="js/jquery-1.8.0.min.js"></script>
</head>

<body>
<form method="post" class="frm" id="myform" onSubmit="j1">
 <label for="name">name</label>
 <input type="text" name="name" id="name" />
 <input type="submit" value="submit" name="submit" class="sub" id="sub" />
</form>

 <script>

     function j1(){
     var query = $('#myform').serialize();
     var url = 'demo1.php';
     $.post(url, query, function (response) {

         swal({
    title: "Thank You!",
    type: "success",
    confirmButtonColor: '#DD6B55',
    confirmButtonText: 'Okay'
    },
    function(isConfirm){
    if (isConfirm){
    window.location.replace('demo.php');
    } 
   });

   });

    }
   $("#myform").submit(function(){
    return false;
   });

  </script>
</body>
</html>

demo1.php:

<?php
   $a=mysql_connect("localhost","root","");
   $a1=mysql_select_db("test",$a);      

                 $name=$_POST['name'];
     $i=mysql_query("insert into student (name) values ('$name')");

 ?>

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