running js code in php
仅有的幸福
仅有的幸福 2017-05-16 13:07:03
0
3
641

How to run js code in php

1. In this case, ok will not pop up

<?php
session_start();
if($_POST){
    if($_POST['mobile']!=$_SESSION['mobile'] or $_POST['mobile_code']!=$_SESSION['mobile_code'] or empty($_POST['mobile']) or empty($ _POST['mobile_code'])){
        echo '<script type="text/javascript">';
        echo 'alert("ok");';
        echo '</script>';
    }else{
        $_SESSION['mobile'] = '';
        $_SESSION['mobile_code'] = '';

}
?>

2. In this case, you can pop up ok

<?php
        echo '<script language="javascript">';
        echo 'alert("ok");';
        echo '</script>';
?>

Is there any difference between the two? Please tell me, thank you

仅有的幸福
仅有的幸福

reply all(3)
phpcn_u1582

There is no difference. In the first case, you did not enter the if statement, so there is no pop-up box. You can try removing the outer if ($_POST), and you can pop up the box. I guess you haven't sent a post request at all, so you can't enter the if code block.

Ty80

Dear, this does not run in php.
php -> Output html+js -> The browser receives and executes js -> Result

For your first example, it is the web page returned when making a POST request. Generally, you use a browser to enter the URL, which is a GET request

滿天的星座

There is no difference if you look at the code alone, but it may be different if you look at the actual situation. Only if it is judged that there is no entry, the pop-up box will not pop up. You should use ajax in JQ. It’s not difficult to check it out on Baidu

Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template