dpp66953
2016-05-16 17:53
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如何在提交表单时将PHP变量传递给Javascript cookie?

I am setting up a custom form to be used in my wordpress site. What I would like to do is grab the AUTO_INCREMENT ID of that submission and pass it into a Javascript cookie when the form is submitted. Currently the ID is working, but because the ID isn't grabbed until after the form is submitted, I'm having trouble figuring out how I can pass the variable into the cookie immediately after the form has been submitted(currently I have to submit the form twice before the cookie with the ID is created, and its 1 number lower than it should be since it's a submission behind).

Here's what I have currently:

<?php
if(isset($_POST["submit"])) {
    $name = $_POST['full_name'];
    $city = $_POST['city'];
    $state = $_POST['state'];
    $email = $_POST['email'];

    $wpdb->insert(
        'reps',
        array(
            'name' => stripslashes($name),
            'city' => stripslashes($city),
            'state' => stripslashes($state),
            'email' => stripslashes($email)
        )
    );

    $lastid = $wpdb->insert_id;
}
?>
<script>
$('#dealer-form').submit(function() {
    var repID = '<?php echo $lastid ?>';
    setCookie('ID', repID);
});
</script>

As I mentioned this code works on the second submission(since the variable has no value on the first submission), and the ID is 1 number behind because it is grabbing the ID of the previous submission before reassigning the variable value.

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我正在设置一个自定义表单,用于我的wordpress网站。 我想要做的是获取该提交的AUTO_INCREMENT ID,并在提交表单时将其传递给Javascript cookie。 目前ID正在运行,但由于在提交表单之前不会抓取ID,因此我无法确定如何在表单提交后立即将变量传递到cookie中(目前我必须提交 在创建带有ID的cookie之前两次使用该表单,并且它的1位数应该低于它应该是因为它是后面提交的。)

这是我目前所拥有的: \ n

 &lt;?php 
if(isset($ _ POST [“submit”])){
 $ name = $ _POST ['full_name']; 
 $ city = $ _POST ['  city']; 
 $ state = $ _POST ['state']; 
 $ email = $ _POST ['email']; 
 
 $ wpdb-&gt; insert(
'reps',
 array  (
'name'=&gt; stripslashes($ name),
'city'=&gt; stripslashes($ city),
'state'=&gt; stripslashes($ state),
'email'=&gt  ; stripslashes($ email)
)
); 
 
 $ lastid = $ wpdb-&gt; insert_id; 
} 
?&gt; 
&lt; script&gt; 
 $('#dealer-form'  ).submit(function(){
 var repID ='&lt;?php ec  ho $ lastid?&gt;'; 
 setCookie('ID',repID); 
}); 
&lt; / script&gt; 
   
 
 

正如我所提到的 此代码适用于第二次提交(因为变量在第一次提交时没有值),并且ID后面是1个数字,因为它在重新分配变量值之前抓取先前提交的ID。

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1条回答 默认 最新

  • dsb0003795 2016-05-16 18:02
    已采纳

    PHP, being a server-side scripting language, is executed before the data is sent to your browser. JavaScript, a client-side scripting language, is executed as soon as the script is encountered by the browser.

    Your approach is forgetting this separation between front- and back-end.

    To accomplish what you're trying to do, simply output the setCookie call when you've submitted your form in php:

    <?php
        if(isset($_POST["submit"])) {
            $name = $_POST['full_name'];
            $city = $_POST['city'];
            $state = $_POST['state'];
            $email = $_POST['email'];
    
            $wpdb->insert(
                'reps',
                array(
                    'name' => stripslashes($name),
                    'city' => stripslashes($city),
                    'state' => stripslashes($state),
                    'email' => stripslashes($email)
                )
            );
    
            $lastid = $wpdb->insert_id;
    
            printf( '<script>setCookie("ID", %d);</script>', $lastid );
        }
    ?>
    
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